Mixing
A continuous function $f$ from a closed bounded interval $[a, b]$ into $mathbb{R}$ is uniformly continuous
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I am reading (as a supplement) the book Basic Real Analysis, by Anthony Knapp. Before I proceed into reading a proof, I want to be sure that the result seems obvious. Yet, I am having trouble seeing through this one. It annoys me too much in order to disregard it:
Theorem.
A continuous function $f$ from a closed bounded interval $[a, b]$ into $mathbb{R}$ is uniformly continuous.
What gives? Why can't we provide the counterexample $f(x)=x^2$ and $[a,b] subset [1,+infty)$, for $b < +infty$ sufficiently large and show that the theorem is incorrect?
Doesn't it seem to be an insufficient statement?
I'm having trouble picturing it, is all.
Hints are fine.
analysis proof-writing
edited Dec 29 '16 at 19:52
TripleA
737935
asked Mar 17 '13 at 20:41
MikhailSchmokloffMikhailSchmokloff
128116
$endgroup$
add a comment |
$begingroup$
I am reading (as a supplement) the book Basic Real Analysis, by Anthony Knapp. Before I proceed into reading a proof, I want to be sure that the result seems obvious. Yet, I am having trouble seeing through this one. It annoys me too much in order to disregard it:
Theorem.
A continuous function $f$ from a closed bounded interval $[a, b]$ into $mathbb{R}$ is uniformly continuous.
What gives? Why can't we provide the counterexample $f(x)=x^2$ and $[a,b] subset [1,+infty)$, for $b < +infty$ sufficiently large and show that the theorem is incorrect?
Doesn't it seem to be an insufficient statement?
I'm having trouble picturing it, is all.
Hints are fine.
analysis proof-writing
edited Dec 29 '16 at 19:52
TripleA
737935
asked Mar 17 '13 at 20:41
MikhailSchmokloffMikhailSchmokloff
128116
$endgroup$
1
$begingroup$
Because even if you take $b$ as large as you want, it is still finite, and you still have $|f(x)-f(y)| leq C|x-y|$ for a big constant $C$. (I'm talking about the function f(x)=x^2$ here)
$endgroup$
– Beni Bogosel
Mar 17 '13 at 20:44
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Heine-Cantor Theorem.
$endgroup$
– Julien
Mar 17 '13 at 20:55
2
$begingroup$
This theorem illustrates the importance of distinguishing very large and infinite. Your example gives a direct illustration of this, when coupled with the theorem.
$endgroup$
– Kieran Cooney
Oct 26 '13 at 18:51
add a comment |
$begingroup$
I am reading (as a supplement) the book Basic Real Analysis, by Anthony Knapp. Before I proceed into reading a proof, I want to be sure that the result seems obvious. Yet, I am having trouble seeing through this one. It annoys me too much in order to disregard it:
Theorem.
A continuous function $f$ from a closed bounded interval $[a, b]$ into $mathbb{R}$ is uniformly continuous.
What gives? Why can't we provide the counterexample $f(x)=x^2$ and $[a,b] subset [1,+infty)$, for $b < +infty$ sufficiently large and show that the theorem is incorrect?
Doesn't it seem to be an insufficient statement?
I'm having trouble picturing it, is all.
Hints are fine.
analysis proof-writing
edited Dec 29 '16 at 19:52
TripleA
737935
asked Mar 17 '13 at 20:41
MikhailSchmokloffMikhailSchmokloff
128116
$endgroup$
I am reading (as a supplement) the book Basic Real Analysis, by Anthony Knapp. Before I proceed into reading a proof, I want to be sure that the result seems obvious. Yet, I am having trouble seeing through this one. It annoys me too much in order to disregard it:
Theorem.
A continuous function $f$ from a closed bounded interval $[a, b]$ into $mathbb{R}$ is uniformly continuous.
What gives? Why can't we provide the counterexample $f(x)=x^2$ and $[a,b] subset [1,+infty)$, for $b < +infty$ sufficiently large and show that the theorem is incorrect?
Doesn't it seem to be an insufficient statement?
I'm having trouble picturing it, is all.
Hints are fine.
analysis proof-writing
analysis proof-writing
edited Dec 29 '16 at 19:52
TripleA
737935
asked Mar 17 '13 at 20:41
MikhailSchmokloffMikhailSchmokloff
128116
edited Dec 29 '16 at 19:52
TripleA
737935
asked Mar 17 '13 at 20:41
MikhailSchmokloffMikhailSchmokloff
128116
edited Dec 29 '16 at 19:52
TripleA
737935
edited Dec 29 '16 at 19:52
TripleA
737935
edited Dec 29 '16 at 19:52
TripleA
737935
737935
asked Mar 17 '13 at 20:41
MikhailSchmokloffMikhailSchmokloff
128116
asked Mar 17 '13 at 20:41
MikhailSchmokloffMikhailSchmokloff
128116
asked Mar 17 '13 at 20:41
MikhailSchmokloffMikhailSchmokloff
128116
128116
1
$begingroup$
Because even if you take $b$ as large as you want, it is still finite, and you still have $|f(x)-f(y)| leq C|x-y|$ for a big constant $C$. (I'm talking about the function f(x)=x^2$ here)
$endgroup$
– Beni Bogosel
Mar 17 '13 at 20:44
$begingroup$
Heine-Cantor Theorem.
$endgroup$
– Julien
Mar 17 '13 at 20:55
2
$begingroup$
This theorem illustrates the importance of distinguishing very large and infinite. Your example gives a direct illustration of this, when coupled with the theorem.
$endgroup$
– Kieran Cooney
Oct 26 '13 at 18:51
add a comment |
1
$begingroup$
Because even if you take $b$ as large as you want, it is still finite, and you still have $|f(x)-f(y)| leq C|x-y|$ for a big constant $C$. (I'm talking about the function f(x)=x^2$ here)
$endgroup$
– Beni Bogosel
Mar 17 '13 at 20:44
$begingroup$
Heine-Cantor Theorem.
$endgroup$
– Julien
Mar 17 '13 at 20:55
2
$begingroup$
This theorem illustrates the importance of distinguishing very large and infinite. Your example gives a direct illustration of this, when coupled with the theorem.
$endgroup$
– Kieran Cooney
Oct 26 '13 at 18:51
1
1
$begingroup$
Because even if you take $b$ as large as you want, it is still finite, and you still have $|f(x)-f(y)| leq C|x-y|$ for a big constant $C$. (I'm talking about the function f(x)=x^2$ here)
$endgroup$
– Beni Bogosel
Mar 17 '13 at 20:44
$begingroup$
Because even if you take $b$ as large as you want, it is still finite, and you still have $|f(x)-f(y)| leq C|x-y|$ for a big constant $C$. (I'm talking about the function f(x)=x^2$ here)
$endgroup$
– Beni Bogosel
Mar 17 '13 at 20:44
$begingroup$
Heine-Cantor Theorem.
$endgroup$
– Julien
Mar 17 '13 at 20:55
$begingroup$
Heine-Cantor Theorem.
$endgroup$
– Julien
Mar 17 '13 at 20:55
2
2
$begingroup$
This theorem illustrates the importance of distinguishing very large and infinite. Your example gives a direct illustration of this, when coupled with the theorem.
$endgroup$
– Kieran Cooney
Oct 26 '13 at 18:51
$begingroup$
This theorem illustrates the importance of distinguishing very large and infinite. Your example gives a direct illustration of this, when coupled with the theorem.
$endgroup$
– Kieran Cooney
Oct 26 '13 at 18:51
add a comment |
2 Answers
2
active
oldest
votes
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I can provide you with a proof. We use the lemma that $[a,b]$ is compact. The general statement is that if $f:Xto Y$ is continuous and $X$ is a compact metric space, then i$f$ is uniformly continuous. This is usually known as the Heine Cantor theorem, while the fact that $[a,b]$is compact might be found as Borel's Lemma, if memory serves. So
THEOREM (Spivak) Let $f:[a,b]to Bbb R$ be continuous. Then it is uniformly continuous.
We first prove the
LEMMA Let $f$ be a continuous function defined on $[a,c]$. If, given $epsilon >0$, there exists $delta_1>0$ such that, for each pair $$x,yin[a,b]text{ ; } |x-y|<delta_1 implies |f(x)-f(y)|<epsilon$$ and $delta_2>0$ such that for each
$$x,yin[b,c]text{ ; } |x-y|<delta_2 implies |f(x)-f(y)|<epsilon$$
Then there exists $delta $ such that for each
$$x,yin[a,c]text{ ; } |x-y|<delta implies |f(x)-f(y)|<epsilon$$
P
Since $f$ is continuous at $x=b$, there exists a $delta_3$ such that for every $x$ with $|b-x|<delta_3$, we have $|f(b)-f(x)|<frac{epsilon}2$.
Thus, whenever $|x-b|<delta_3$ and $|y-b|<delta_3$ we will certainly have $$|f(x)-f(y)|<epsilon$$
We take $delta=min{delta_1,delta_2,delta_3}$. Then $delta$ works: indeed, consider any pair $x,yin[a,c]$. If $x,yin[a,b]$ or $x,yin[b,c]$, we're done. If $x<b<y$ or $y<b<x$. In any case, since $|x-y|<delta$, we must have $|x-b|,|y-b|<delta$, so that $|f(x)-f(y)|<epsilon$, as claimed.
PROOF1 Fix $epsilon >0$. Let's agree to call $f$ $epsilon$-good on an interval $[a,b]$ if for this $epsilon$ there exists a $delta$ such that for any $x,yin[a,b]$, $|x-y|<deltaimplies |f(x)-f(y)|<epsilon$. We thus want to prove that $f$ is $epsilon$-good on $[a,b]$ for any $epsilon >0$. Let $epsilon >0$ be given, and consider the set $$A(epsilon)={xin[a,b]:f text{ is } epsilon text{-good on}: [a,x]}$$ Then $Aneq varnothing$ for $ain A(epsilon)$, and $A(epsilon)$ is bounded above by $b$. Thus $sup A=alpha $ exists. Suppose that $alpha <b$. Since $f$ is continuous at $alpha$ there exists a $delta'$ such that $|y-alpha|<delta'$ implies $|f(y)-f(alpha)|<epsilon/2$. Thus, if $|y-alpha|,|x-alpha|<delta'$, we'll have $|f(y)-f(x)|<epsilon$. Thus $f$ is $epsilon$-good on $[alpha-delta,alpha+delta]$. Since $alpha=sup A(epsilon)$, it is clear $f$ is $epsilon$-good on $[a,alpha+delta]$, which is absurd. Thus $alphageq b$, which means $alpha =b$. It suffices to show that $b$ is also an element of $A(epsilon)$. But since $f$ is continuous on $b$, there exists a $delta_0$ such that $|b-y|<delta_0$ implies $|f(b)-f(y)|<epsilon/2$. Thus, $f$ is $epsilon$-good on $[b-delta_0,b]$. The lemma implies $f$ is $epsilon$-good on $[a,b]$. Since $epsilon$ was arbitrary, the result follows. $blacktriangle$
PROOF2 Let $epsilon >0$ be given. Assign, to each $xin [a,b]$ a $delta_x>0$ such that for each $yin(x-2delta_x,x+2delta_x)$, we have $|f(x)-f(y)|< epsilon/2$, to obtain a open cover of $[a,b]$, namely the set $mathcal O$ of intervals $(x-delta_x,x+delta_x)$. This is possible since $f$ is continuous at each $x$. Since $[a,b]$ is compact, there is a finite number of $x_iin [a,b]$ such that $$bigcup_{i=1}^n (x_i-delta_{x_i},x_i+delta_{x_i})supset [a,b]$$
Choose now $delta =min{delta_{x_i}}$, and let $x,yin [a,b]$ with $ |y-x|<delta$. Since $mathcal O$ is a cover, for some $x_i$ we have that $|x-x_i|<delta_{x_i}$. Then, we'll have $$|y-x_i|leq |y-x|+|x-x_i|<delta+delta_ileq 2delta_i$$ It follows that $$|f(x_i)-f(x)|<epsilon/2$$
$$|f(y)-f(x)|<epsilon/2$$
which means by the triangle inequality that $$|f(x)-f(y)|<epsilon$$
Then for any $x,yin[a,b]$, $|x-y|<delta$ will imply $|f(x)-f(y)|<epsilon$; and $f$ is uniformly continuous. $blacktriangle$
There is yet another way of proving this.
LEMMA (Mendelson) Let $X$ be a metric space such that every infinite subset of $X$ has an accumulation point in $X$. Then for each covering $mathcal O={O_beta}_{betain I}$ there exists a positive $epsilon$ such that each ball $B(x;epsilon)$ is contained in an element $O_beta$ of this covering.
PROOF If it wasn't the case, we'd obtain for each $n$ an point $x_n$ and an open ball $B(x_n;1/n)notsubseteq O_beta$ for each $beta in I$. Let $A={x_1,dots}$. If $A$ is finite, $x_n=x$ infinitely often for some $xin X$. Since $mathcal O$ is a cover, $xin O_alpha$ for some $alpha$. Since the cover is open, there is a $delta >0$ for which $B(x;delta)subseteq O_alpha$. We can take $n$ such that $1/n<delta$ and $x_n=x$, in whichcase we get a contradiction $$Bleft(x;frac 1n right)subseteq Bleft(x;deltaright)subseteq O_alpha$$ If $A$ is infinite, there is an accumulation point $xin X$. Thus $xin O_beta$ for some index, and there are infinitely many points of $A$ in $B(x:delta /2)subseteq O_beta$. We can take $n$ such that $1/n<delta /2$ and we'd have $B(x_n;1/n)subseteq B(x;delta)subseteq O_beta$, a contradiction.
After having proven that for metric spaces, the existence of accumulation points for infinite subsets is equivalent to compactness.
PROOF3 Let $f:Xto Y$ be a continuous function from a compactum $X$ to a metric space $Y$. Then $f$ is uniformly continuous.
PROOF Given $epsilon >0$, for each $xin X$ there is a $delta_x>0$ such that if $yin B(x:delta_x)$, $f(y)in Bleft(f(x);epsilon /2right)$. These balls are an open cover for $X$, thus there exists such a number $delta_L$ as in the previous lemma (usually called a Lebesgue number). Choose $delta$ to be positive yet smaller than $delta_L$. If $z,z'in X$ and $d(z,z')<delta$ (so that $z,z'$ are in a ball of radius less than $delta$), we have $z,z'in B(x,delta_x)$ for some $xin X$. In that case $f(z),f(z')in B(f(x),epsilon/2)$ so $d'(f(z),f(z'))<epsilon$ by the triangle inequality. $blacktriangle$.
edited Jan 10 at 14:29
Richard Clare
1,066314
answered Mar 17 '13 at 20:56
Pedro Tamaroff♦Pedro Tamaroff
96.8k10153297
$endgroup$
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I think you need to elaborate more on the last step. $delta_x$ and $delta_y$ could be smaller than the chosen $delta$. You need to find the $x_i$ for which $x in (x_i - delta_{x_i}, x_i + delta_{x_i})$, and make sure that $y$ is close enough to $x$ so that it also belongs to this interval.
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– Ayman Hourieh
Mar 18 '13 at 9:33
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@Ayman I see what you mean.
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– Pedro Tamaroff♦
Mar 18 '13 at 13:11
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Thanks. The elaboration needed for your original proof isn't big. It's something along the lines of my previous comment. But it is necessary; otherwise it's unclear why $left|f(x) - f(y)right| < epsilon$.
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– Ayman Hourieh
Mar 18 '13 at 17:40
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@ayman After some reading I have found something that fixes my proof try. I will try and digest it, and if possible add it tomorrow.
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– Pedro Tamaroff♦
Mar 26 '13 at 2:29
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The fix I had in mind was similar to proof 1 at proofwiki.org/wiki/Heine-Cantor_Theorem#Proof_1.
$endgroup$
– Ayman Hourieh
Mar 26 '13 at 15:43
|
show 7 more comments
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In this case, intuitively, no matter how large the interval is, as long as it is closed and bounded, the function is still controlled. This is a special case of a more general result: a continuous function on a compact metric space is uniformly continuous, and we know that a subset of Euclidean space is compact if and only if it is closed and bounded.
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2 Answers
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I can provide you with a proof. We use the lemma that $[a,b]$ is compact. The general statement is that if $f:Xto Y$ is continuous and $X$ is a compact metric space, then i$f$ is uniformly continuous. This is usually known as the Heine Cantor theorem, while the fact that $[a,b]$is compact might be found as Borel's Lemma, if memory serves. So
THEOREM (Spivak) Let $f:[a,b]to Bbb R$ be continuous. Then it is uniformly continuous.
We first prove the
LEMMA Let $f$ be a continuous function defined on $[a,c]$. If, given $epsilon >0$, there exists $delta_1>0$ such that, for each pair $$x,yin[a,b]text{ ; } |x-y|<delta_1 implies |f(x)-f(y)|<epsilon$$ and $delta_2>0$ such that for each
$$x,yin[b,c]text{ ; } |x-y|<delta_2 implies |f(x)-f(y)|<epsilon$$
Then there exists $delta $ such that for each
$$x,yin[a,c]text{ ; } |x-y|<delta implies |f(x)-f(y)|<epsilon$$
P
Since $f$ is continuous at $x=b$, there exists a $delta_3$ such that for every $x$ with $|b-x|<delta_3$, we have $|f(b)-f(x)|<frac{epsilon}2$.
Thus, whenever $|x-b|<delta_3$ and $|y-b|<delta_3$ we will certainly have $$|f(x)-f(y)|<epsilon$$
We take $delta=min{delta_1,delta_2,delta_3}$. Then $delta$ works: indeed, consider any pair $x,yin[a,c]$. If $x,yin[a,b]$ or $x,yin[b,c]$, we're done. If $x<b<y$ or $y<b<x$. In any case, since $|x-y|<delta$, we must have $|x-b|,|y-b|<delta$, so that $|f(x)-f(y)|<epsilon$, as claimed.
PROOF1 Fix $epsilon >0$. Let's agree to call $f$ $epsilon$-good on an interval $[a,b]$ if for this $epsilon$ there exists a $delta$ such that for any $x,yin[a,b]$, $|x-y|<deltaimplies |f(x)-f(y)|<epsilon$. We thus want to prove that $f$ is $epsilon$-good on $[a,b]$ for any $epsilon >0$. Let $epsilon >0$ be given, and consider the set $$A(epsilon)={xin[a,b]:f text{ is } epsilon text{-good on}: [a,x]}$$ Then $Aneq varnothing$ for $ain A(epsilon)$, and $A(epsilon)$ is bounded above by $b$. Thus $sup A=alpha $ exists. Suppose that $alpha <b$. Since $f$ is continuous at $alpha$ there exists a $delta'$ such that $|y-alpha|<delta'$ implies $|f(y)-f(alpha)|<epsilon/2$. Thus, if $|y-alpha|,|x-alpha|<delta'$, we'll have $|f(y)-f(x)|<epsilon$. Thus $f$ is $epsilon$-good on $[alpha-delta,alpha+delta]$. Since $alpha=sup A(epsilon)$, it is clear $f$ is $epsilon$-good on $[a,alpha+delta]$, which is absurd. Thus $alphageq b$, which means $alpha =b$. It suffices to show that $b$ is also an element of $A(epsilon)$. But since $f$ is continuous on $b$, there exists a $delta_0$ such that $|b-y|<delta_0$ implies $|f(b)-f(y)|<epsilon/2$. Thus, $f$ is $epsilon$-good on $[b-delta_0,b]$. The lemma implies $f$ is $epsilon$-good on $[a,b]$. Since $epsilon$ was arbitrary, the result follows. $blacktriangle$
PROOF2 Let $epsilon >0$ be given. Assign, to each $xin [a,b]$ a $delta_x>0$ such that for each $yin(x-2delta_x,x+2delta_x)$, we have $|f(x)-f(y)|< epsilon/2$, to obtain a open cover of $[a,b]$, namely the set $mathcal O$ of intervals $(x-delta_x,x+delta_x)$. This is possible since $f$ is continuous at each $x$. Since $[a,b]$ is compact, there is a finite number of $x_iin [a,b]$ such that $$bigcup_{i=1}^n (x_i-delta_{x_i},x_i+delta_{x_i})supset [a,b]$$
Choose now $delta =min{delta_{x_i}}$, and let $x,yin [a,b]$ with $ |y-x|<delta$. Since $mathcal O$ is a cover, for some $x_i$ we have that $|x-x_i|<delta_{x_i}$. Then, we'll have $$|y-x_i|leq |y-x|+|x-x_i|<delta+delta_ileq 2delta_i$$ It follows that $$|f(x_i)-f(x)|<epsilon/2$$
$$|f(y)-f(x)|<epsilon/2$$
which means by the triangle inequality that $$|f(x)-f(y)|<epsilon$$
Then for any $x,yin[a,b]$, $|x-y|<delta$ will imply $|f(x)-f(y)|<epsilon$; and $f$ is uniformly continuous. $blacktriangle$
There is yet another way of proving this.
LEMMA (Mendelson) Let $X$ be a metric space such that every infinite subset of $X$ has an accumulation point in $X$. Then for each covering $mathcal O={O_beta}_{betain I}$ there exists a positive $epsilon$ such that each ball $B(x;epsilon)$ is contained in an element $O_beta$ of this covering.
PROOF If it wasn't the case, we'd obtain for each $n$ an point $x_n$ and an open ball $B(x_n;1/n)notsubseteq O_beta$ for each $beta in I$. Let $A={x_1,dots}$. If $A$ is finite, $x_n=x$ infinitely often for some $xin X$. Since $mathcal O$ is a cover, $xin O_alpha$ for some $alpha$. Since the cover is open, there is a $delta >0$ for which $B(x;delta)subseteq O_alpha$. We can take $n$ such that $1/n<delta$ and $x_n=x$, in whichcase we get a contradiction $$Bleft(x;frac 1n right)subseteq Bleft(x;deltaright)subseteq O_alpha$$ If $A$ is infinite, there is an accumulation point $xin X$. Thus $xin O_beta$ for some index, and there are infinitely many points of $A$ in $B(x:delta /2)subseteq O_beta$. We can take $n$ such that $1/n<delta /2$ and we'd have $B(x_n;1/n)subseteq B(x;delta)subseteq O_beta$, a contradiction.
After having proven that for metric spaces, the existence of accumulation points for infinite subsets is equivalent to compactness.
PROOF3 Let $f:Xto Y$ be a continuous function from a compactum $X$ to a metric space $Y$. Then $f$ is uniformly continuous.
PROOF Given $epsilon >0$, for each $xin X$ there is a $delta_x>0$ such that if $yin B(x:delta_x)$, $f(y)in Bleft(f(x);epsilon /2right)$. These balls are an open cover for $X$, thus there exists such a number $delta_L$ as in the previous lemma (usually called a Lebesgue number). Choose $delta$ to be positive yet smaller than $delta_L$. If $z,z'in X$ and $d(z,z')<delta$ (so that $z,z'$ are in a ball of radius less than $delta$), we have $z,z'in B(x,delta_x)$ for some $xin X$. In that case $f(z),f(z')in B(f(x),epsilon/2)$ so $d'(f(z),f(z'))<epsilon$ by the triangle inequality. $blacktriangle$.
edited Jan 10 at 14:29
Richard Clare
1,066314
answered Mar 17 '13 at 20:56
Pedro Tamaroff♦Pedro Tamaroff
96.8k10153297
$endgroup$
$begingroup$
I think you need to elaborate more on the last step. $delta_x$ and $delta_y$ could be smaller than the chosen $delta$. You need to find the $x_i$ for which $x in (x_i - delta_{x_i}, x_i + delta_{x_i})$, and make sure that $y$ is close enough to $x$ so that it also belongs to this interval.
$endgroup$
– Ayman Hourieh
Mar 18 '13 at 9:33
$begingroup$
@Ayman I see what you mean.
$endgroup$
– Pedro Tamaroff♦
Mar 18 '13 at 13:11
$begingroup$
Thanks. The elaboration needed for your original proof isn't big. It's something along the lines of my previous comment. But it is necessary; otherwise it's unclear why $left|f(x) - f(y)right| < epsilon$.
$endgroup$
– Ayman Hourieh
Mar 18 '13 at 17:40
$begingroup$
@ayman After some reading I have found something that fixes my proof try. I will try and digest it, and if possible add it tomorrow.
$endgroup$
– Pedro Tamaroff♦
Mar 26 '13 at 2:29
$begingroup$
The fix I had in mind was similar to proof 1 at proofwiki.org/wiki/Heine-Cantor_Theorem#Proof_1.
$endgroup$
– Ayman Hourieh
Mar 26 '13 at 15:43
|
show 7 more comments
$begingroup$
I can provide you with a proof. We use the lemma that $[a,b]$ is compact. The general statement is that if $f:Xto Y$ is continuous and $X$ is a compact metric space, then i$f$ is uniformly continuous. This is usually known as the Heine Cantor theorem, while the fact that $[a,b]$is compact might be found as Borel's Lemma, if memory serves. So
THEOREM (Spivak) Let $f:[a,b]to Bbb R$ be continuous. Then it is uniformly continuous.
We first prove the
LEMMA Let $f$ be a continuous function defined on $[a,c]$. If, given $epsilon >0$, there exists $delta_1>0$ such that, for each pair $$x,yin[a,b]text{ ; } |x-y|<delta_1 implies |f(x)-f(y)|<epsilon$$ and $delta_2>0$ such that for each
$$x,yin[b,c]text{ ; } |x-y|<delta_2 implies |f(x)-f(y)|<epsilon$$
Then there exists $delta $ such that for each
$$x,yin[a,c]text{ ; } |x-y|<delta implies |f(x)-f(y)|<epsilon$$
P
Since $f$ is continuous at $x=b$, there exists a $delta_3$ such that for every $x$ with $|b-x|<delta_3$, we have $|f(b)-f(x)|<frac{epsilon}2$.
Thus, whenever $|x-b|<delta_3$ and $|y-b|<delta_3$ we will certainly have $$|f(x)-f(y)|<epsilon$$
We take $delta=min{delta_1,delta_2,delta_3}$. Then $delta$ works: indeed, consider any pair $x,yin[a,c]$. If $x,yin[a,b]$ or $x,yin[b,c]$, we're done. If $x<b<y$ or $y<b<x$. In any case, since $|x-y|<delta$, we must have $|x-b|,|y-b|<delta$, so that $|f(x)-f(y)|<epsilon$, as claimed.
PROOF1 Fix $epsilon >0$. Let's agree to call $f$ $epsilon$-good on an interval $[a,b]$ if for this $epsilon$ there exists a $delta$ such that for any $x,yin[a,b]$, $|x-y|<deltaimplies |f(x)-f(y)|<epsilon$. We thus want to prove that $f$ is $epsilon$-good on $[a,b]$ for any $epsilon >0$. Let $epsilon >0$ be given, and consider the set $$A(epsilon)={xin[a,b]:f text{ is } epsilon text{-good on}: [a,x]}$$ Then $Aneq varnothing$ for $ain A(epsilon)$, and $A(epsilon)$ is bounded above by $b$. Thus $sup A=alpha $ exists. Suppose that $alpha <b$. Since $f$ is continuous at $alpha$ there exists a $delta'$ such that $|y-alpha|<delta'$ implies $|f(y)-f(alpha)|<epsilon/2$. Thus, if $|y-alpha|,|x-alpha|<delta'$, we'll have $|f(y)-f(x)|<epsilon$. Thus $f$ is $epsilon$-good on $[alpha-delta,alpha+delta]$. Since $alpha=sup A(epsilon)$, it is clear $f$ is $epsilon$-good on $[a,alpha+delta]$, which is absurd. Thus $alphageq b$, which means $alpha =b$. It suffices to show that $b$ is also an element of $A(epsilon)$. But since $f$ is continuous on $b$, there exists a $delta_0$ such that $|b-y|<delta_0$ implies $|f(b)-f(y)|<epsilon/2$. Thus, $f$ is $epsilon$-good on $[b-delta_0,b]$. The lemma implies $f$ is $epsilon$-good on $[a,b]$. Since $epsilon$ was arbitrary, the result follows. $blacktriangle$
PROOF2 Let $epsilon >0$ be given. Assign, to each $xin [a,b]$ a $delta_x>0$ such that for each $yin(x-2delta_x,x+2delta_x)$, we have $|f(x)-f(y)|< epsilon/2$, to obtain a open cover of $[a,b]$, namely the set $mathcal O$ of intervals $(x-delta_x,x+delta_x)$. This is possible since $f$ is continuous at each $x$. Since $[a,b]$ is compact, there is a finite number of $x_iin [a,b]$ such that $$bigcup_{i=1}^n (x_i-delta_{x_i},x_i+delta_{x_i})supset [a,b]$$
Choose now $delta =min{delta_{x_i}}$, and let $x,yin [a,b]$ with $ |y-x|<delta$. Since $mathcal O$ is a cover, for some $x_i$ we have that $|x-x_i|<delta_{x_i}$. Then, we'll have $$|y-x_i|leq |y-x|+|x-x_i|<delta+delta_ileq 2delta_i$$ It follows that $$|f(x_i)-f(x)|<epsilon/2$$
$$|f(y)-f(x)|<epsilon/2$$
which means by the triangle inequality that $$|f(x)-f(y)|<epsilon$$
Then for any $x,yin[a,b]$, $|x-y|<delta$ will imply $|f(x)-f(y)|<epsilon$; and $f$ is uniformly continuous. $blacktriangle$
There is yet another way of proving this.
LEMMA (Mendelson) Let $X$ be a metric space such that every infinite subset of $X$ has an accumulation point in $X$. Then for each covering $mathcal O={O_beta}_{betain I}$ there exists a positive $epsilon$ such that each ball $B(x;epsilon)$ is contained in an element $O_beta$ of this covering.
PROOF If it wasn't the case, we'd obtain for each $n$ an point $x_n$ and an open ball $B(x_n;1/n)notsubseteq O_beta$ for each $beta in I$. Let $A={x_1,dots}$. If $A$ is finite, $x_n=x$ infinitely often for some $xin X$. Since $mathcal O$ is a cover, $xin O_alpha$ for some $alpha$. Since the cover is open, there is a $delta >0$ for which $B(x;delta)subseteq O_alpha$. We can take $n$ such that $1/n<delta$ and $x_n=x$, in whichcase we get a contradiction $$Bleft(x;frac 1n right)subseteq Bleft(x;deltaright)subseteq O_alpha$$ If $A$ is infinite, there is an accumulation point $xin X$. Thus $xin O_beta$ for some index, and there are infinitely many points of $A$ in $B(x:delta /2)subseteq O_beta$. We can take $n$ such that $1/n<delta /2$ and we'd have $B(x_n;1/n)subseteq B(x;delta)subseteq O_beta$, a contradiction.
After having proven that for metric spaces, the existence of accumulation points for infinite subsets is equivalent to compactness.
PROOF3 Let $f:Xto Y$ be a continuous function from a compactum $X$ to a metric space $Y$. Then $f$ is uniformly continuous.
PROOF Given $epsilon >0$, for each $xin X$ there is a $delta_x>0$ such that if $yin B(x:delta_x)$, $f(y)in Bleft(f(x);epsilon /2right)$. These balls are an open cover for $X$, thus there exists such a number $delta_L$ as in the previous lemma (usually called a Lebesgue number). Choose $delta$ to be positive yet smaller than $delta_L$. If $z,z'in X$ and $d(z,z')<delta$ (so that $z,z'$ are in a ball of radius less than $delta$), we have $z,z'in B(x,delta_x)$ for some $xin X$. In that case $f(z),f(z')in B(f(x),epsilon/2)$ so $d'(f(z),f(z'))<epsilon$ by the triangle inequality. $blacktriangle$.
edited Jan 10 at 14:29
Richard Clare
1,066314
answered Mar 17 '13 at 20:56
Pedro Tamaroff♦Pedro Tamaroff
96.8k10153297
$endgroup$
$begingroup$
I think you need to elaborate more on the last step. $delta_x$ and $delta_y$ could be smaller than the chosen $delta$. You need to find the $x_i$ for which $x in (x_i - delta_{x_i}, x_i + delta_{x_i})$, and make sure that $y$ is close enough to $x$ so that it also belongs to this interval.
$endgroup$
– Ayman Hourieh
Mar 18 '13 at 9:33
$begingroup$
@Ayman I see what you mean.
$endgroup$
– Pedro Tamaroff♦
Mar 18 '13 at 13:11
$begingroup$
Thanks. The elaboration needed for your original proof isn't big. It's something along the lines of my previous comment. But it is necessary; otherwise it's unclear why $left|f(x) - f(y)right| < epsilon$.
$endgroup$
– Ayman Hourieh
Mar 18 '13 at 17:40
$begingroup$
@ayman After some reading I have found something that fixes my proof try. I will try and digest it, and if possible add it tomorrow.
$endgroup$
– Pedro Tamaroff♦
Mar 26 '13 at 2:29
$begingroup$
The fix I had in mind was similar to proof 1 at proofwiki.org/wiki/Heine-Cantor_Theorem#Proof_1.
$endgroup$
– Ayman Hourieh
Mar 26 '13 at 15:43
|
show 7 more comments
$begingroup$
I can provide you with a proof. We use the lemma that $[a,b]$ is compact. The general statement is that if $f:Xto Y$ is continuous and $X$ is a compact metric space, then i$f$ is uniformly continuous. This is usually known as the Heine Cantor theorem, while the fact that $[a,b]$is compact might be found as Borel's Lemma, if memory serves. So
THEOREM (Spivak) Let $f:[a,b]to Bbb R$ be continuous. Then it is uniformly continuous.
We first prove the
LEMMA Let $f$ be a continuous function defined on $[a,c]$. If, given $epsilon >0$, there exists $delta_1>0$ such that, for each pair $$x,yin[a,b]text{ ; } |x-y|<delta_1 implies |f(x)-f(y)|<epsilon$$ and $delta_2>0$ such that for each
$$x,yin[b,c]text{ ; } |x-y|<delta_2 implies |f(x)-f(y)|<epsilon$$
Then there exists $delta $ such that for each
$$x,yin[a,c]text{ ; } |x-y|<delta implies |f(x)-f(y)|<epsilon$$
P
Since $f$ is continuous at $x=b$, there exists a $delta_3$ such that for every $x$ with $|b-x|<delta_3$, we have $|f(b)-f(x)|<frac{epsilon}2$.
Thus, whenever $|x-b|<delta_3$ and $|y-b|<delta_3$ we will certainly have $$|f(x)-f(y)|<epsilon$$
We take $delta=min{delta_1,delta_2,delta_3}$. Then $delta$ works: indeed, consider any pair $x,yin[a,c]$. If $x,yin[a,b]$ or $x,yin[b,c]$, we're done. If $x<b<y$ or $y<b<x$. In any case, since $|x-y|<delta$, we must have $|x-b|,|y-b|<delta$, so that $|f(x)-f(y)|<epsilon$, as claimed.
PROOF1 Fix $epsilon >0$. Let's agree to call $f$ $epsilon$-good on an interval $[a,b]$ if for this $epsilon$ there exists a $delta$ such that for any $x,yin[a,b]$, $|x-y|<deltaimplies |f(x)-f(y)|<epsilon$. We thus want to prove that $f$ is $epsilon$-good on $[a,b]$ for any $epsilon >0$. Let $epsilon >0$ be given, and consider the set $$A(epsilon)={xin[a,b]:f text{ is } epsilon text{-good on}: [a,x]}$$ Then $Aneq varnothing$ for $ain A(epsilon)$, and $A(epsilon)$ is bounded above by $b$. Thus $sup A=alpha $ exists. Suppose that $alpha <b$. Since $f$ is continuous at $alpha$ there exists a $delta'$ such that $|y-alpha|<delta'$ implies $|f(y)-f(alpha)|<epsilon/2$. Thus, if $|y-alpha|,|x-alpha|<delta'$, we'll have $|f(y)-f(x)|<epsilon$. Thus $f$ is $epsilon$-good on $[alpha-delta,alpha+delta]$. Since $alpha=sup A(epsilon)$, it is clear $f$ is $epsilon$-good on $[a,alpha+delta]$, which is absurd. Thus $alphageq b$, which means $alpha =b$. It suffices to show that $b$ is also an element of $A(epsilon)$. But since $f$ is continuous on $b$, there exists a $delta_0$ such that $|b-y|<delta_0$ implies $|f(b)-f(y)|<epsilon/2$. Thus, $f$ is $epsilon$-good on $[b-delta_0,b]$. The lemma implies $f$ is $epsilon$-good on $[a,b]$. Since $epsilon$ was arbitrary, the result follows. $blacktriangle$
PROOF2 Let $epsilon >0$ be given. Assign, to each $xin [a,b]$ a $delta_x>0$ such that for each $yin(x-2delta_x,x+2delta_x)$, we have $|f(x)-f(y)|< epsilon/2$, to obtain a open cover of $[a,b]$, namely the set $mathcal O$ of intervals $(x-delta_x,x+delta_x)$. This is possible since $f$ is continuous at each $x$. Since $[a,b]$ is compact, there is a finite number of $x_iin [a,b]$ such that $$bigcup_{i=1}^n (x_i-delta_{x_i},x_i+delta_{x_i})supset [a,b]$$
Choose now $delta =min{delta_{x_i}}$, and let $x,yin [a,b]$ with $ |y-x|<delta$. Since $mathcal O$ is a cover, for some $x_i$ we have that $|x-x_i|<delta_{x_i}$. Then, we'll have $$|y-x_i|leq |y-x|+|x-x_i|<delta+delta_ileq 2delta_i$$ It follows that $$|f(x_i)-f(x)|<epsilon/2$$
$$|f(y)-f(x)|<epsilon/2$$
which means by the triangle inequality that $$|f(x)-f(y)|<epsilon$$
Then for any $x,yin[a,b]$, $|x-y|<delta$ will imply $|f(x)-f(y)|<epsilon$; and $f$ is uniformly continuous. $blacktriangle$
There is yet another way of proving this.
LEMMA (Mendelson) Let $X$ be a metric space such that every infinite subset of $X$ has an accumulation point in $X$. Then for each covering $mathcal O={O_beta}_{betain I}$ there exists a positive $epsilon$ such that each ball $B(x;epsilon)$ is contained in an element $O_beta$ of this covering.
PROOF If it wasn't the case, we'd obtain for each $n$ an point $x_n$ and an open ball $B(x_n;1/n)notsubseteq O_beta$ for each $beta in I$. Let $A={x_1,dots}$. If $A$ is finite, $x_n=x$ infinitely often for some $xin X$. Since $mathcal O$ is a cover, $xin O_alpha$ for some $alpha$. Since the cover is open, there is a $delta >0$ for which $B(x;delta)subseteq O_alpha$. We can take $n$ such that $1/n<delta$ and $x_n=x$, in whichcase we get a contradiction $$Bleft(x;frac 1n right)subseteq Bleft(x;deltaright)subseteq O_alpha$$ If $A$ is infinite, there is an accumulation point $xin X$. Thus $xin O_beta$ for some index, and there are infinitely many points of $A$ in $B(x:delta /2)subseteq O_beta$. We can take $n$ such that $1/n<delta /2$ and we'd have $B(x_n;1/n)subseteq B(x;delta)subseteq O_beta$, a contradiction.
After having proven that for metric spaces, the existence of accumulation points for infinite subsets is equivalent to compactness.
PROOF3 Let $f:Xto Y$ be a continuous function from a compactum $X$ to a metric space $Y$. Then $f$ is uniformly continuous.
PROOF Given $epsilon >0$, for each $xin X$ there is a $delta_x>0$ such that if $yin B(x:delta_x)$, $f(y)in Bleft(f(x);epsilon /2right)$. These balls are an open cover for $X$, thus there exists such a number $delta_L$ as in the previous lemma (usually called a Lebesgue number). Choose $delta$ to be positive yet smaller than $delta_L$. If $z,z'in X$ and $d(z,z')<delta$ (so that $z,z'$ are in a ball of radius less than $delta$), we have $z,z'in B(x,delta_x)$ for some $xin X$. In that case $f(z),f(z')in B(f(x),epsilon/2)$ so $d'(f(z),f(z'))<epsilon$ by the triangle inequality. $blacktriangle$.
edited Jan 10 at 14:29
Richard Clare
1,066314
answered Mar 17 '13 at 20:56
Pedro Tamaroff♦Pedro Tamaroff
96.8k10153297
$endgroup$
I can provide you with a proof. We use the lemma that $[a,b]$ is compact. The general statement is that if $f:Xto Y$ is continuous and $X$ is a compact metric space, then i$f$ is uniformly continuous. This is usually known as the Heine Cantor theorem, while the fact that $[a,b]$is compact might be found as Borel's Lemma, if memory serves. So
THEOREM (Spivak) Let $f:[a,b]to Bbb R$ be continuous. Then it is uniformly continuous.
We first prove the
LEMMA Let $f$ be a continuous function defined on $[a,c]$. If, given $epsilon >0$, there exists $delta_1>0$ such that, for each pair $$x,yin[a,b]text{ ; } |x-y|<delta_1 implies |f(x)-f(y)|<epsilon$$ and $delta_2>0$ such that for each
$$x,yin[b,c]text{ ; } |x-y|<delta_2 implies |f(x)-f(y)|<epsilon$$
Then there exists $delta $ such that for each
$$x,yin[a,c]text{ ; } |x-y|<delta implies |f(x)-f(y)|<epsilon$$
P
Since $f$ is continuous at $x=b$, there exists a $delta_3$ such that for every $x$ with $|b-x|<delta_3$, we have $|f(b)-f(x)|<frac{epsilon}2$.
Thus, whenever $|x-b|<delta_3$ and $|y-b|<delta_3$ we will certainly have $$|f(x)-f(y)|<epsilon$$
We take $delta=min{delta_1,delta_2,delta_3}$. Then $delta$ works: indeed, consider any pair $x,yin[a,c]$. If $x,yin[a,b]$ or $x,yin[b,c]$, we're done. If $x<b<y$ or $y<b<x$. In any case, since $|x-y|<delta$, we must have $|x-b|,|y-b|<delta$, so that $|f(x)-f(y)|<epsilon$, as claimed.
PROOF1 Fix $epsilon >0$. Let's agree to call $f$ $epsilon$-good on an interval $[a,b]$ if for this $epsilon$ there exists a $delta$ such that for any $x,yin[a,b]$, $|x-y|<deltaimplies |f(x)-f(y)|<epsilon$. We thus want to prove that $f$ is $epsilon$-good on $[a,b]$ for any $epsilon >0$. Let $epsilon >0$ be given, and consider the set $$A(epsilon)={xin[a,b]:f text{ is } epsilon text{-good on}: [a,x]}$$ Then $Aneq varnothing$ for $ain A(epsilon)$, and $A(epsilon)$ is bounded above by $b$. Thus $sup A=alpha $ exists. Suppose that $alpha <b$. Since $f$ is continuous at $alpha$ there exists a $delta'$ such that $|y-alpha|<delta'$ implies $|f(y)-f(alpha)|<epsilon/2$. Thus, if $|y-alpha|,|x-alpha|<delta'$, we'll have $|f(y)-f(x)|<epsilon$. Thus $f$ is $epsilon$-good on $[alpha-delta,alpha+delta]$. Since $alpha=sup A(epsilon)$, it is clear $f$ is $epsilon$-good on $[a,alpha+delta]$, which is absurd. Thus $alphageq b$, which means $alpha =b$. It suffices to show that $b$ is also an element of $A(epsilon)$. But since $f$ is continuous on $b$, there exists a $delta_0$ such that $|b-y|<delta_0$ implies $|f(b)-f(y)|<epsilon/2$. Thus, $f$ is $epsilon$-good on $[b-delta_0,b]$. The lemma implies $f$ is $epsilon$-good on $[a,b]$. Since $epsilon$ was arbitrary, the result follows. $blacktriangle$
PROOF2 Let $epsilon >0$ be given. Assign, to each $xin [a,b]$ a $delta_x>0$ such that for each $yin(x-2delta_x,x+2delta_x)$, we have $|f(x)-f(y)|< epsilon/2$, to obtain a open cover of $[a,b]$, namely the set $mathcal O$ of intervals $(x-delta_x,x+delta_x)$. This is possible since $f$ is continuous at each $x$. Since $[a,b]$ is compact, there is a finite number of $x_iin [a,b]$ such that $$bigcup_{i=1}^n (x_i-delta_{x_i},x_i+delta_{x_i})supset [a,b]$$
Choose now $delta =min{delta_{x_i}}$, and let $x,yin [a,b]$ with $ |y-x|<delta$. Since $mathcal O$ is a cover, for some $x_i$ we have that $|x-x_i|<delta_{x_i}$. Then, we'll have $$|y-x_i|leq |y-x|+|x-x_i|<delta+delta_ileq 2delta_i$$ It follows that $$|f(x_i)-f(x)|<epsilon/2$$
$$|f(y)-f(x)|<epsilon/2$$
which means by the triangle inequality that $$|f(x)-f(y)|<epsilon$$
Then for any $x,yin[a,b]$, $|x-y|<delta$ will imply $|f(x)-f(y)|<epsilon$; and $f$ is uniformly continuous. $blacktriangle$
There is yet another way of proving this.
LEMMA (Mendelson) Let $X$ be a metric space such that every infinite subset of $X$ has an accumulation point in $X$. Then for each covering $mathcal O={O_beta}_{betain I}$ there exists a positive $epsilon$ such that each ball $B(x;epsilon)$ is contained in an element $O_beta$ of this covering.
PROOF If it wasn't the case, we'd obtain for each $n$ an point $x_n$ and an open ball $B(x_n;1/n)notsubseteq O_beta$ for each $beta in I$. Let $A={x_1,dots}$. If $A$ is finite, $x_n=x$ infinitely often for some $xin X$. Since $mathcal O$ is a cover, $xin O_alpha$ for some $alpha$. Since the cover is open, there is a $delta >0$ for which $B(x;delta)subseteq O_alpha$. We can take $n$ such that $1/n<delta$ and $x_n=x$, in whichcase we get a contradiction $$Bleft(x;frac 1n right)subseteq Bleft(x;deltaright)subseteq O_alpha$$ If $A$ is infinite, there is an accumulation point $xin X$. Thus $xin O_beta$ for some index, and there are infinitely many points of $A$ in $B(x:delta /2)subseteq O_beta$. We can take $n$ such that $1/n<delta /2$ and we'd have $B(x_n;1/n)subseteq B(x;delta)subseteq O_beta$, a contradiction.
After having proven that for metric spaces, the existence of accumulation points for infinite subsets is equivalent to compactness.
PROOF3 Let $f:Xto Y$ be a continuous function from a compactum $X$ to a metric space $Y$. Then $f$ is uniformly continuous.
PROOF Given $epsilon >0$, for each $xin X$ there is a $delta_x>0$ such that if $yin B(x:delta_x)$, $f(y)in Bleft(f(x);epsilon /2right)$. These balls are an open cover for $X$, thus there exists such a number $delta_L$ as in the previous lemma (usually called a Lebesgue number). Choose $delta$ to be positive yet smaller than $delta_L$. If $z,z'in X$ and $d(z,z')<delta$ (so that $z,z'$ are in a ball of radius less than $delta$), we have $z,z'in B(x,delta_x)$ for some $xin X$. In that case $f(z),f(z')in B(f(x),epsilon/2)$ so $d'(f(z),f(z'))<epsilon$ by the triangle inequality. $blacktriangle$.
edited Jan 10 at 14:29
Richard Clare
1,066314
answered Mar 17 '13 at 20:56
Pedro Tamaroff♦Pedro Tamaroff
96.8k10153297
edited Jan 10 at 14:29
Richard Clare
1,066314
edited Jan 10 at 14:29
Richard Clare
1,066314
edited Jan 10 at 14:29
Richard Clare
1,066314
1,066314
answered Mar 17 '13 at 20:56
Pedro Tamaroff♦Pedro Tamaroff
96.8k10153297
answered Mar 17 '13 at 20:56
Pedro Tamaroff♦Pedro Tamaroff
96.8k10153297
answered Mar 17 '13 at 20:56
Pedro Tamaroff♦Pedro Tamaroff
96.8k10153297
96.8k10153297
$begingroup$
I think you need to elaborate more on the last step. $delta_x$ and $delta_y$ could be smaller than the chosen $delta$. You need to find the $x_i$ for which $x in (x_i - delta_{x_i}, x_i + delta_{x_i})$, and make sure that $y$ is close enough to $x$ so that it also belongs to this interval.
$endgroup$
– Ayman Hourieh
Mar 18 '13 at 9:33
$begingroup$
@Ayman I see what you mean.
$endgroup$
– Pedro Tamaroff♦
Mar 18 '13 at 13:11
$begingroup$
Thanks. The elaboration needed for your original proof isn't big. It's something along the lines of my previous comment. But it is necessary; otherwise it's unclear why $left|f(x) - f(y)right| < epsilon$.
$endgroup$
– Ayman Hourieh
Mar 18 '13 at 17:40
$begingroup$
@ayman After some reading I have found something that fixes my proof try. I will try and digest it, and if possible add it tomorrow.
$endgroup$
– Pedro Tamaroff♦
Mar 26 '13 at 2:29
$begingroup$
The fix I had in mind was similar to proof 1 at proofwiki.org/wiki/Heine-Cantor_Theorem#Proof_1.
$endgroup$
– Ayman Hourieh
Mar 26 '13 at 15:43
|
show 7 more comments
$begingroup$
I think you need to elaborate more on the last step. $delta_x$ and $delta_y$ could be smaller than the chosen $delta$. You need to find the $x_i$ for which $x in (x_i - delta_{x_i}, x_i + delta_{x_i})$, and make sure that $y$ is close enough to $x$ so that it also belongs to this interval.
$endgroup$
– Ayman Hourieh
Mar 18 '13 at 9:33
$begingroup$
@Ayman I see what you mean.
$endgroup$
– Pedro Tamaroff♦
Mar 18 '13 at 13:11
$begingroup$
Thanks. The elaboration needed for your original proof isn't big. It's something along the lines of my previous comment. But it is necessary; otherwise it's unclear why $left|f(x) - f(y)right| < epsilon$.
$endgroup$
– Ayman Hourieh
Mar 18 '13 at 17:40
$begingroup$
@ayman After some reading I have found something that fixes my proof try. I will try and digest it, and if possible add it tomorrow.
$endgroup$
– Pedro Tamaroff♦
Mar 26 '13 at 2:29
$begingroup$
The fix I had in mind was similar to proof 1 at proofwiki.org/wiki/Heine-Cantor_Theorem#Proof_1.
$endgroup$
– Ayman Hourieh
Mar 26 '13 at 15:43
$begingroup$
I think you need to elaborate more on the last step. $delta_x$ and $delta_y$ could be smaller than the chosen $delta$. You need to find the $x_i$ for which $x in (x_i - delta_{x_i}, x_i + delta_{x_i})$, and make sure that $y$ is close enough to $x$ so that it also belongs to this interval.
$endgroup$
– Ayman Hourieh
Mar 18 '13 at 9:33
$begingroup$
I think you need to elaborate more on the last step. $delta_x$ and $delta_y$ could be smaller than the chosen $delta$. You need to find the $x_i$ for which $x in (x_i - delta_{x_i}, x_i + delta_{x_i})$, and make sure that $y$ is close enough to $x$ so that it also belongs to this interval.
$endgroup$
– Ayman Hourieh
Mar 18 '13 at 9:33
$begingroup$
@Ayman I see what you mean.
$endgroup$
– Pedro Tamaroff♦
Mar 18 '13 at 13:11
$begingroup$
@Ayman I see what you mean.
$endgroup$
– Pedro Tamaroff♦
Mar 18 '13 at 13:11
$begingroup$
Thanks. The elaboration needed for your original proof isn't big. It's something along the lines of my previous comment. But it is necessary; otherwise it's unclear why $left|f(x) - f(y)right| < epsilon$.
$endgroup$
– Ayman Hourieh
Mar 18 '13 at 17:40
$begingroup$
Thanks. The elaboration needed for your original proof isn't big. It's something along the lines of my previous comment. But it is necessary; otherwise it's unclear why $left|f(x) - f(y)right| < epsilon$.
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– Ayman Hourieh
Mar 18 '13 at 17:40
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@ayman After some reading I have found something that fixes my proof try. I will try and digest it, and if possible add it tomorrow.
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– Pedro Tamaroff♦
Mar 26 '13 at 2:29
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@ayman After some reading I have found something that fixes my proof try. I will try and digest it, and if possible add it tomorrow.
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– Pedro Tamaroff♦
Mar 26 '13 at 2:29
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The fix I had in mind was similar to proof 1 at proofwiki.org/wiki/Heine-Cantor_Theorem#Proof_1.
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– Ayman Hourieh
Mar 26 '13 at 15:43
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The fix I had in mind was similar to proof 1 at proofwiki.org/wiki/Heine-Cantor_Theorem#Proof_1.
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– Ayman Hourieh
Mar 26 '13 at 15:43
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show 7 more comments
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In this case, intuitively, no matter how large the interval is, as long as it is closed and bounded, the function is still controlled. This is a special case of a more general result: a continuous function on a compact metric space is uniformly continuous, and we know that a subset of Euclidean space is compact if and only if it is closed and bounded.
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add a comment |
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In this case, intuitively, no matter how large the interval is, as long as it is closed and bounded, the function is still controlled. This is a special case of a more general result: a continuous function on a compact metric space is uniformly continuous, and we know that a subset of Euclidean space is compact if and only if it is closed and bounded.
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add a comment |
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In this case, intuitively, no matter how large the interval is, as long as it is closed and bounded, the function is still controlled. This is a special case of a more general result: a continuous function on a compact metric space is uniformly continuous, and we know that a subset of Euclidean space is compact if and only if it is closed and bounded.
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In this case, intuitively, no matter how large the interval is, as long as it is closed and bounded, the function is still controlled. This is a special case of a more general result: a continuous function on a compact metric space is uniformly continuous, and we know that a subset of Euclidean space is compact if and only if it is closed and bounded.
edited Mar 17 '13 at 20:54
answered Mar 17 '13 at 20:49
user4594
add a comment |
add a comment |
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Because even if you take $b$ as large as you want, it is still finite, and you still have $|f(x)-f(y)| leq C|x-y|$ for a big constant $C$. (I'm talking about the function f(x)=x^2$ here)
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– Beni Bogosel
Mar 17 '13 at 20:44
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Heine-Cantor Theorem.
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– Julien
Mar 17 '13 at 20:55
2
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This theorem illustrates the importance of distinguishing very large and infinite. Your example gives a direct illustration of this, when coupled with the theorem.
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– Kieran Cooney
Oct 26 '13 at 18:51