Mixing
Is my proof that this limit does not exist correct?
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Proposition:
Let $f$ be a function defined as $f(x) = x$ if $x$ is rational and $f(x) = 0$ otherwise. The limit of $f(x)$ as $x$ approaches any number $a neq 0$ does not exist.
My proof:
Suppose by contradiction that there is a limit $L$ for $f(x)$, as $x$ approaches $a neq 0$.
By definition of limit, there is a $delta$-neighborhood of $a$ in which, as $x$ gets closer to $a$, in a $epsilon$-neighborhood of $L$, $f(x)$ gets closer to $L$.
Let's take a irrational number $x$ in such an neighborhood of $a$. That would imply $|L| < epsilon$.
We know that we can find a $y$ even closer to $a$ which implies that $f(y)$ is even closer to $L$ than $f(x)$ was.
Since there are infinitely many numbers like that who happen to be rational, we have that $|y - L| < |L|$.
But we also have infinitely many irrational numbers $z$, that are even closer to $a$ such that $f(z)$ is closer to $L$ than $f(y)$ was.
So we must have $|L| < |y - L| < |L|$ which is absurd.
We conclude then that there is no limit $L$. ■
calculus limits functions proof-verification proof-writing
asked Jan 24 at 13:29
ArielKArielK
386
$endgroup$
add a comment |
$begingroup$
Proposition:
Let $f$ be a function defined as $f(x) = x$ if $x$ is rational and $f(x) = 0$ otherwise. The limit of $f(x)$ as $x$ approaches any number $a neq 0$ does not exist.
My proof:
Suppose by contradiction that there is a limit $L$ for $f(x)$, as $x$ approaches $a neq 0$.
By definition of limit, there is a $delta$-neighborhood of $a$ in which, as $x$ gets closer to $a$, in a $epsilon$-neighborhood of $L$, $f(x)$ gets closer to $L$.
Let's take a irrational number $x$ in such an neighborhood of $a$. That would imply $|L| < epsilon$.
We know that we can find a $y$ even closer to $a$ which implies that $f(y)$ is even closer to $L$ than $f(x)$ was.
Since there are infinitely many numbers like that who happen to be rational, we have that $|y - L| < |L|$.
But we also have infinitely many irrational numbers $z$, that are even closer to $a$ such that $f(z)$ is closer to $L$ than $f(y)$ was.
So we must have $|L| < |y - L| < |L|$ which is absurd.
We conclude then that there is no limit $L$. ■
calculus limits functions proof-verification proof-writing
asked Jan 24 at 13:29
ArielKArielK
386
$endgroup$
add a comment |
$begingroup$
Proposition:
Let $f$ be a function defined as $f(x) = x$ if $x$ is rational and $f(x) = 0$ otherwise. The limit of $f(x)$ as $x$ approaches any number $a neq 0$ does not exist.
My proof:
Suppose by contradiction that there is a limit $L$ for $f(x)$, as $x$ approaches $a neq 0$.
By definition of limit, there is a $delta$-neighborhood of $a$ in which, as $x$ gets closer to $a$, in a $epsilon$-neighborhood of $L$, $f(x)$ gets closer to $L$.
Let's take a irrational number $x$ in such an neighborhood of $a$. That would imply $|L| < epsilon$.
We know that we can find a $y$ even closer to $a$ which implies that $f(y)$ is even closer to $L$ than $f(x)$ was.
Since there are infinitely many numbers like that who happen to be rational, we have that $|y - L| < |L|$.
But we also have infinitely many irrational numbers $z$, that are even closer to $a$ such that $f(z)$ is closer to $L$ than $f(y)$ was.
So we must have $|L| < |y - L| < |L|$ which is absurd.
We conclude then that there is no limit $L$. ■
calculus limits functions proof-verification proof-writing
asked Jan 24 at 13:29
ArielKArielK
386
$endgroup$
Proposition:
Let $f$ be a function defined as $f(x) = x$ if $x$ is rational and $f(x) = 0$ otherwise. The limit of $f(x)$ as $x$ approaches any number $a neq 0$ does not exist.
My proof:
Suppose by contradiction that there is a limit $L$ for $f(x)$, as $x$ approaches $a neq 0$.
By definition of limit, there is a $delta$-neighborhood of $a$ in which, as $x$ gets closer to $a$, in a $epsilon$-neighborhood of $L$, $f(x)$ gets closer to $L$.
Let's take a irrational number $x$ in such an neighborhood of $a$. That would imply $|L| < epsilon$.
We know that we can find a $y$ even closer to $a$ which implies that $f(y)$ is even closer to $L$ than $f(x)$ was.
Since there are infinitely many numbers like that who happen to be rational, we have that $|y - L| < |L|$.
But we also have infinitely many irrational numbers $z$, that are even closer to $a$ such that $f(z)$ is closer to $L$ than $f(y)$ was.
So we must have $|L| < |y - L| < |L|$ which is absurd.
We conclude then that there is no limit $L$. ■
calculus limits functions proof-verification proof-writing
calculus limits functions proof-verification proof-writing
asked Jan 24 at 13:29
ArielKArielK
386
asked Jan 24 at 13:29
ArielKArielK
386
asked Jan 24 at 13:29
ArielKArielK
386
asked Jan 24 at 13:29
ArielKArielK
386
asked Jan 24 at 13:29
ArielKArielK
386
386
add a comment |
add a comment |
2 Answers
2
active
oldest
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$begingroup$
I'm not sure I follow your proof entirely. you're using vague notions like "getting closer", and your logic is a bit wacky.
You want to disprove that there is a limit at $a$. The definition of limit at $a$ is
$lim_{xto a}f(x)$ exists $iff$ there exists an $L$ such that for any $epsilon>0$, there is a $delta>0$ such that for any $x$ with $0<|x-a|<delta$ we have $|f(x) - L|<epsilon$
If you want to disprove this, then you have to prove the negation. We get:
$lim_{xto a}f(x)$ doesn't exist $iff$ for any $L$, there exists an $epsilon > 0$ such that for any $delta > 0$ there is an $x$ with $0<|x-a|<delta$ such that $|f(x) - L|geq epsilon$
Wow, that's a mouthful. However, the key part here is this: "there exists an $epsilon >0$". You have to show that some $epsilon$ exists. The easiest way of doing that is to pick one (cleverly), and then use that. Once I saw that you hadn't done that, I was almost certain that your proof wouldn't be correct.
For instance, a proof might go like this:
Let $L$ be arbitrary. Set $epsilon = frac{|a|}3$. Let $delta$ be arbitrary. Now pick a rational $x_1$ and irrational $x_2$ such that $0<|x_i-a|<delta$ for $i = 1, 2$ and at the same time $|x_1|>frac23|a|$. (For small $delta$'s this last condition won't matter, as $|x_1-a|<delta$ is a strictly stronger condidion, but one has to take into account that a devil's advocate might pick our arbitrary $delta$ to be larger than $frac13|a|$.) I will now show that $|f(x_1) - L|<epsilon$ and $|f(x_2) - L|<epsilon$ cannot both hold (and thus one of them must be $!{}geqepsilon$).
Assume for contradiction that $|f(x_1) - L|<epsilon$ and $|f(x_2) - L|<epsilon$. We know that $x_2$ is irrational, which gives us $|L|<frac{|a|}3$. We know that $x_1$ is rational, giving us $|x_1-L|<frac{|a|}3$. Thus
$$
|x_1 - L| + |L| < frac23|a|\
|x_1 - L + L| < frac23|a|\
|x_1| < frac23|a|
$$
which is a contradiction.
answered Jan 24 at 14:11
ArthurArthur
118k7118201
$endgroup$
$begingroup$
But why $|x_1|>frac23|a|$?
$endgroup$
– ArielK
Jan 24 at 14:22
1
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@ArielK Because we need it for the contradiction later.
$endgroup$
– Arthur
Jan 24 at 14:23
1
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@ArielK I'm not saying $|x_1| = frac23|a|$. I'm saying $|x_1|> frac23 |a|$. Regardless of whehter $a$ is rational or not, there are plenty of rational numbers to choose from for $x_1$.
$endgroup$
– Arthur
Jan 24 at 14:29
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Thanks, just realized that :P sorry for my stupidity
$endgroup$
– ArielK
Jan 24 at 14:33
1
$begingroup$
@ArielK These proofs are often thought through before the final version is written down. Things need constant tweaking as you figure out what you need. For instance, I started with $epsilon = frac{|a|}2$, and then thought "but what if $x_1$ is chosen to be smaller than $a$?", so I changed it to $frac{|a|}3$. Then I thought "What if $delta$ is really large, and $x_1$ is chosen to be really close to $0$?", so I made the requirement that $|x_1|>frac23|a|$. Maybe I could've saved myself all the trouble and just required that $|x_1|>|a|$ in the first place. The proof might've been neater.
$endgroup$
– Arthur
Jan 24 at 14:36
|
show 2 more comments
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Your solution is not acceptable.
E.g. you use terminology like "closer" or "getting closer" without making clear what is meant by that. This makes it impossible to check your proof.
Observe that $mathbb Q^{complement}$ and $mathbb Q$ in $mathbb R$ are both dense subsets of $mathbb R$.
From that conclude that there are a sequences $(p_n)_n$ and $(q_n)_n$ both converging to $a$ with $p_ninmathbb Q^{complement}$ and $q_ninmathbb Q$ for every $n$.
If a limit $L$ exists then $L=lim_{ntoinfty}f(p_n)=a$ and $L=lim_{ntoinfty}f(q_n)=0$ which is not possible if $aneq0$.
answered Jan 24 at 13:51
drhabdrhab
103k545136
$endgroup$
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Why should "$y$ even closer to $a$ than $x$" imply that "$f(y)$ is even closer to $L$" than $f(x)$"? Such $y$ will exist, but must be chosen. It is not enough to let it rest only on the (vague) fact that $y$ is closer to $a$. Further to judge informal arguments is more difficult for me than to judge formal proofs.
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– drhab
Jan 24 at 14:06
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Because the definition of limit implies that there is such an neighborhood of a for which we can make f(x) as close as we wish from the limit by taking sufficiently closer to a
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– ArielK
Jan 24 at 14:10
1
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Yes, and the condition must be that $y$ belongs to such neighborhood. Not only that $y$ is closer to $a$ than $x$. That is not enough on its own.
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– drhab
Jan 24 at 14:14
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But I stated that $y$ not only is closer but is also in that neighborhood, how could it be out of the neighborhood if it's closer to $a$ than $x$?
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– ArielK
Jan 24 at 14:31
1
$begingroup$
Where did you state that $y$ is in that neighborhood? I do not encounter that. Only that $x$ and $y$ are in some $epsilon$-neighborhood of $L$ and that $y$ ix closer to $a$ than $x$. Again: that is not enough to ensure that $f(y)$ is closer to $L$.
$endgroup$
– drhab
Jan 24 at 14:37
add a comment |
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2 Answers
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2 Answers
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$begingroup$
I'm not sure I follow your proof entirely. you're using vague notions like "getting closer", and your logic is a bit wacky.
You want to disprove that there is a limit at $a$. The definition of limit at $a$ is
$lim_{xto a}f(x)$ exists $iff$ there exists an $L$ such that for any $epsilon>0$, there is a $delta>0$ such that for any $x$ with $0<|x-a|<delta$ we have $|f(x) - L|<epsilon$
If you want to disprove this, then you have to prove the negation. We get:
$lim_{xto a}f(x)$ doesn't exist $iff$ for any $L$, there exists an $epsilon > 0$ such that for any $delta > 0$ there is an $x$ with $0<|x-a|<delta$ such that $|f(x) - L|geq epsilon$
Wow, that's a mouthful. However, the key part here is this: "there exists an $epsilon >0$". You have to show that some $epsilon$ exists. The easiest way of doing that is to pick one (cleverly), and then use that. Once I saw that you hadn't done that, I was almost certain that your proof wouldn't be correct.
For instance, a proof might go like this:
Let $L$ be arbitrary. Set $epsilon = frac{|a|}3$. Let $delta$ be arbitrary. Now pick a rational $x_1$ and irrational $x_2$ such that $0<|x_i-a|<delta$ for $i = 1, 2$ and at the same time $|x_1|>frac23|a|$. (For small $delta$'s this last condition won't matter, as $|x_1-a|<delta$ is a strictly stronger condidion, but one has to take into account that a devil's advocate might pick our arbitrary $delta$ to be larger than $frac13|a|$.) I will now show that $|f(x_1) - L|<epsilon$ and $|f(x_2) - L|<epsilon$ cannot both hold (and thus one of them must be $!{}geqepsilon$).
Assume for contradiction that $|f(x_1) - L|<epsilon$ and $|f(x_2) - L|<epsilon$. We know that $x_2$ is irrational, which gives us $|L|<frac{|a|}3$. We know that $x_1$ is rational, giving us $|x_1-L|<frac{|a|}3$. Thus
$$
|x_1 - L| + |L| < frac23|a|\
|x_1 - L + L| < frac23|a|\
|x_1| < frac23|a|
$$
which is a contradiction.
answered Jan 24 at 14:11
ArthurArthur
118k7118201
$endgroup$
$begingroup$
But why $|x_1|>frac23|a|$?
$endgroup$
– ArielK
Jan 24 at 14:22
1
$begingroup$
@ArielK Because we need it for the contradiction later.
$endgroup$
– Arthur
Jan 24 at 14:23
1
$begingroup$
@ArielK I'm not saying $|x_1| = frac23|a|$. I'm saying $|x_1|> frac23 |a|$. Regardless of whehter $a$ is rational or not, there are plenty of rational numbers to choose from for $x_1$.
$endgroup$
– Arthur
Jan 24 at 14:29
$begingroup$
Thanks, just realized that :P sorry for my stupidity
$endgroup$
– ArielK
Jan 24 at 14:33
1
$begingroup$
@ArielK These proofs are often thought through before the final version is written down. Things need constant tweaking as you figure out what you need. For instance, I started with $epsilon = frac{|a|}2$, and then thought "but what if $x_1$ is chosen to be smaller than $a$?", so I changed it to $frac{|a|}3$. Then I thought "What if $delta$ is really large, and $x_1$ is chosen to be really close to $0$?", so I made the requirement that $|x_1|>frac23|a|$. Maybe I could've saved myself all the trouble and just required that $|x_1|>|a|$ in the first place. The proof might've been neater.
$endgroup$
– Arthur
Jan 24 at 14:36
|
show 2 more comments
$begingroup$
I'm not sure I follow your proof entirely. you're using vague notions like "getting closer", and your logic is a bit wacky.
You want to disprove that there is a limit at $a$. The definition of limit at $a$ is
$lim_{xto a}f(x)$ exists $iff$ there exists an $L$ such that for any $epsilon>0$, there is a $delta>0$ such that for any $x$ with $0<|x-a|<delta$ we have $|f(x) - L|<epsilon$
If you want to disprove this, then you have to prove the negation. We get:
$lim_{xto a}f(x)$ doesn't exist $iff$ for any $L$, there exists an $epsilon > 0$ such that for any $delta > 0$ there is an $x$ with $0<|x-a|<delta$ such that $|f(x) - L|geq epsilon$
Wow, that's a mouthful. However, the key part here is this: "there exists an $epsilon >0$". You have to show that some $epsilon$ exists. The easiest way of doing that is to pick one (cleverly), and then use that. Once I saw that you hadn't done that, I was almost certain that your proof wouldn't be correct.
For instance, a proof might go like this:
Let $L$ be arbitrary. Set $epsilon = frac{|a|}3$. Let $delta$ be arbitrary. Now pick a rational $x_1$ and irrational $x_2$ such that $0<|x_i-a|<delta$ for $i = 1, 2$ and at the same time $|x_1|>frac23|a|$. (For small $delta$'s this last condition won't matter, as $|x_1-a|<delta$ is a strictly stronger condidion, but one has to take into account that a devil's advocate might pick our arbitrary $delta$ to be larger than $frac13|a|$.) I will now show that $|f(x_1) - L|<epsilon$ and $|f(x_2) - L|<epsilon$ cannot both hold (and thus one of them must be $!{}geqepsilon$).
Assume for contradiction that $|f(x_1) - L|<epsilon$ and $|f(x_2) - L|<epsilon$. We know that $x_2$ is irrational, which gives us $|L|<frac{|a|}3$. We know that $x_1$ is rational, giving us $|x_1-L|<frac{|a|}3$. Thus
$$
|x_1 - L| + |L| < frac23|a|\
|x_1 - L + L| < frac23|a|\
|x_1| < frac23|a|
$$
which is a contradiction.
answered Jan 24 at 14:11
ArthurArthur
118k7118201
$endgroup$
$begingroup$
But why $|x_1|>frac23|a|$?
$endgroup$
– ArielK
Jan 24 at 14:22
1
$begingroup$
@ArielK Because we need it for the contradiction later.
$endgroup$
– Arthur
Jan 24 at 14:23
1
$begingroup$
@ArielK I'm not saying $|x_1| = frac23|a|$. I'm saying $|x_1|> frac23 |a|$. Regardless of whehter $a$ is rational or not, there are plenty of rational numbers to choose from for $x_1$.
$endgroup$
– Arthur
Jan 24 at 14:29
$begingroup$
Thanks, just realized that :P sorry for my stupidity
$endgroup$
– ArielK
Jan 24 at 14:33
1
$begingroup$
@ArielK These proofs are often thought through before the final version is written down. Things need constant tweaking as you figure out what you need. For instance, I started with $epsilon = frac{|a|}2$, and then thought "but what if $x_1$ is chosen to be smaller than $a$?", so I changed it to $frac{|a|}3$. Then I thought "What if $delta$ is really large, and $x_1$ is chosen to be really close to $0$?", so I made the requirement that $|x_1|>frac23|a|$. Maybe I could've saved myself all the trouble and just required that $|x_1|>|a|$ in the first place. The proof might've been neater.
$endgroup$
– Arthur
Jan 24 at 14:36
|
show 2 more comments
$begingroup$
I'm not sure I follow your proof entirely. you're using vague notions like "getting closer", and your logic is a bit wacky.
You want to disprove that there is a limit at $a$. The definition of limit at $a$ is
$lim_{xto a}f(x)$ exists $iff$ there exists an $L$ such that for any $epsilon>0$, there is a $delta>0$ such that for any $x$ with $0<|x-a|<delta$ we have $|f(x) - L|<epsilon$
If you want to disprove this, then you have to prove the negation. We get:
$lim_{xto a}f(x)$ doesn't exist $iff$ for any $L$, there exists an $epsilon > 0$ such that for any $delta > 0$ there is an $x$ with $0<|x-a|<delta$ such that $|f(x) - L|geq epsilon$
Wow, that's a mouthful. However, the key part here is this: "there exists an $epsilon >0$". You have to show that some $epsilon$ exists. The easiest way of doing that is to pick one (cleverly), and then use that. Once I saw that you hadn't done that, I was almost certain that your proof wouldn't be correct.
For instance, a proof might go like this:
Let $L$ be arbitrary. Set $epsilon = frac{|a|}3$. Let $delta$ be arbitrary. Now pick a rational $x_1$ and irrational $x_2$ such that $0<|x_i-a|<delta$ for $i = 1, 2$ and at the same time $|x_1|>frac23|a|$. (For small $delta$'s this last condition won't matter, as $|x_1-a|<delta$ is a strictly stronger condidion, but one has to take into account that a devil's advocate might pick our arbitrary $delta$ to be larger than $frac13|a|$.) I will now show that $|f(x_1) - L|<epsilon$ and $|f(x_2) - L|<epsilon$ cannot both hold (and thus one of them must be $!{}geqepsilon$).
Assume for contradiction that $|f(x_1) - L|<epsilon$ and $|f(x_2) - L|<epsilon$. We know that $x_2$ is irrational, which gives us $|L|<frac{|a|}3$. We know that $x_1$ is rational, giving us $|x_1-L|<frac{|a|}3$. Thus
$$
|x_1 - L| + |L| < frac23|a|\
|x_1 - L + L| < frac23|a|\
|x_1| < frac23|a|
$$
which is a contradiction.
answered Jan 24 at 14:11
ArthurArthur
118k7118201
$endgroup$
I'm not sure I follow your proof entirely. you're using vague notions like "getting closer", and your logic is a bit wacky.
You want to disprove that there is a limit at $a$. The definition of limit at $a$ is
$lim_{xto a}f(x)$ exists $iff$ there exists an $L$ such that for any $epsilon>0$, there is a $delta>0$ such that for any $x$ with $0<|x-a|<delta$ we have $|f(x) - L|<epsilon$
If you want to disprove this, then you have to prove the negation. We get:
$lim_{xto a}f(x)$ doesn't exist $iff$ for any $L$, there exists an $epsilon > 0$ such that for any $delta > 0$ there is an $x$ with $0<|x-a|<delta$ such that $|f(x) - L|geq epsilon$
Wow, that's a mouthful. However, the key part here is this: "there exists an $epsilon >0$". You have to show that some $epsilon$ exists. The easiest way of doing that is to pick one (cleverly), and then use that. Once I saw that you hadn't done that, I was almost certain that your proof wouldn't be correct.
For instance, a proof might go like this:
Let $L$ be arbitrary. Set $epsilon = frac{|a|}3$. Let $delta$ be arbitrary. Now pick a rational $x_1$ and irrational $x_2$ such that $0<|x_i-a|<delta$ for $i = 1, 2$ and at the same time $|x_1|>frac23|a|$. (For small $delta$'s this last condition won't matter, as $|x_1-a|<delta$ is a strictly stronger condidion, but one has to take into account that a devil's advocate might pick our arbitrary $delta$ to be larger than $frac13|a|$.) I will now show that $|f(x_1) - L|<epsilon$ and $|f(x_2) - L|<epsilon$ cannot both hold (and thus one of them must be $!{}geqepsilon$).
Assume for contradiction that $|f(x_1) - L|<epsilon$ and $|f(x_2) - L|<epsilon$. We know that $x_2$ is irrational, which gives us $|L|<frac{|a|}3$. We know that $x_1$ is rational, giving us $|x_1-L|<frac{|a|}3$. Thus
$$
|x_1 - L| + |L| < frac23|a|\
|x_1 - L + L| < frac23|a|\
|x_1| < frac23|a|
$$
which is a contradiction.
answered Jan 24 at 14:11
ArthurArthur
118k7118201
answered Jan 24 at 14:11
ArthurArthur
118k7118201
answered Jan 24 at 14:11
ArthurArthur
118k7118201
answered Jan 24 at 14:11
ArthurArthur
118k7118201
118k7118201
$begingroup$
But why $|x_1|>frac23|a|$?
$endgroup$
– ArielK
Jan 24 at 14:22
1
$begingroup$
@ArielK Because we need it for the contradiction later.
$endgroup$
– Arthur
Jan 24 at 14:23
1
$begingroup$
@ArielK I'm not saying $|x_1| = frac23|a|$. I'm saying $|x_1|> frac23 |a|$. Regardless of whehter $a$ is rational or not, there are plenty of rational numbers to choose from for $x_1$.
$endgroup$
– Arthur
Jan 24 at 14:29
$begingroup$
Thanks, just realized that :P sorry for my stupidity
$endgroup$
– ArielK
Jan 24 at 14:33
1
$begingroup$
@ArielK These proofs are often thought through before the final version is written down. Things need constant tweaking as you figure out what you need. For instance, I started with $epsilon = frac{|a|}2$, and then thought "but what if $x_1$ is chosen to be smaller than $a$?", so I changed it to $frac{|a|}3$. Then I thought "What if $delta$ is really large, and $x_1$ is chosen to be really close to $0$?", so I made the requirement that $|x_1|>frac23|a|$. Maybe I could've saved myself all the trouble and just required that $|x_1|>|a|$ in the first place. The proof might've been neater.
$endgroup$
– Arthur
Jan 24 at 14:36
|
show 2 more comments
$begingroup$
But why $|x_1|>frac23|a|$?
$endgroup$
– ArielK
Jan 24 at 14:22
1
$begingroup$
@ArielK Because we need it for the contradiction later.
$endgroup$
– Arthur
Jan 24 at 14:23
1
$begingroup$
@ArielK I'm not saying $|x_1| = frac23|a|$. I'm saying $|x_1|> frac23 |a|$. Regardless of whehter $a$ is rational or not, there are plenty of rational numbers to choose from for $x_1$.
$endgroup$
– Arthur
Jan 24 at 14:29
$begingroup$
Thanks, just realized that :P sorry for my stupidity
$endgroup$
– ArielK
Jan 24 at 14:33
1
$begingroup$
@ArielK These proofs are often thought through before the final version is written down. Things need constant tweaking as you figure out what you need. For instance, I started with $epsilon = frac{|a|}2$, and then thought "but what if $x_1$ is chosen to be smaller than $a$?", so I changed it to $frac{|a|}3$. Then I thought "What if $delta$ is really large, and $x_1$ is chosen to be really close to $0$?", so I made the requirement that $|x_1|>frac23|a|$. Maybe I could've saved myself all the trouble and just required that $|x_1|>|a|$ in the first place. The proof might've been neater.
$endgroup$
– Arthur
Jan 24 at 14:36
$begingroup$
But why $|x_1|>frac23|a|$?
$endgroup$
– ArielK
Jan 24 at 14:22
$begingroup$
But why $|x_1|>frac23|a|$?
$endgroup$
– ArielK
Jan 24 at 14:22
1
1
$begingroup$
@ArielK Because we need it for the contradiction later.
$endgroup$
– Arthur
Jan 24 at 14:23
$begingroup$
@ArielK Because we need it for the contradiction later.
$endgroup$
– Arthur
Jan 24 at 14:23
1
1
$begingroup$
@ArielK I'm not saying $|x_1| = frac23|a|$. I'm saying $|x_1|> frac23 |a|$. Regardless of whehter $a$ is rational or not, there are plenty of rational numbers to choose from for $x_1$.
$endgroup$
– Arthur
Jan 24 at 14:29
$begingroup$
@ArielK I'm not saying $|x_1| = frac23|a|$. I'm saying $|x_1|> frac23 |a|$. Regardless of whehter $a$ is rational or not, there are plenty of rational numbers to choose from for $x_1$.
$endgroup$
– Arthur
Jan 24 at 14:29
$begingroup$
Thanks, just realized that :P sorry for my stupidity
$endgroup$
– ArielK
Jan 24 at 14:33
$begingroup$
Thanks, just realized that :P sorry for my stupidity
$endgroup$
– ArielK
Jan 24 at 14:33
1
1
$begingroup$
@ArielK These proofs are often thought through before the final version is written down. Things need constant tweaking as you figure out what you need. For instance, I started with $epsilon = frac{|a|}2$, and then thought "but what if $x_1$ is chosen to be smaller than $a$?", so I changed it to $frac{|a|}3$. Then I thought "What if $delta$ is really large, and $x_1$ is chosen to be really close to $0$?", so I made the requirement that $|x_1|>frac23|a|$. Maybe I could've saved myself all the trouble and just required that $|x_1|>|a|$ in the first place. The proof might've been neater.
$endgroup$
– Arthur
Jan 24 at 14:36
$begingroup$
@ArielK These proofs are often thought through before the final version is written down. Things need constant tweaking as you figure out what you need. For instance, I started with $epsilon = frac{|a|}2$, and then thought "but what if $x_1$ is chosen to be smaller than $a$?", so I changed it to $frac{|a|}3$. Then I thought "What if $delta$ is really large, and $x_1$ is chosen to be really close to $0$?", so I made the requirement that $|x_1|>frac23|a|$. Maybe I could've saved myself all the trouble and just required that $|x_1|>|a|$ in the first place. The proof might've been neater.
$endgroup$
– Arthur
Jan 24 at 14:36
|
show 2 more comments
$begingroup$
Your solution is not acceptable.
E.g. you use terminology like "closer" or "getting closer" without making clear what is meant by that. This makes it impossible to check your proof.
Observe that $mathbb Q^{complement}$ and $mathbb Q$ in $mathbb R$ are both dense subsets of $mathbb R$.
From that conclude that there are a sequences $(p_n)_n$ and $(q_n)_n$ both converging to $a$ with $p_ninmathbb Q^{complement}$ and $q_ninmathbb Q$ for every $n$.
If a limit $L$ exists then $L=lim_{ntoinfty}f(p_n)=a$ and $L=lim_{ntoinfty}f(q_n)=0$ which is not possible if $aneq0$.
answered Jan 24 at 13:51
drhabdrhab
103k545136
$endgroup$
$begingroup$
Why should "$y$ even closer to $a$ than $x$" imply that "$f(y)$ is even closer to $L$" than $f(x)$"? Such $y$ will exist, but must be chosen. It is not enough to let it rest only on the (vague) fact that $y$ is closer to $a$. Further to judge informal arguments is more difficult for me than to judge formal proofs.
$endgroup$
– drhab
Jan 24 at 14:06
$begingroup$
Because the definition of limit implies that there is such an neighborhood of a for which we can make f(x) as close as we wish from the limit by taking sufficiently closer to a
$endgroup$
– ArielK
Jan 24 at 14:10
1
$begingroup$
Yes, and the condition must be that $y$ belongs to such neighborhood. Not only that $y$ is closer to $a$ than $x$. That is not enough on its own.
$endgroup$
– drhab
Jan 24 at 14:14
$begingroup$
But I stated that $y$ not only is closer but is also in that neighborhood, how could it be out of the neighborhood if it's closer to $a$ than $x$?
$endgroup$
– ArielK
Jan 24 at 14:31
1
$begingroup$
Where did you state that $y$ is in that neighborhood? I do not encounter that. Only that $x$ and $y$ are in some $epsilon$-neighborhood of $L$ and that $y$ ix closer to $a$ than $x$. Again: that is not enough to ensure that $f(y)$ is closer to $L$.
$endgroup$
– drhab
Jan 24 at 14:37
add a comment |
$begingroup$
Your solution is not acceptable.
E.g. you use terminology like "closer" or "getting closer" without making clear what is meant by that. This makes it impossible to check your proof.
Observe that $mathbb Q^{complement}$ and $mathbb Q$ in $mathbb R$ are both dense subsets of $mathbb R$.
From that conclude that there are a sequences $(p_n)_n$ and $(q_n)_n$ both converging to $a$ with $p_ninmathbb Q^{complement}$ and $q_ninmathbb Q$ for every $n$.
If a limit $L$ exists then $L=lim_{ntoinfty}f(p_n)=a$ and $L=lim_{ntoinfty}f(q_n)=0$ which is not possible if $aneq0$.
answered Jan 24 at 13:51
drhabdrhab
103k545136
$endgroup$
$begingroup$
Why should "$y$ even closer to $a$ than $x$" imply that "$f(y)$ is even closer to $L$" than $f(x)$"? Such $y$ will exist, but must be chosen. It is not enough to let it rest only on the (vague) fact that $y$ is closer to $a$. Further to judge informal arguments is more difficult for me than to judge formal proofs.
$endgroup$
– drhab
Jan 24 at 14:06
$begingroup$
Because the definition of limit implies that there is such an neighborhood of a for which we can make f(x) as close as we wish from the limit by taking sufficiently closer to a
$endgroup$
– ArielK
Jan 24 at 14:10
1
$begingroup$
Yes, and the condition must be that $y$ belongs to such neighborhood. Not only that $y$ is closer to $a$ than $x$. That is not enough on its own.
$endgroup$
– drhab
Jan 24 at 14:14
$begingroup$
But I stated that $y$ not only is closer but is also in that neighborhood, how could it be out of the neighborhood if it's closer to $a$ than $x$?
$endgroup$
– ArielK
Jan 24 at 14:31
1
$begingroup$
Where did you state that $y$ is in that neighborhood? I do not encounter that. Only that $x$ and $y$ are in some $epsilon$-neighborhood of $L$ and that $y$ ix closer to $a$ than $x$. Again: that is not enough to ensure that $f(y)$ is closer to $L$.
$endgroup$
– drhab
Jan 24 at 14:37
add a comment |
$begingroup$
Your solution is not acceptable.
E.g. you use terminology like "closer" or "getting closer" without making clear what is meant by that. This makes it impossible to check your proof.
Observe that $mathbb Q^{complement}$ and $mathbb Q$ in $mathbb R$ are both dense subsets of $mathbb R$.
From that conclude that there are a sequences $(p_n)_n$ and $(q_n)_n$ both converging to $a$ with $p_ninmathbb Q^{complement}$ and $q_ninmathbb Q$ for every $n$.
If a limit $L$ exists then $L=lim_{ntoinfty}f(p_n)=a$ and $L=lim_{ntoinfty}f(q_n)=0$ which is not possible if $aneq0$.
answered Jan 24 at 13:51
drhabdrhab
103k545136
$endgroup$
Your solution is not acceptable.
E.g. you use terminology like "closer" or "getting closer" without making clear what is meant by that. This makes it impossible to check your proof.
Observe that $mathbb Q^{complement}$ and $mathbb Q$ in $mathbb R$ are both dense subsets of $mathbb R$.
From that conclude that there are a sequences $(p_n)_n$ and $(q_n)_n$ both converging to $a$ with $p_ninmathbb Q^{complement}$ and $q_ninmathbb Q$ for every $n$.
If a limit $L$ exists then $L=lim_{ntoinfty}f(p_n)=a$ and $L=lim_{ntoinfty}f(q_n)=0$ which is not possible if $aneq0$.
answered Jan 24 at 13:51
drhabdrhab
103k545136
answered Jan 24 at 13:51
drhabdrhab
103k545136
answered Jan 24 at 13:51
drhabdrhab
103k545136
answered Jan 24 at 13:51
drhabdrhab
103k545136
103k545136
$begingroup$
Why should "$y$ even closer to $a$ than $x$" imply that "$f(y)$ is even closer to $L$" than $f(x)$"? Such $y$ will exist, but must be chosen. It is not enough to let it rest only on the (vague) fact that $y$ is closer to $a$. Further to judge informal arguments is more difficult for me than to judge formal proofs.
$endgroup$
– drhab
Jan 24 at 14:06
$begingroup$
Because the definition of limit implies that there is such an neighborhood of a for which we can make f(x) as close as we wish from the limit by taking sufficiently closer to a
$endgroup$
– ArielK
Jan 24 at 14:10
1
$begingroup$
Yes, and the condition must be that $y$ belongs to such neighborhood. Not only that $y$ is closer to $a$ than $x$. That is not enough on its own.
$endgroup$
– drhab
Jan 24 at 14:14
$begingroup$
But I stated that $y$ not only is closer but is also in that neighborhood, how could it be out of the neighborhood if it's closer to $a$ than $x$?
$endgroup$
– ArielK
Jan 24 at 14:31
1
$begingroup$
Where did you state that $y$ is in that neighborhood? I do not encounter that. Only that $x$ and $y$ are in some $epsilon$-neighborhood of $L$ and that $y$ ix closer to $a$ than $x$. Again: that is not enough to ensure that $f(y)$ is closer to $L$.
$endgroup$
– drhab
Jan 24 at 14:37
add a comment |
$begingroup$
Why should "$y$ even closer to $a$ than $x$" imply that "$f(y)$ is even closer to $L$" than $f(x)$"? Such $y$ will exist, but must be chosen. It is not enough to let it rest only on the (vague) fact that $y$ is closer to $a$. Further to judge informal arguments is more difficult for me than to judge formal proofs.
$endgroup$
– drhab
Jan 24 at 14:06
$begingroup$
Because the definition of limit implies that there is such an neighborhood of a for which we can make f(x) as close as we wish from the limit by taking sufficiently closer to a
$endgroup$
– ArielK
Jan 24 at 14:10
1
$begingroup$
Yes, and the condition must be that $y$ belongs to such neighborhood. Not only that $y$ is closer to $a$ than $x$. That is not enough on its own.
$endgroup$
– drhab
Jan 24 at 14:14
$begingroup$
But I stated that $y$ not only is closer but is also in that neighborhood, how could it be out of the neighborhood if it's closer to $a$ than $x$?
$endgroup$
– ArielK
Jan 24 at 14:31
1
$begingroup$
Where did you state that $y$ is in that neighborhood? I do not encounter that. Only that $x$ and $y$ are in some $epsilon$-neighborhood of $L$ and that $y$ ix closer to $a$ than $x$. Again: that is not enough to ensure that $f(y)$ is closer to $L$.
$endgroup$
– drhab
Jan 24 at 14:37
$begingroup$
Why should "$y$ even closer to $a$ than $x$" imply that "$f(y)$ is even closer to $L$" than $f(x)$"? Such $y$ will exist, but must be chosen. It is not enough to let it rest only on the (vague) fact that $y$ is closer to $a$. Further to judge informal arguments is more difficult for me than to judge formal proofs.
$endgroup$
– drhab
Jan 24 at 14:06
$begingroup$
Why should "$y$ even closer to $a$ than $x$" imply that "$f(y)$ is even closer to $L$" than $f(x)$"? Such $y$ will exist, but must be chosen. It is not enough to let it rest only on the (vague) fact that $y$ is closer to $a$. Further to judge informal arguments is more difficult for me than to judge formal proofs.
$endgroup$
– drhab
Jan 24 at 14:06
$begingroup$
Because the definition of limit implies that there is such an neighborhood of a for which we can make f(x) as close as we wish from the limit by taking sufficiently closer to a
$endgroup$
– ArielK
Jan 24 at 14:10
$begingroup$
Because the definition of limit implies that there is such an neighborhood of a for which we can make f(x) as close as we wish from the limit by taking sufficiently closer to a
$endgroup$
– ArielK
Jan 24 at 14:10
1
1
$begingroup$
Yes, and the condition must be that $y$ belongs to such neighborhood. Not only that $y$ is closer to $a$ than $x$. That is not enough on its own.
$endgroup$
– drhab
Jan 24 at 14:14
$begingroup$
Yes, and the condition must be that $y$ belongs to such neighborhood. Not only that $y$ is closer to $a$ than $x$. That is not enough on its own.
$endgroup$
– drhab
Jan 24 at 14:14
$begingroup$
But I stated that $y$ not only is closer but is also in that neighborhood, how could it be out of the neighborhood if it's closer to $a$ than $x$?
$endgroup$
– ArielK
Jan 24 at 14:31
$begingroup$
But I stated that $y$ not only is closer but is also in that neighborhood, how could it be out of the neighborhood if it's closer to $a$ than $x$?
$endgroup$
– ArielK
Jan 24 at 14:31
1
1
$begingroup$
Where did you state that $y$ is in that neighborhood? I do not encounter that. Only that $x$ and $y$ are in some $epsilon$-neighborhood of $L$ and that $y$ ix closer to $a$ than $x$. Again: that is not enough to ensure that $f(y)$ is closer to $L$.
$endgroup$
– drhab
Jan 24 at 14:37
$begingroup$
Where did you state that $y$ is in that neighborhood? I do not encounter that. Only that $x$ and $y$ are in some $epsilon$-neighborhood of $L$ and that $y$ ix closer to $a$ than $x$. Again: that is not enough to ensure that $f(y)$ is closer to $L$.
$endgroup$
– drhab
Jan 24 at 14:37
add a comment |
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