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If $f$ is uniformly continuous on $[1,4]$ and uniformly continuous on $[2,5]$ is $f$ uniformly continuous on...
If $f$ is uniformly continuous on $[1,4]$ and uniformly continuous on $[2,5]$ is $f$ uniformly continuous on $[1,5]$?
I know that the answer is that it is true, but I am interested as to why my proof is incorrect. What I did was:
Since
$exists delta_1 s.t. |x-y|<delta_1 to |f(x)-f(y)|<frac{epsilon}{2}$ for $x,y in [1,2]$ and $exists delta_2 s.t. |x-y|<delta_2 to |f(x)-f(y)|<frac{epsilon}{2}$ for $x,y in [2,5]$
Then set $delta_3 = min{delta_1,delta_2}$
and you get (by triangle inequality)
$|f(x)-f(y)|=|f(x)-f(3)+f(3)-f(y)|leq |f(x)-f(3)| + |f(y)-f(3)|leqfrac{epsilon}{2}+frac{epsilon}{2}=epsilon$
as long as $|x-y|<delta_3$
What's wrong with this proof and how can I fix it?
real-analysis proof-verification epsilon-delta
asked Nov 20 '18 at 20:24
Riley H
896
|
show 3 more comments
If $f$ is uniformly continuous on $[1,4]$ and uniformly continuous on $[2,5]$ is $f$ uniformly continuous on $[1,5]$?
I know that the answer is that it is true, but I am interested as to why my proof is incorrect. What I did was:
Since
$exists delta_1 s.t. |x-y|<delta_1 to |f(x)-f(y)|<frac{epsilon}{2}$ for $x,y in [1,2]$ and $exists delta_2 s.t. |x-y|<delta_2 to |f(x)-f(y)|<frac{epsilon}{2}$ for $x,y in [2,5]$
Then set $delta_3 = min{delta_1,delta_2}$
and you get (by triangle inequality)
$|f(x)-f(y)|=|f(x)-f(3)+f(3)-f(y)|leq |f(x)-f(3)| + |f(y)-f(3)|leqfrac{epsilon}{2}+frac{epsilon}{2}=epsilon$
as long as $|x-y|<delta_3$
What's wrong with this proof and how can I fix it?
real-analysis proof-verification epsilon-delta
asked Nov 20 '18 at 20:24
Riley H
896
What does $3$ have to do with it? In general $|f(x) - f(3)|$ and $|f(y) - f(3)|$ are not small.
– Robert Israel
Nov 20 '18 at 20:27
@RobertIsrael because 3 is in both intervals, so we know that there exists a delta that makes $|f(x)-f(3)|<epsilon$, right?
– Riley H
Nov 20 '18 at 20:31
Would there be any other changes if you used continuity at 2? @hamam_Abdallah
– Riley H
Nov 20 '18 at 20:32
2
Here is a bizarre proof: If $f$ is continuous on $[1,4]$ and on $[2,5]$, then it is continuous on $[1,5]$ by the pasting lemma. Then, being continuous on a compact set, it is uniformly continuous.
– Sangchul Lee
Nov 20 '18 at 20:35
1
For the standard proof, let $delta_1, delta_2$ be as in your setting and let $delta = min{delta_1, delta_2, 2}$. If $x,yin[1,5]$ satisfy $|x-y|<delta$, argue that either $x,yin[1,4]$ or $x,yin[2,5]$ holds, so that the defining property of $delta_i$'s kicks in.
– Sangchul Lee
Nov 20 '18 at 20:39
|
show 3 more comments
If $f$ is uniformly continuous on $[1,4]$ and uniformly continuous on $[2,5]$ is $f$ uniformly continuous on $[1,5]$?
I know that the answer is that it is true, but I am interested as to why my proof is incorrect. What I did was:
Since
$exists delta_1 s.t. |x-y|<delta_1 to |f(x)-f(y)|<frac{epsilon}{2}$ for $x,y in [1,2]$ and $exists delta_2 s.t. |x-y|<delta_2 to |f(x)-f(y)|<frac{epsilon}{2}$ for $x,y in [2,5]$
Then set $delta_3 = min{delta_1,delta_2}$
and you get (by triangle inequality)
$|f(x)-f(y)|=|f(x)-f(3)+f(3)-f(y)|leq |f(x)-f(3)| + |f(y)-f(3)|leqfrac{epsilon}{2}+frac{epsilon}{2}=epsilon$
as long as $|x-y|<delta_3$
What's wrong with this proof and how can I fix it?
real-analysis proof-verification epsilon-delta
asked Nov 20 '18 at 20:24
Riley H
896
If $f$ is uniformly continuous on $[1,4]$ and uniformly continuous on $[2,5]$ is $f$ uniformly continuous on $[1,5]$?
I know that the answer is that it is true, but I am interested as to why my proof is incorrect. What I did was:
Since
$exists delta_1 s.t. |x-y|<delta_1 to |f(x)-f(y)|<frac{epsilon}{2}$ for $x,y in [1,2]$ and $exists delta_2 s.t. |x-y|<delta_2 to |f(x)-f(y)|<frac{epsilon}{2}$ for $x,y in [2,5]$
Then set $delta_3 = min{delta_1,delta_2}$
and you get (by triangle inequality)
$|f(x)-f(y)|=|f(x)-f(3)+f(3)-f(y)|leq |f(x)-f(3)| + |f(y)-f(3)|leqfrac{epsilon}{2}+frac{epsilon}{2}=epsilon$
as long as $|x-y|<delta_3$
What's wrong with this proof and how can I fix it?
real-analysis proof-verification epsilon-delta
real-analysis proof-verification epsilon-delta
asked Nov 20 '18 at 20:24
Riley H
896
asked Nov 20 '18 at 20:24
Riley H
896
asked Nov 20 '18 at 20:24
Riley H
896
asked Nov 20 '18 at 20:24
Riley H
896
asked Nov 20 '18 at 20:24
Riley H
896
896
What does $3$ have to do with it? In general $|f(x) - f(3)|$ and $|f(y) - f(3)|$ are not small.
– Robert Israel
Nov 20 '18 at 20:27
@RobertIsrael because 3 is in both intervals, so we know that there exists a delta that makes $|f(x)-f(3)|<epsilon$, right?
– Riley H
Nov 20 '18 at 20:31
Would there be any other changes if you used continuity at 2? @hamam_Abdallah
– Riley H
Nov 20 '18 at 20:32
2
Here is a bizarre proof: If $f$ is continuous on $[1,4]$ and on $[2,5]$, then it is continuous on $[1,5]$ by the pasting lemma. Then, being continuous on a compact set, it is uniformly continuous.
– Sangchul Lee
Nov 20 '18 at 20:35
1
For the standard proof, let $delta_1, delta_2$ be as in your setting and let $delta = min{delta_1, delta_2, 2}$. If $x,yin[1,5]$ satisfy $|x-y|<delta$, argue that either $x,yin[1,4]$ or $x,yin[2,5]$ holds, so that the defining property of $delta_i$'s kicks in.
– Sangchul Lee
Nov 20 '18 at 20:39
|
show 3 more comments
What does $3$ have to do with it? In general $|f(x) - f(3)|$ and $|f(y) - f(3)|$ are not small.
– Robert Israel
Nov 20 '18 at 20:27
@RobertIsrael because 3 is in both intervals, so we know that there exists a delta that makes $|f(x)-f(3)|<epsilon$, right?
– Riley H
Nov 20 '18 at 20:31
Would there be any other changes if you used continuity at 2? @hamam_Abdallah
– Riley H
Nov 20 '18 at 20:32
2
Here is a bizarre proof: If $f$ is continuous on $[1,4]$ and on $[2,5]$, then it is continuous on $[1,5]$ by the pasting lemma. Then, being continuous on a compact set, it is uniformly continuous.
– Sangchul Lee
Nov 20 '18 at 20:35
1
For the standard proof, let $delta_1, delta_2$ be as in your setting and let $delta = min{delta_1, delta_2, 2}$. If $x,yin[1,5]$ satisfy $|x-y|<delta$, argue that either $x,yin[1,4]$ or $x,yin[2,5]$ holds, so that the defining property of $delta_i$'s kicks in.
– Sangchul Lee
Nov 20 '18 at 20:39
What does $3$ have to do with it? In general $|f(x) - f(3)|$ and $|f(y) - f(3)|$ are not small.
– Robert Israel
Nov 20 '18 at 20:27
What does $3$ have to do with it? In general $|f(x) - f(3)|$ and $|f(y) - f(3)|$ are not small.
– Robert Israel
Nov 20 '18 at 20:27
@RobertIsrael because 3 is in both intervals, so we know that there exists a delta that makes $|f(x)-f(3)|<epsilon$, right?
– Riley H
Nov 20 '18 at 20:31
@RobertIsrael because 3 is in both intervals, so we know that there exists a delta that makes $|f(x)-f(3)|<epsilon$, right?
– Riley H
Nov 20 '18 at 20:31
Would there be any other changes if you used continuity at 2? @hamam_Abdallah
– Riley H
Nov 20 '18 at 20:32
Would there be any other changes if you used continuity at 2? @hamam_Abdallah
– Riley H
Nov 20 '18 at 20:32
2
2
Here is a bizarre proof: If $f$ is continuous on $[1,4]$ and on $[2,5]$, then it is continuous on $[1,5]$ by the pasting lemma. Then, being continuous on a compact set, it is uniformly continuous.
– Sangchul Lee
Nov 20 '18 at 20:35
Here is a bizarre proof: If $f$ is continuous on $[1,4]$ and on $[2,5]$, then it is continuous on $[1,5]$ by the pasting lemma. Then, being continuous on a compact set, it is uniformly continuous.
– Sangchul Lee
Nov 20 '18 at 20:35
1
1
For the standard proof, let $delta_1, delta_2$ be as in your setting and let $delta = min{delta_1, delta_2, 2}$. If $x,yin[1,5]$ satisfy $|x-y|<delta$, argue that either $x,yin[1,4]$ or $x,yin[2,5]$ holds, so that the defining property of $delta_i$'s kicks in.
– Sangchul Lee
Nov 20 '18 at 20:39
For the standard proof, let $delta_1, delta_2$ be as in your setting and let $delta = min{delta_1, delta_2, 2}$. If $x,yin[1,5]$ satisfy $|x-y|<delta$, argue that either $x,yin[1,4]$ or $x,yin[2,5]$ holds, so that the defining property of $delta_i$'s kicks in.
– Sangchul Lee
Nov 20 '18 at 20:39
|
show 3 more comments
1 Answer
1
active
oldest
votes
hint
The uniform continuity at $[1,4]$ gives $delta_1$
the UC at $[4,5]$ will give $delta_2$.
the continuity at $x=4$ gives $delta_3$ such that
$$|x-4|<delta_3 implies |f(x)-f(4)|<frac{epsilon}{2}$$
Take $delta=min(delta_i,i=1,2,3).$
If $xin[1,4]$ and $yin[4,5]$ are such $|x-y|<delta$ then
$|x-4|<delta_3$ and $|y-4|<delta_3$
thus
$$|f(x)-f(y)|=$$
$$=|f(x)-f(4)+f(4)-f(y)|<epsilon$$
answered Nov 20 '18 at 20:39
hamam_Abdallah
37.9k21634
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
hint
The uniform continuity at $[1,4]$ gives $delta_1$
the UC at $[4,5]$ will give $delta_2$.
the continuity at $x=4$ gives $delta_3$ such that
$$|x-4|<delta_3 implies |f(x)-f(4)|<frac{epsilon}{2}$$
Take $delta=min(delta_i,i=1,2,3).$
If $xin[1,4]$ and $yin[4,5]$ are such $|x-y|<delta$ then
$|x-4|<delta_3$ and $|y-4|<delta_3$
thus
$$|f(x)-f(y)|=$$
$$=|f(x)-f(4)+f(4)-f(y)|<epsilon$$
answered Nov 20 '18 at 20:39
hamam_Abdallah
37.9k21634
add a comment |
hint
The uniform continuity at $[1,4]$ gives $delta_1$
the UC at $[4,5]$ will give $delta_2$.
the continuity at $x=4$ gives $delta_3$ such that
$$|x-4|<delta_3 implies |f(x)-f(4)|<frac{epsilon}{2}$$
Take $delta=min(delta_i,i=1,2,3).$
If $xin[1,4]$ and $yin[4,5]$ are such $|x-y|<delta$ then
$|x-4|<delta_3$ and $|y-4|<delta_3$
thus
$$|f(x)-f(y)|=$$
$$=|f(x)-f(4)+f(4)-f(y)|<epsilon$$
answered Nov 20 '18 at 20:39
hamam_Abdallah
37.9k21634
add a comment |
hint
The uniform continuity at $[1,4]$ gives $delta_1$
the UC at $[4,5]$ will give $delta_2$.
the continuity at $x=4$ gives $delta_3$ such that
$$|x-4|<delta_3 implies |f(x)-f(4)|<frac{epsilon}{2}$$
Take $delta=min(delta_i,i=1,2,3).$
If $xin[1,4]$ and $yin[4,5]$ are such $|x-y|<delta$ then
$|x-4|<delta_3$ and $|y-4|<delta_3$
thus
$$|f(x)-f(y)|=$$
$$=|f(x)-f(4)+f(4)-f(y)|<epsilon$$
answered Nov 20 '18 at 20:39
hamam_Abdallah
37.9k21634
hint
The uniform continuity at $[1,4]$ gives $delta_1$
the UC at $[4,5]$ will give $delta_2$.
the continuity at $x=4$ gives $delta_3$ such that
$$|x-4|<delta_3 implies |f(x)-f(4)|<frac{epsilon}{2}$$
Take $delta=min(delta_i,i=1,2,3).$
If $xin[1,4]$ and $yin[4,5]$ are such $|x-y|<delta$ then
$|x-4|<delta_3$ and $|y-4|<delta_3$
thus
$$|f(x)-f(y)|=$$
$$=|f(x)-f(4)+f(4)-f(y)|<epsilon$$
answered Nov 20 '18 at 20:39
hamam_Abdallah
37.9k21634
edited Nov 20 '18 at 20:44
answered Nov 20 '18 at 20:39
hamam_Abdallah
37.9k21634
answered Nov 20 '18 at 20:39
hamam_Abdallah
37.9k21634
answered Nov 20 '18 at 20:39
hamam_Abdallah
37.9k21634
37.9k21634
add a comment |
add a comment |
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What does $3$ have to do with it? In general $|f(x) - f(3)|$ and $|f(y) - f(3)|$ are not small.
– Robert Israel
Nov 20 '18 at 20:27
@RobertIsrael because 3 is in both intervals, so we know that there exists a delta that makes $|f(x)-f(3)|<epsilon$, right?
– Riley H
Nov 20 '18 at 20:31
Would there be any other changes if you used continuity at 2? @hamam_Abdallah
– Riley H
Nov 20 '18 at 20:32
2
Here is a bizarre proof: If $f$ is continuous on $[1,4]$ and on $[2,5]$, then it is continuous on $[1,5]$ by the pasting lemma. Then, being continuous on a compact set, it is uniformly continuous.
– Sangchul Lee
Nov 20 '18 at 20:35
1
For the standard proof, let $delta_1, delta_2$ be as in your setting and let $delta = min{delta_1, delta_2, 2}$. If $x,yin[1,5]$ satisfy $|x-y|<delta$, argue that either $x,yin[1,4]$ or $x,yin[2,5]$ holds, so that the defining property of $delta_i$'s kicks in.
– Sangchul Lee
Nov 20 '18 at 20:39