Showing continuous function on a closed interval is bounded via uniform continuity?
In Analysis I, Tao writes:
Remark 9.9.17. One should compare Lemma 9.6.3, Proposition 9.9.15, and Theorem 9.9.16 with each other. No two of these results imply the third, but they are all consistent with each other.
For reference, here are the three results:
Lemma 9.6.3. Let $a < b$ be real numbers, and let $f : [a,b] to mathbf R$ be a function continuous on $[a,b]$. Then $f$ is a bounded function.
Proposition 9.9.15. Let $X$ be a subset of $mathbf R$, and let $f : X to mathbf R$ be a uniformly continuous function. Suppose that $E$ is a bounded subset of $X$. Then $f(E)$ is also bounded.
Theorem 9.9.16. Let $a < b$ be real numbers, and let $f : [a,b] to mathbf R$ be a function which is continuous on $[a,b]$. Then $f$ is also uniformly continuous.
I am confused as to why Proposition 9.9.15 and Theorem 9.9.16 don't imply Lemma 9.6.3. The reasoning goes like this: let $f : [a,b] to mathbf R$ be continuous on $[a,b]$. Then by Theorem 9.9.16, $f$ is uniformly continuous. Since $f$ is uniformly continuous and $[a,b] subset mathbf R$ is bounded, by Proposition 9.9.15 $f([a,b])$ is bounded. But this means $f$ is bounded, so we have Lemma 9.6.3.
What is wrong with the reasoning above? Alternatively, is the remark wrong?
real-analysis
asked Nov 29 at 3:26
riceissa
38447
add a comment |
In Analysis I, Tao writes:
Remark 9.9.17. One should compare Lemma 9.6.3, Proposition 9.9.15, and Theorem 9.9.16 with each other. No two of these results imply the third, but they are all consistent with each other.
For reference, here are the three results:
Lemma 9.6.3. Let $a < b$ be real numbers, and let $f : [a,b] to mathbf R$ be a function continuous on $[a,b]$. Then $f$ is a bounded function.
Proposition 9.9.15. Let $X$ be a subset of $mathbf R$, and let $f : X to mathbf R$ be a uniformly continuous function. Suppose that $E$ is a bounded subset of $X$. Then $f(E)$ is also bounded.
Theorem 9.9.16. Let $a < b$ be real numbers, and let $f : [a,b] to mathbf R$ be a function which is continuous on $[a,b]$. Then $f$ is also uniformly continuous.
I am confused as to why Proposition 9.9.15 and Theorem 9.9.16 don't imply Lemma 9.6.3. The reasoning goes like this: let $f : [a,b] to mathbf R$ be continuous on $[a,b]$. Then by Theorem 9.9.16, $f$ is uniformly continuous. Since $f$ is uniformly continuous and $[a,b] subset mathbf R$ is bounded, by Proposition 9.9.15 $f([a,b])$ is bounded. But this means $f$ is bounded, so we have Lemma 9.6.3.
What is wrong with the reasoning above? Alternatively, is the remark wrong?
real-analysis
asked Nov 29 at 3:26
riceissa
38447
1
Nothing wrong at all in your argument. The remark is wrong.
– Kavi Rama Murthy
Nov 29 at 5:55
add a comment |
In Analysis I, Tao writes:
Remark 9.9.17. One should compare Lemma 9.6.3, Proposition 9.9.15, and Theorem 9.9.16 with each other. No two of these results imply the third, but they are all consistent with each other.
For reference, here are the three results:
Lemma 9.6.3. Let $a < b$ be real numbers, and let $f : [a,b] to mathbf R$ be a function continuous on $[a,b]$. Then $f$ is a bounded function.
Proposition 9.9.15. Let $X$ be a subset of $mathbf R$, and let $f : X to mathbf R$ be a uniformly continuous function. Suppose that $E$ is a bounded subset of $X$. Then $f(E)$ is also bounded.
Theorem 9.9.16. Let $a < b$ be real numbers, and let $f : [a,b] to mathbf R$ be a function which is continuous on $[a,b]$. Then $f$ is also uniformly continuous.
I am confused as to why Proposition 9.9.15 and Theorem 9.9.16 don't imply Lemma 9.6.3. The reasoning goes like this: let $f : [a,b] to mathbf R$ be continuous on $[a,b]$. Then by Theorem 9.9.16, $f$ is uniformly continuous. Since $f$ is uniformly continuous and $[a,b] subset mathbf R$ is bounded, by Proposition 9.9.15 $f([a,b])$ is bounded. But this means $f$ is bounded, so we have Lemma 9.6.3.
What is wrong with the reasoning above? Alternatively, is the remark wrong?
real-analysis
asked Nov 29 at 3:26
riceissa
38447
In Analysis I, Tao writes:
Remark 9.9.17. One should compare Lemma 9.6.3, Proposition 9.9.15, and Theorem 9.9.16 with each other. No two of these results imply the third, but they are all consistent with each other.
For reference, here are the three results:
Lemma 9.6.3. Let $a < b$ be real numbers, and let $f : [a,b] to mathbf R$ be a function continuous on $[a,b]$. Then $f$ is a bounded function.
Proposition 9.9.15. Let $X$ be a subset of $mathbf R$, and let $f : X to mathbf R$ be a uniformly continuous function. Suppose that $E$ is a bounded subset of $X$. Then $f(E)$ is also bounded.
Theorem 9.9.16. Let $a < b$ be real numbers, and let $f : [a,b] to mathbf R$ be a function which is continuous on $[a,b]$. Then $f$ is also uniformly continuous.
I am confused as to why Proposition 9.9.15 and Theorem 9.9.16 don't imply Lemma 9.6.3. The reasoning goes like this: let $f : [a,b] to mathbf R$ be continuous on $[a,b]$. Then by Theorem 9.9.16, $f$ is uniformly continuous. Since $f$ is uniformly continuous and $[a,b] subset mathbf R$ is bounded, by Proposition 9.9.15 $f([a,b])$ is bounded. But this means $f$ is bounded, so we have Lemma 9.6.3.
What is wrong with the reasoning above? Alternatively, is the remark wrong?
real-analysis
real-analysis
asked Nov 29 at 3:26
riceissa
38447
asked Nov 29 at 3:26
riceissa
38447
asked Nov 29 at 3:26
riceissa
38447
asked Nov 29 at 3:26
riceissa
38447
asked Nov 29 at 3:26
riceissa
38447
38447
1
Nothing wrong at all in your argument. The remark is wrong.
– Kavi Rama Murthy
Nov 29 at 5:55
add a comment |
1
Nothing wrong at all in your argument. The remark is wrong.
– Kavi Rama Murthy
Nov 29 at 5:55
1
1
Nothing wrong at all in your argument. The remark is wrong.
– Kavi Rama Murthy
Nov 29 at 5:55
Nothing wrong at all in your argument. The remark is wrong.
– Kavi Rama Murthy
Nov 29 at 5:55
add a comment |
1 Answer
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The remark was wrong, and has now been fixed in the book's errata.
answered Dec 13 at 5:52
riceissa
38447
add a comment |
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The remark was wrong, and has now been fixed in the book's errata.
answered Dec 13 at 5:52
riceissa
38447
add a comment |
The remark was wrong, and has now been fixed in the book's errata.
answered Dec 13 at 5:52
riceissa
38447
add a comment |
The remark was wrong, and has now been fixed in the book's errata.
answered Dec 13 at 5:52
riceissa
38447
The remark was wrong, and has now been fixed in the book's errata.
answered Dec 13 at 5:52
riceissa
38447
answered Dec 13 at 5:52
riceissa
38447
answered Dec 13 at 5:52
riceissa
38447
answered Dec 13 at 5:52
riceissa
38447
38447
add a comment |
add a comment |
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Nothing wrong at all in your argument. The remark is wrong.
– Kavi Rama Murthy
Nov 29 at 5:55