Proof of Arzela's Theorem
I am doing problem 3 from section 45 in Munkres. The problem is Prove Arzela's Theorem, which states:
Let $X$ be compact: let $f_n in mathcal{C}(X,mathbb{R}^k)$. If the collection ${f_n}$ is pointwise bounded and equicontinuous, then the sequence $f_n$ has a uniformly convergent subsequence. Here is a sketch of my proof:
Let X is compact and ${f_n}subseteq mathcal{C}(X,mathbb{R}^k)$. Since ${f_n}$ is pointwise bounded and equicontinuous by Ascoli's Theorem $overline{{f_n}}$ is compact. Since $overline{{f_n}}$ is a compact subset of a complete metric space it's complete. Since $overline{{f_n}}$ is compact then the sequence ${f_n}$ has a convergent subsequence ${f_{n_i}} to f$. Since we are in the uniform metric, this subsequence converges uniformly.
real-analysis general-topology arzela-ascoli
asked Dec 9 at 0:50
Issacg628496
132
|
show 3 more comments
I am doing problem 3 from section 45 in Munkres. The problem is Prove Arzela's Theorem, which states:
Let $X$ be compact: let $f_n in mathcal{C}(X,mathbb{R}^k)$. If the collection ${f_n}$ is pointwise bounded and equicontinuous, then the sequence $f_n$ has a uniformly convergent subsequence. Here is a sketch of my proof:
Let X is compact and ${f_n}subseteq mathcal{C}(X,mathbb{R}^k)$. Since ${f_n}$ is pointwise bounded and equicontinuous by Ascoli's Theorem $overline{{f_n}}$ is compact. Since $overline{{f_n}}$ is a compact subset of a complete metric space it's complete. Since $overline{{f_n}}$ is compact then the sequence ${f_n}$ has a convergent subsequence ${f_{n_i}} to f$. Since we are in the uniform metric, this subsequence converges uniformly.
real-analysis general-topology arzela-ascoli
asked Dec 9 at 0:50
Issacg628496
132
3
Arzela's theorem is often called Arzela-Ascoli's theorem. So what you refered to Ascoli's theorem in your proof is actually Arzela's theorem in disguise. It is circular reasoning and cannot be a valid proof.
– Song
Dec 9 at 0:55
I figured. I find it odd the that Munkres proves Ascoli's then makes you prove Arzela's even though they are equivalent.
– Issacg628496
Dec 9 at 1:08
So, should I instead show that Arzela's Theorem implies Ascoli's?
– Issacg628496
Dec 9 at 1:17
@Song How is it necessarily circular? As long as you prove one version from first principles, proving the second version as a corollary of the first is totally legitimate, no? I feel that I am missing something here.
– Chill2Macht
Dec 9 at 1:18
@Chill2Macht You're right. I found what I was missing after I browsed the book. One version of the theorem is stated in terms of general $mathcal{F}subset C(X,mathbb{R}^k)$, and the problem requires the sequence version.
– Song
Dec 9 at 1:30
|
show 3 more comments
I am doing problem 3 from section 45 in Munkres. The problem is Prove Arzela's Theorem, which states:
Let $X$ be compact: let $f_n in mathcal{C}(X,mathbb{R}^k)$. If the collection ${f_n}$ is pointwise bounded and equicontinuous, then the sequence $f_n$ has a uniformly convergent subsequence. Here is a sketch of my proof:
Let X is compact and ${f_n}subseteq mathcal{C}(X,mathbb{R}^k)$. Since ${f_n}$ is pointwise bounded and equicontinuous by Ascoli's Theorem $overline{{f_n}}$ is compact. Since $overline{{f_n}}$ is a compact subset of a complete metric space it's complete. Since $overline{{f_n}}$ is compact then the sequence ${f_n}$ has a convergent subsequence ${f_{n_i}} to f$. Since we are in the uniform metric, this subsequence converges uniformly.
real-analysis general-topology arzela-ascoli
asked Dec 9 at 0:50
Issacg628496
132
I am doing problem 3 from section 45 in Munkres. The problem is Prove Arzela's Theorem, which states:
Let $X$ be compact: let $f_n in mathcal{C}(X,mathbb{R}^k)$. If the collection ${f_n}$ is pointwise bounded and equicontinuous, then the sequence $f_n$ has a uniformly convergent subsequence. Here is a sketch of my proof:
Let X is compact and ${f_n}subseteq mathcal{C}(X,mathbb{R}^k)$. Since ${f_n}$ is pointwise bounded and equicontinuous by Ascoli's Theorem $overline{{f_n}}$ is compact. Since $overline{{f_n}}$ is a compact subset of a complete metric space it's complete. Since $overline{{f_n}}$ is compact then the sequence ${f_n}$ has a convergent subsequence ${f_{n_i}} to f$. Since we are in the uniform metric, this subsequence converges uniformly.
real-analysis general-topology arzela-ascoli
real-analysis general-topology arzela-ascoli
asked Dec 9 at 0:50
Issacg628496
132
asked Dec 9 at 0:50
Issacg628496
132
asked Dec 9 at 0:50
Issacg628496
132
asked Dec 9 at 0:50
Issacg628496
132
asked Dec 9 at 0:50
Issacg628496
132
132
3
Arzela's theorem is often called Arzela-Ascoli's theorem. So what you refered to Ascoli's theorem in your proof is actually Arzela's theorem in disguise. It is circular reasoning and cannot be a valid proof.
– Song
Dec 9 at 0:55
I figured. I find it odd the that Munkres proves Ascoli's then makes you prove Arzela's even though they are equivalent.
– Issacg628496
Dec 9 at 1:08
So, should I instead show that Arzela's Theorem implies Ascoli's?
– Issacg628496
Dec 9 at 1:17
@Song How is it necessarily circular? As long as you prove one version from first principles, proving the second version as a corollary of the first is totally legitimate, no? I feel that I am missing something here.
– Chill2Macht
Dec 9 at 1:18
@Chill2Macht You're right. I found what I was missing after I browsed the book. One version of the theorem is stated in terms of general $mathcal{F}subset C(X,mathbb{R}^k)$, and the problem requires the sequence version.
– Song
Dec 9 at 1:30
|
show 3 more comments
3
Arzela's theorem is often called Arzela-Ascoli's theorem. So what you refered to Ascoli's theorem in your proof is actually Arzela's theorem in disguise. It is circular reasoning and cannot be a valid proof.
– Song
Dec 9 at 0:55
I figured. I find it odd the that Munkres proves Ascoli's then makes you prove Arzela's even though they are equivalent.
– Issacg628496
Dec 9 at 1:08
So, should I instead show that Arzela's Theorem implies Ascoli's?
– Issacg628496
Dec 9 at 1:17
@Song How is it necessarily circular? As long as you prove one version from first principles, proving the second version as a corollary of the first is totally legitimate, no? I feel that I am missing something here.
– Chill2Macht
Dec 9 at 1:18
@Chill2Macht You're right. I found what I was missing after I browsed the book. One version of the theorem is stated in terms of general $mathcal{F}subset C(X,mathbb{R}^k)$, and the problem requires the sequence version.
– Song
Dec 9 at 1:30
3
3
Arzela's theorem is often called Arzela-Ascoli's theorem. So what you refered to Ascoli's theorem in your proof is actually Arzela's theorem in disguise. It is circular reasoning and cannot be a valid proof.
– Song
Dec 9 at 0:55
Arzela's theorem is often called Arzela-Ascoli's theorem. So what you refered to Ascoli's theorem in your proof is actually Arzela's theorem in disguise. It is circular reasoning and cannot be a valid proof.
– Song
Dec 9 at 0:55
I figured. I find it odd the that Munkres proves Ascoli's then makes you prove Arzela's even though they are equivalent.
– Issacg628496
Dec 9 at 1:08
I figured. I find it odd the that Munkres proves Ascoli's then makes you prove Arzela's even though they are equivalent.
– Issacg628496
Dec 9 at 1:08
So, should I instead show that Arzela's Theorem implies Ascoli's?
– Issacg628496
Dec 9 at 1:17
So, should I instead show that Arzela's Theorem implies Ascoli's?
– Issacg628496
Dec 9 at 1:17
@Song How is it necessarily circular? As long as you prove one version from first principles, proving the second version as a corollary of the first is totally legitimate, no? I feel that I am missing something here.
– Chill2Macht
Dec 9 at 1:18
@Song How is it necessarily circular? As long as you prove one version from first principles, proving the second version as a corollary of the first is totally legitimate, no? I feel that I am missing something here.
– Chill2Macht
Dec 9 at 1:18
@Chill2Macht You're right. I found what I was missing after I browsed the book. One version of the theorem is stated in terms of general $mathcal{F}subset C(X,mathbb{R}^k)$, and the problem requires the sequence version.
– Song
Dec 9 at 1:30
@Chill2Macht You're right. I found what I was missing after I browsed the book. One version of the theorem is stated in terms of general $mathcal{F}subset C(X,mathbb{R}^k)$, and the problem requires the sequence version.
– Song
Dec 9 at 1:30
|
show 3 more comments
1 Answer
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The argument should say: as we have a compact subset of a metric space, the subset is sequentially compact and so we have a convergent subsequence for the sequence $(f_n)_n$ (completeness is irrelevant as we don't have a Cauchy sequence). The rest seems correct.
answered Dec 9 at 8:08
Henno Brandsma
105k346113
add a comment |
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The argument should say: as we have a compact subset of a metric space, the subset is sequentially compact and so we have a convergent subsequence for the sequence $(f_n)_n$ (completeness is irrelevant as we don't have a Cauchy sequence). The rest seems correct.
answered Dec 9 at 8:08
Henno Brandsma
105k346113
add a comment |
The argument should say: as we have a compact subset of a metric space, the subset is sequentially compact and so we have a convergent subsequence for the sequence $(f_n)_n$ (completeness is irrelevant as we don't have a Cauchy sequence). The rest seems correct.
answered Dec 9 at 8:08
Henno Brandsma
105k346113
add a comment |
The argument should say: as we have a compact subset of a metric space, the subset is sequentially compact and so we have a convergent subsequence for the sequence $(f_n)_n$ (completeness is irrelevant as we don't have a Cauchy sequence). The rest seems correct.
answered Dec 9 at 8:08
Henno Brandsma
105k346113
The argument should say: as we have a compact subset of a metric space, the subset is sequentially compact and so we have a convergent subsequence for the sequence $(f_n)_n$ (completeness is irrelevant as we don't have a Cauchy sequence). The rest seems correct.
answered Dec 9 at 8:08
Henno Brandsma
105k346113
edited Dec 9 at 17:22
answered Dec 9 at 8:08
Henno Brandsma
105k346113
answered Dec 9 at 8:08
Henno Brandsma
105k346113
answered Dec 9 at 8:08
Henno Brandsma
105k346113
105k346113
add a comment |
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3
Arzela's theorem is often called Arzela-Ascoli's theorem. So what you refered to Ascoli's theorem in your proof is actually Arzela's theorem in disguise. It is circular reasoning and cannot be a valid proof.
– Song
Dec 9 at 0:55
I figured. I find it odd the that Munkres proves Ascoli's then makes you prove Arzela's even though they are equivalent.
– Issacg628496
Dec 9 at 1:08
So, should I instead show that Arzela's Theorem implies Ascoli's?
– Issacg628496
Dec 9 at 1:17
@Song How is it necessarily circular? As long as you prove one version from first principles, proving the second version as a corollary of the first is totally legitimate, no? I feel that I am missing something here.
– Chill2Macht
Dec 9 at 1:18
@Chill2Macht You're right. I found what I was missing after I browsed the book. One version of the theorem is stated in terms of general $mathcal{F}subset C(X,mathbb{R}^k)$, and the problem requires the sequence version.
– Song
Dec 9 at 1:30