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I would like to know as many relations as possible to get a better picture.

I know that if $f$ is continuous and $(X,d)$ is complete, then $f(X)$ is complete $\iff$ closed.

Question:However, are complete sets closed? Since complete sets are when Cauchy sequences converge in those sets, it would make sense for them to be closed (since they contain their limit points)?

Question: I know that compact sets are closed. Take for example $\{0\cup\frac{1}{n}\}$. If I take $x_n=1/n$ I see that $x_n\rightarrow 0$ which means that it's closed. But is it compact? What cover can I take? Will $B(0,r=1)$ will do? If yes, then it's bounded and I could ?possibly? use Heine Borel to show it's compact?

Example: Is $C[0,1]$ compact? Could you provide me with an example of interesting non-compact spaces? It seems I am only working with intervals

GRS
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  • What definition of compactness do you use? And also, what do you mean by ${0 \cup 1/n}$? Do you mean ${0}\cup{1/n\mid n\in\mathbb{N}_{0}}$? I do not know what is your level, but if you look for interesting non-compact subspaces, be aware that there is this famous theorem: "A normed vector space is finite dimensional if and only if its unit ball is compact". If you look for trivial example, $\mathbb{R}$ equipped with the standard topology is non-compact. – MoebiusCorzer May 10 '16 at 19:17
  • @MoebiusCorzer I use that every open cover $V_\alpha$ has a finite subcover $\cup^N V_\alpha$ , and yes – GRS May 10 '16 at 19:18

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Question 1: Any convergent sequence of some metric space is a Cauchy sequence in this metric space. If any Cauchy sequence in this space has a limit in this space (i.e. if the metric space is complete), it means in particular that any convergent sequence has its limit in this space, hence the metric space is closed.

Question 2: I assume you consider $\{0\}\cup\{1/n\mid n\in\mathbb{N}_{0}\}$ as a subset of $\mathbb{R}$ equipped with the euclidean distance (and seen as a topological space for the topology induced by this distance) with the induced topology. In this case, as your subset $\{0\}\cup\{1/n\mid n\in\mathbb{N}_{0}\}$ is bounded and closed, by Heine-Borel you know it is indeed compact.

Question 3: I assume you look for bounded closed subsets of a topological space that are not compact in this topological space (because if we can take unbounded closed subsets, taking $\mathbb{R}$ with the classical topology induced by the euclidean distance, it is non-compact). I do not know if you have knowledge of measure theory and functional analysis, but in these fields arise infinite dimensional normed vector spaces whose unit ball is not compact. As I said in the comments, a normed vector space is finite dimensional if and only if its closed unit ball is compact (the proof is not complex).

For your particular example, with which topology do you equip $C[0,1]$?

For an easier example, take the space $\ell^{2}$ of all sequences of reals $(x_{n})_{n}$ such that

$$\Vert (x_{n})_{n}\Vert_{\ell^{2}}=\sqrt{\sum_{i=1}^{\infty}\vert x_{i}\vert^{2}}<\infty$$

Such a sequence is said to be "square-summable" and the function:

$$\ell^{2}\to\mathbb{R}^{+}:(x_{n})_{n}\mapsto\Vert (x_{n})_{n}\Vert_{\ell^{2}}=\sqrt{\sum_{i=1}^{\infty}\vert x_{i}\vert^{2}}$$

defines a norm on $\ell^{2}$, which induces a natural structure of metric space (and thus of topological space) with distance $d(x,y)=\Vert x-y\Vert_{\ell^{2}}$ where $x,y$ are two sequences of $\ell^{2}$.

One can show that $\ell^{2}$ is complete. However, the closed unit ball

$$\overline{B(0,1)}=\left\{(x_{n})_{n}\in\ell^{2}\mid \Vert (x_{n})_{n}\Vert_{\ell^{2}}\le 1\right\}$$

is not compact (and, in particular, it not anymore holds that being bounded and closed is equivalent to be compact!).

MoebiusCorzer
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  • So complete metric spaces are indeed closed. I'm surprised my book doesn't mention such facts. For q2, doest the ball $B(0,1)$ work to be the finite cover? I usually use $\sup |f,g|$ for the $C[0,1]$ (while solving banach fixed points on complete spaces). Thanks a lot for your help – GRS May 10 '16 at 19:47
  • @GRS I am not familiar with this topology, but this question has very interesting (and easy to understand) answers regarding its compactness. Concerning the finite subcover: it depends on the cover you took. Here, your proof does not use the definition of compactness: it uses the fact that any bounded closed subset of $\mathbb{R}$ is compact, which is, as you said, Heine-Borel. And I should add that, in general, it is not practical to prove compactness with the definition. – MoebiusCorzer May 10 '16 at 19:50
  • In general, the definition is used to prove that something is not compact, by constructing a cover which admits no finite subcover. Also, in a metric space, it is often easier to work with this characterization of compactness: a subset $S$ of a metric space $(X,d)$ is compact if and only if any bounded sequence admits a convergent subsequence. This characterization of compactness in metric spaces is often referred to as the Bolzano-Weierstrass theorem when the metric space is $\mathbb{R}^{n}$. – MoebiusCorzer May 10 '16 at 19:56
  • Does it mean that all subsets of $\mathbb R$ are compact? that is including the open ones? I never knew that I could only use Bolzano-Weierstrass on compact spaces. – GRS May 10 '16 at 20:01
  • @GRS Maybe am I not clear or made a mistake. I shall be clearer: a subset $S$ of a metric space $(X,d)$ is compact if any bounded subsequence in $S$ admits a converging subsequence in $S$ (having its limit in $S$). For example, $]0,1]$ is not compact. Take $x_{n}=1/n$ as a sequence. It is bounded and convergent but its limit is not in $]0,1]$, so that it is not compact. Actually, this characterization shows that in order to be compact in $\mathbb{R}^{n}$, you must be closed. – MoebiusCorzer May 10 '16 at 20:09
  • Let us continue this discussion in chat. – MoebiusCorzer May 10 '16 at 20:12