# A closed, bounded set of a complete metric space can fail to be compact

I have a problem in which I have a complete metric space $$(X;d)$$, and a closed, bounded subset $$Y \subset X$$.

What are the possible proof strategies to show that Y is not compact? Why is $$E$$ not compact at Example here

Can you define what "compact" means? I just roughly know it means closed and bounded but clearly this is not precise.

I am also confused if there is a difference when we talk about a compact metric space vs a compact subset of a metric space?

The usual definition of compactness involves open (sub)covers. Suppose we have a topological space $$X$$ (or metric space, if you want), with a subset $$Y \subseteq X$$. We call a family $$\{A_\lambda\}_{\lambda \in \Lambda} \subseteq X$$ a cover of $$Y$$ if $$Y \subseteq \bigcup_{\lambda \in \Lambda} A_\lambda.$$ Note, this is not necessarily a sequence here, but a family of sets indexed by some arbitrary (possibly uncountable) set $$\Lambda$$. When $$A_\lambda$$ is open in the topology for all $$\lambda \in \Lambda$$, we say $$\{A_\lambda\}_{\lambda \in \Lambda}$$ is an open cover of $$Y$$.

Covers can often have a lot of unnecessary sets. If $$\Lambda' \subseteq \Lambda$$ such that $$\{A_{\lambda}\}_{\lambda \in \Lambda'}$$ is still a cover of $$Y$$ (i.e. $$Y$$ is still contained in the union of the remaining sets), then we call $$\{A_{\lambda}\}_{\lambda \in \Lambda'}$$ a subcover. Further, if $$\Lambda'$$ is finite, then we call it a finite subcover.

We call a subset $$Y$$ compact if every open cover admits a finite subcover.

As a non-example, consider the sets $$A_n = (-n, n)$$ where $$n \in \Bbb{N}$$, which cover $$\Bbb{R}$$ and are open in the Euclidean topology. Note that taking finitely many of them will result in a bounded union (as each $$A_n$$ is bounded), which cannot possibly cover $$\Bbb{R}$$, so no finite subcover exists. This provides a counterexample for the compactness of $$\Bbb{R}$$.

On the other hand, if we consider a convergent sequence $$x_n \to x$$, then the set $$Y = \{x_n : n \in \Bbb{N}\} \cup \{x\}$$ is compact. To see this, suppose $$\{A_\lambda\}_{\lambda \in \Lambda}$$ is an open cover. Since $$x \in Y$$, there must exist some $$\lambda_0 \in \Lambda$$ such that $$x \in A_{\lambda_0}$$. Since $$A_{\lambda_0}$$ is open and contains the limit $$x$$, then all but finitely many points of $$(x_n)$$ must occur in $$A_{\lambda_0}$$. Each of these finitely many points must occur in one of the sets in the open cover. Collecting a set for each point (removing any repeats), as well as the set $$A_{\lambda_0}$$, forms a finite cover for $$Y$$, proving $$Y$$ is compact.

There is also another, often easier, notion of compactness, called sequential compactness. A subset $$Y$$ of a topological space $$X$$ is sequentially compact if every sequence $$(y_n)$$ of points in $$Y$$ admits a convergent subsequence $$(y_{n_k})$$ which converges to some $$y \in Y$$.

This definition is neither stronger nor weaker than compactness in topological spaces; one notion of compactness does not imply the other. However, in the case of metric spaces, it turns out that both of these notions are equivalent! Often, in practice, sequential compactness is easier to work with.

As for your final question, a topological space (or metric space) $$X$$ is compact if the subset $$X \subseteq X$$ is a compact subset in the topological space $$X$$ (in the usual way). Equivalently, a topological space $$X$$ is compact if every family of open subsets $$\{A_\lambda\}_{\lambda \in \Lambda}$$ such that $$X = \bigcup_{\lambda \in \Lambda} A_\lambda,$$ there exists a finite subfamily of $$\{A_\lambda\}_{\lambda \in \Lambda}$$ whose union also equals $$X$$. We can obtain $$=$$ instead of $$\subseteq$$ because $$X$$ is the whole space, and each $$A_\lambda \subseteq X$$.