# Definition of Compactness

I am revising for my exam in Metric Spaces. Within this course we cover closed and complete sets also. However I'm really struggled to wrap my head around the definition of compactness. Definition I have been given is: $A$ a subset of metric space $X$, $A$ is compact if every sequence in $A$ has a subsequence that converges to a point of $A$. My main lack of understanding is how would I prove a set is compact? Also I know that compact sets are complete and closed but this isn't an obvious result to me.

• That is not obvious to most people, really. But surely a proof of that was part of your course! Similarly, it would have been a rather heterodox course on metric spaces if you had not had several examples of proving that some sets are compact! Jan 27 '18 at 22:03
• Has your professor given you any examples of compact spaces, and proofs that they are compact? Jan 27 '18 at 22:03
• I have got many consequential properties of being compact e.g Heine Borel but I'm finding these proofs hard to follow do to lack of initial examples of compact sets.
– user509405
Jan 27 '18 at 22:07
• There is the complete & totally bounded equivalence in metric spaces which is useful. I have often found the open set definition to be directly applicable. Jan 27 '18 at 22:10

## 4 Answers

Well, if you have an arbitrary sequence, you should find some subsequence that converges within that set using some property of that set.

Example: $X$ some metric space and $A$ finite, then you know that any sequence would some value in $A$ take infinitely many times: choose these $n$ as $n_k$ of your subsequence and you get a constant, hence convergent subsequence.

Take any Cauchy sequence in $A$, then it has a convergent subsequence in $A$. As we have a convergent subsequence of a Cauchy sequence, this implies that the Cauchy sequence converges to the same limit as its subsequence: hence $A$ is complete. Take any convergent sequence in $A$ (with limit in $X$), then it is a subsequence in $A$ with limit in $A$, so this sequence converges within $A$: hence $A$ is closed.

In a general metric space, we can make the following statement: if $K$ is compact then $K$ is closed and bounded. The closedness arises because otherwise you can pick a sequence that converges to a point not in $K$, which cannot have a subsequence converging to a point in $K$. The boundedness arises because in an unbounded set you can take a sequence that is blowing up monotonically, which cannot have a convergent subsequence.

In $\mathbb{R}^n$ we have the Bolzano-Weierstrass theorem that says that these conditions are also sufficient. Thus the "classic" compact sets are finite sets and closed intervals.

In a general metric space, there are sets which are complete, bounded, and not compact. The sufficient condition in a general metric space replaces "bounded" with "totally bounded", which means that you can cover $K$ with finitely many arbitrarily small balls (the number depending on the size but remaining finite as long as the balls have finite size).

To understand the definition of compactness, first you need to understand the definition of covering.

A covering of A in X is a set of open sets in X such that A is contained in the union of this open sets.

A set A is said to be compact if, for each covering of A, there exists a finite subcovering of A.

With this definition, if you know the topology of the space, you can prove that a set is compact.

Another definition of compactness is that every open cover has a finite subcover...

But, using your definition, take an arbitrary sequence and show it has a convergent subsequence, with the limit point in $A$...

We also have Heine-Borel, namely, in $\mathbb R^n$, $A$ compact $\iff A$ closed and bounded...