# Compactness is a topological property, proof without open covers

An exercise in the book wants me to show that compactness is a topological property. This chapter is about topology of $R^n$ and we are working with an assumed metric. We haven't come to definition of topological spaces yet. We defined; open sets, closed sets, limit points, sequences, continuous functions, bounded sets, homeomorphisms and topological properties etc.

Here is the definition of compactness that the book uses.

Defn: A set $S$ is compact if every infinite sequence contained in $S$ has a limit point contained in $S$.

Now I am trying to prove the following.

Proposition: Compactness is a topological property.

I am trying to show that if $A \subset R^n$ is compact and there exists a homeomorphism $f: A \to B$, then $B$ is compact.

Strategy: Assume $A$ is compact, and $f$ is continuous with a continuous inverse. Go with proof by contradiction. Suppose $B$ is not compact. Then by Heine-Borel, $B$ is not (closed and bounded). So $B$ is either not closed or not bounded.

I showed that if $B$ is not closed, then $A$ is not compact and found a contradiction. I am trying to also show that if $B$ is not bounded, a contradiction arises. Then I think, I can conclude $B$ must be compact because all roads lead to a contradiction.

I need to show if $B$ is not bounded, a contradiction arises. Possible candidate is to show that $A$ is not compact, which contradicts the hypothesis.

• $B$ is already compact from being equal to $f[A]$. A continuous image of compact is compact. – Henno Brandsma Aug 21 '17 at 22:03
• This is precisely what the OP wants to prove. – Francesco Polizzi Aug 21 '17 at 22:04
• @HennoBrandsma That's what OP is trying to prove without using open covers. – Sahiba Arora Aug 21 '17 at 22:04
• You can also use that compact is the same as every infinite subset has a strong limit point, or every sequence has a convergent subsequence. (for metric spaces). – Henno Brandsma Aug 21 '17 at 22:05
• @meguli What is your definition of compactness? Are you required to prove this using Heine-Borel or is that just an idea? – Sahiba Arora Aug 21 '17 at 22:07

If $A$ is closed then $B$ is closed because a homeomorphism is a closed map. Now we prove that $B$ is also bounded, by exploiting the fact that a subset of $\mathbb{R}^n$ is compact if and only if is sequentially compact.

So, assume by contradiction that $B$ is not bounded. Then we can find a sequence $\{b_n \} \subset B$ such that $||b_n|| > n$ for all $n \in \mathbb{N}$. This sequence clearly does not admit any convergent subsequence.

On the other hand, we have $b_n = f(a_n)$ with $a_n \in A$, and by compactness of $A$ we can find a subsequence $\{a_{n_k}\}$ of $\{a_n\}$ converging to $a \in A$. So, by continuity of $f$, the subsequence $\{b_{n_k}\} = \{f(a_{n_k})\}$ should converge to $f(a)$, contradiction.

Then $B$ is also bounded, hence compact by Heine-Borel.

You have to use open covers (for all spaces not just $\mathbb{R}^n$): a homeomorphism need not preserve boundedness at all.

• But isn't it true that it will preserve total boundedness? – AJY Aug 21 '17 at 21:58
• @AJY no, only uniformly continuous ones do. – Henno Brandsma Aug 21 '17 at 21:59
• Continuous function on a compact domain is uniformly continuous. – Sahiba Arora Aug 21 '17 at 22:00
• Then wouldn't that entail that homeomorphisms don't preserve compactness? – AJY Aug 21 '17 at 22:00
• But for complete spaces, total boundedness is just "closure is compact" – Henno Brandsma Aug 21 '17 at 22:00

Let $$(y_n)$$ be an infinite sequence in $$B.$$ As $$f$$ is surjective, $$y_n=f(x_n)$$ where $$(x_n)$$ is s sequence in $$A.$$ Injectivity implies that $$(x_n)$$ is also an infinite sequence. Since, $$A$$ is compact, therefore $$(x_n)$$ has a convergent subsequence (denoted again by $$(x_n)$$, for convenience) in $$A,$$ to say $$x$$. Thus, we have $$x_n \to x\text{ as } n\to \infty$$ Continuity of $$f$$ implies $$y_n=f(x_n) \to f(x)\text{ as } n\to \infty$$ Thus, $$f(x)\in B$$ is a limit point of $$(y_n).$$

• Why would it be so that $(x_n)$ converges. Hint: it is not so. – Jens Renders May 26 at 17:59
• @JensRenders Edited. – Sahiba Arora May 26 at 18:18

The definition of compactness you have is what is in general topological spaces known as sequential compactness. It turns out that it is equivalent to compactness defined by open covers in metric spaces.

Now, the proof should use definition directly. If you have $f\colon A\to B$ a homeomorphism, and assume that $A$ is compact, then take any sequence $(b_n)$ in $B$. Sequence $(f^{-1}(b_n))$ is sequence in $A$, so it has convergent subsequence $(f^{-1}(b_{p(n)}))$ with limit $a$. Since $f$ is continuous, we have that $$f(a) = \lim_n f(f^{-1}(b_{p(n)})) = \lim_n b_{p(n)},$$ so $(b_n)$ has convergent subsequence.

• Sequences in a compact set need to have a convergent subsequent because the sequence need to have a limit point in the set and a subsequent must converge to this limit point, this is how I got it. – meguli Aug 21 '17 at 22:28
• @meguli, well yes, limit point of a sequence is the same thing as limit of convergent subsequence. Is that what you wanted to check? – Ennar Aug 21 '17 at 22:35
• I didn't know they are the same, book defines limit points without reference to convergent subsequences. But I needed a little thought to infer that they are the same after your answer. This different definitions for samely named concepts at different level of abstractions are what's tripping me. – meguli Aug 21 '17 at 22:39
• @meguli, do you know how to rigorously show it now? – Ennar Aug 21 '17 at 22:42
• Yes, by construcrion. – meguli Aug 21 '17 at 23:44