# Let $X$ and $Y$ be metric spaces, $X$ compact, and $T: X \to Y$ bijective and continuous. Show that $T$ is a homeomorphism. [duplicate]

I never used the fact the $$X$$ is compact in my proof below. Which makes me worry if my proof is complete.

Let $$X$$ and $$Y$$ be metric spaces, $$X$$ compact, and $$T: X \to Y$$ bijective and continuous. To show that $$T^{-1}$$ is continuous, we will proceed by contradiction. That is, suppose that $$T^{-1}$$ is discontinuous. Since $$T$$ is bijective, then every element $$y \in Y$$ is uniquely determined by an element $$x \in X$$, \emph{viz.}, $$Tx = y$$ or equivalently $$x = T^{-1}y$$. By continuity of $$T$$, if $$(x_n)$$ is a convergent sequence in $$X$$ with limit point $$x_0 \in X$$, then $$Tx_n = y_n \to Tx_0 = y_0$$. However, this would imply that if $$y_n \to y_0$$, then $$T^{-1}y_n = x_n \to T^{-1}y_0$$. A contradiction, since $$T^{-1}$$ was assumed to be discontinuous.

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• Do you really want to put the tag of functional analysis?? Is there some linear map between linear spaces somewhere? – Praphulla Koushik Sep 25 at 18:47
• I put it since it came from my Functional Analysis book by Kreyszig. I can add a linear algebra tag. – Taylor McMillan Sep 25 at 18:52
• I don’t think that’s how it works.,, linear algebra is irrelevant as there are no linear map here.. – Praphulla Koushik Sep 25 at 18:53
• That makes sense. – Taylor McMillan Sep 25 at 18:55

You have already assumed $$(T^{-1}(y_n))=(x_n)\rightarrow x_0=T^{-1}(y_0)$$. That’s why it looks that $$T^{-1}$$ is continuous.

What you have to do is, take an arbitrary sequence $$y_n\rightarrow y$$ and prove that $$T^{-1}(y_n)\rightarrow T^{-1}(y)$$.

How to use compactness of $$X$$? Remember that, as $$X$$ is compact, every sequence has convergent subsequence. From $$y_n\rightarrow y$$ you can get a sequence $$(x_n)$$ in $$X$$ with $$f(x_n)=y_n$$ (because of surjective property). This has a convergent subsequence, can you finish from here?

Another hint : In case you can prove $$(x_n)$$ is Cauchy, then you are done. A Cauchy sequence is convergent if it has a convergent subsequence. By above observation, it is clear that it has convergent subsequence. Showing that Cauchy, and using the injective property, you will see that $$(x_n)\rightarrow x$$. Thus $$T^{-1}$$ is continuous.

As $$T$$ is continuous, $$x_{n_k}\rightarrow x_0$$ which imply $$T(x_{n_k})\rightarrow T(x_0)$$; as $$T(x_{n_k})=y_{n_k}\rightarrow y$$, uniqueness of limits says that $$T(x_0)=y_0$$. This says there exists a subsequence of $$(x_n)$$ that converge to inverse image of $$y_0$$. How do you see whole sequence converge?

• I have attempted this. Let $y_n \to y_0 \in Y$. We want to show that $T^{-1}(y_n) \to T^{-1}(y_0)$. By the subjectiveness of $T$ we can pick a sequence $x_n \in X$ with $Tx_n = y_n$. By compactness of $X$, there exists a convergent subsequence $x_{n_k} \to x_0$. By continuity of $T$, we then have that $Tx_{n_k} = y_{n_k} \to Tx_0 = y_0$. I am just not sure if I can then just take the inverse of everything and then say that I am done. – Taylor McMillan Sep 25 at 18:40
• Where did you fail in your attempt? How do you know $T(x_0)=y_0$?. – Praphulla Koushik Sep 25 at 18:40
• Awww. Continuity only tells me that $Tx_{n_k} \to Tx_0$. – Taylor McMillan Sep 25 at 18:50
• Yes.. do you see if it is Cauchy?? – Praphulla Koushik Sep 25 at 18:50
• Trying to work through that now. – Taylor McMillan Sep 25 at 18:51

Your proof is not clear. Yes, if $$x_n\to x_0$$ then $$T(x_n)\to T(x_0)=y_0$$. But it is not necessary an "if and only if" relation. Anyway, you must use the fact that $$X$$ is compact because otherwise the statement is false. (take the identity map from the discrete metric space in $$\mathbb{R}$$ to the standard metric space).

Here is a proof. It is enough to show that if $$F\subseteq X$$ is closed then $$(T^{-1})^{-1}(F)$$ is a closed set in $$Y$$. Since $$X$$ is compact and $$F$$ is a closed subset we know that $$F$$ is compact as well. Continuous functions preserve compact sets, hence $$T(F)$$ is compact in $$Y$$. But $$Y$$ is a Hausdorff space (since it is a metric space), so it follows that $$T(F)$$ is closed in $$Y$$. This means $$(T^{-1})^{-1}(F)$$ is closed.

Your proof is surely incorrect, as you need compactness. Consider the identity function from $$\Bbb R$$ with the discrete topology to $$\Bbb R$$ with the standard topology.

You just need to prove that $$f$$ is open. Let $$U\subset X$$ be open. Then $$U^c$$ is compact. Then $$f(U^c)\subset Y$$ is compact. Since $$Y$$ is a metric space, it's Hausdorff and $$f(U^c)$$ is closed. Then $$f(U)=f(U^c)^c$$ is open.

$$T$$ is closed because $$C \subseteq X$$ closed implies $$C$$ compact, so $$T[C]$$ is compact by continuity of $$T$$ so $$T[C]$$ is closed (as $$Y$$ is metric).

And a closed continuous bijection is a homeomorphism.

Or if you want to go the sequence route: suppose $$y_n \to y$$ in $$Y$$.

Let $$x_n,x$$ be the unique points in $$X$$ such that $$T(x_n)=y_n$$ and $$T(x)=y$$.

As $$X$$ is compact there is $$x_0 \in X$$ and a subsequence $$x_{n_k}$$ such that $$x_{n_k} \to x_0$$, and then continuity of $$T$$ implies $$y_{n_k}=T(x_{n_k}) \to T(x_0)$$ and as $$y_{n_k} \to y$$ (as subsequences have the same limit as the total sequence) and as limits are unique we know that $$T(x_0)=y=T(x)$$ so $$x=x_0$$. So $$T^{-1}(y_n) \to T^{-1}(y)$$ and $$T^{-1}$$ is continuous.