3
$\begingroup$

Here's Problem 10 in Section 2.5 in Introductory Functional Analysis With Applications by Erwin Kreyszig:

Let $X$ and $Y$ be metric spaces, let $X$ be (sequentially) compact, and let the mapping $T \colon X \to Y$ be bijective and continuous. Show that the mapping $T^{-1} \colon Y \to X$ is also continuous.

My effort:

Let $y$ be a point in $Y$, and let $(y_n)$ be a sequence in $Y$ such that $(y_n)$ converges to $y$. We need to show that the sequence $(T^{-1}y_n)$ converges to $T^{-1}y$ in the metric space $X$.

Now, for each $n \in \mathbb{N}$, there exists a unique point $x_n \in X$ such that $Tx_n = y_n$ so that $x_n = T^{-1}y_n$. Thus, $(x_n)$ is a sequence in the compact metric space $X$; so it has a convergent subsequence $(x_{n_k})$, say; let $x$ be the limit of this subsequence.

Then, by the continuity of $T$ we can conclude that the sequence $(Tx_{n_k})$ converges in $Y$ to the point $Tx$.

But $Tx_{n_k} = y_{n_k}$. Thus, $(Tx_{n_k})$ is a subsequence of $(y_n)$ and hence it must also converge to $y$. Hence $Tx = y$; so $x = T^{-1}y$.

Since $X$ is compact and since $T$ is continuous, the image $T(X)$ is also compact. But as $T$ is bijective, this image is all of $Y$. Thus, $Y$ is also a compact metric space.

What next?

How to show from here that the sequence $(T^{-1}y_n)$, which is the same as the sequence $(x_n)$, converges to $T^{-1}y = x$?

Or, is there a better alternative proof of our assertion?

$\endgroup$
2
$\begingroup$

In a metric space, sequential compactness is equivalent to compactness, and all metric spaces are Hausdorff. Then we have the result that continuous bijection from a compact space to a Hausdorff space is a Homeomorphism i.e. it has a continuous inverse.

$\endgroup$
1
  • $\begingroup$ Arun, can you please give a proof of the assertion that a continuous bijection of a compact space to a Hausdorff space is a homeomorphism? But, more importantly, can you please give a proof along the lines I've mapped out above? $\endgroup$ Dec 7 '14 at 8:38
2
$\begingroup$

Your proof looks fine.

If you know a few more facts about compact sets, here is a shorter proof:

To show $T^{-1} : Y \to X$ is continuous, it suffices to show that for each closed $F \subset X$ we have that $(T^{-1})^{-1}(F)$ is closed. Since $T$ is bijective, $(T^{-1})^{-1}(F) = T(F)$. Now since $F$ is closed and $X$ is compact, $F$ is also compact (closed subsets of compact sets are compact), hence $T(F)$ is also compact (continuous image of a compact set is compact). Therefore $T(F)$ is closed (compact subsets of metric spaces are closed).

$\endgroup$
1
  • $\begingroup$ Your proof is beautiful, but I wonder if there's any way of completing the argument I've given above? $\endgroup$ Dec 7 '14 at 8:34
0
$\begingroup$

You can complete your argument as follows. If the sequence $\sigma=\langle x_n:n\in\Bbb N\rangle$ does not converge to $x$, there are an $\epsilon>0$ and a subsequence $\sigma'=\langle x_{m_k}:k\in\Bbb N\rangle$ of $\sigma$ such that $d(x_{m_k},x)\ge\epsilon$ for each $k\in\Bbb N$, where $d$ is the metric in $X$. And $X$ is sequentially compact, so $\sigma'$ has a convergent subsequence $\sigma''$, say with limit $x'$; clearly $d(x',x)\ge\epsilon$. But $Tx'=y=Tx$ by the continuity of $T$, and $T$ is a bijection, so $x'=x$, contradicting the choice of $\sigma''$. Thus, $\sigma$ converges to $x$, and $T^{-1}$ is continuous.

Alternatively, you could avoid the argument by contradiction by observing that the same idea shows that every subsequence of $\sigma$ has a subsequence converging to $x$, and it’s a standard result that this implies that $\sigma$ itself converges to $x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.