Erwin Kreyszig's Introductory Functional Analysis With Applications, Section 2.5, Problem 10

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?

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.

• 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? Dec 7 '14 at 8:38

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).
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 $x$.