A homeomorphism maps neighborhoods to neighborhoods Good night, I'm doing Problem III.3.4 from textbook Analysis I by Amann.



Could you please verify if my proof looks fine or contains logical gaps/errors? Any suggestion is greatly appreciated.
My attempt:
(a)
It suffices to show that $f,f^{-1}$ map open balls to open balls. Let $g= f^{-1}$. For $x_0 \in X$ and $r>0$, let $y_0 = f(x_0)$.
Because $g:Y \to X$ is continuous and $\mathbb B (x_0, r)$ is open in $X$, $g^{-1} [ \mathbb B (x_0, r)] = f [ \mathbb B (x_0, r)]$ is open in $Y$. Because $f: X \to Y$ is continuous and $\mathbb B (y_0, r)$ is open in $Y$, $f^{-1} [ \mathbb B (y_0, r)]$ is open in $X$.
(b)
(i)
Let $A \subseteq X$ be closed in $X$ and $(y_n)$ a sequence in $f[A]$ such that $y_n \to y \in Y$. By Axiom of Countable Choice, there is a sequence $(x_n)$ in $A$ such that $f(x_n) = y_n$. Because $X$ is compact, there is a subsequence $(x_{n_m})$ of $(x_n)$ such that $x_{n_m} \to x \in X$. Because $A$ is closed in $X$, $x\in A$. Because $f$ is continuous, $y_{n_m} = f(x_{n_m}) \to f(x) \in f[A]$. Then the subsequence $(y_{n_m})$ of $(y_n)$ converges to $f(x) \in f[A]$. As such, $(y_n)$ converges to $f(x) \in f[A]$ and thus $f[A]$ is closed.
(ii)
Let $f^{-1} = g:Y \to X$. It remains to show that $g$ is continuous. Let $A \subseteq X$ be closed in $X$. We have $g^{-1}[A] = f[A]$ is closed in $Y$ by (i). As such, $g$ is continuous.
 A: A homeomorphism need not map open balls to open balls in a metric space.
But it does map open sets to open sets, as $f[O]=g^{-1}[O]$ where $g: Y \to X$ is the continuous inverse of $f$.
So if $N \in \mathcal{U}(x)$ it means that some metric ball $B_X(x,r) \subseteq N$ for some $r>0$ (I hope this agrees with your definition). $B_X(x,r)$ is open in $X$ (definition of the metric topology, or a lemma that's often proved right at the start of a topology course) so $f(x) \in f[B_X(x,r)] \subseteq f[N]$ and as that image of an open ball is open we have some ball in $Y$ with $f(x) \in B_Y(f(x),s) \subseteq f[B_X(x,r)] \subseteq f[N]$ so $f[N] \in \mathcal{N}(f(x))$ as required. Simply continuity of $f$ is enough to make $g[N]= f^{-1}[N]$ a neighbourhood of $x$ when $N$ is one for $f(x)$ so that the map $N \to f[N]$ is a bijection between $\mathcal{U}(x)$ and $\mathcal{U}(f(x))$.
b i) can be done more easily: if $A \subseteq X$ is closed, it's compact (being closed in the compact space $X$) and so $f[A]$ is compact by continuity of $f$, and thus closed in $Y$ (should be a well-known fact for you). So $f$ is a closed map. 
And this implies quite easily that the inverse of a continuous bijection on a compact space is also continuous (continuity can also be seen as "the inverse image of a closed set is closed", just as we can for open sets). That part you already saw. But the sequences aren't really needed for b-i. 
