Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and $f:X\to Y$ a function. Then the following statements are equivalent:
I) $f$ is continuous
II) For every open set $V\subseteq Y$ is $f^{-1}(V)\subseteq X$ open.
III) For every closed set $A\subseteq Y$ is $f^{-1}(A)\subseteq X$ closed.
IV) For every subset $S\subseteq X$ holds $f(\overline{S})\subseteq \overline{f(S)}$
In the lecture was shown that I) implies II) implies III) implies IV). Now it is left to show that IV) implies I).
Of course I would like a direct proof of IV) $\Rightarrow$ I), but I can not do it.
So I went the long way:
IV) $\Rightarrow$ III):
Let for every $S\subseteq X$ hold $f(\overline{S})\subseteq \overline{f(S)}$. I want to show that for a closed set $A\subseteq Y$ is $f^{-1}(A)\subseteq X$ closed.
My question is, if you can check my proofs and if you can show from the condition in IV), that $f$ is continuous by definition (with $\varepsilon$-$\delta$)
Proof:
I want to show $f^{-1}(A)=\overline{f^{-1}(A)}$.
$f^{-1}(A)\subseteq\overline{f^{-1}(A)}$ is trivial.
For $\overline{f^{-1}(A)}\subseteq f^{-1}(A)$ we have:
$f(\overline{f^{-1}(A)})\subseteq \overline{f(f^{-1}(A)}\subseteq \overline{A}=A$ since $A$ is closed.
Then $f^{-1}(f(\overline{f^{-1}(A)}))\subseteq f^{-1}(A)$.
Suppose $\overline{f^{-1}(A)}\nsubseteq f^{-1}(A)$. Then exists $x\in\overline{f^{-1}(A)}$ with $x\notin f^{-1}(A)$.
So $f(x)\in f(\overline{f^{-1}(A)})$ and $x\in f^{-1}(f(\overline{f^{-1}(A)}))\subseteq f^{-1}(A)$, which is a contradiction.
So as desired $\overline{f^{-1}(A)}\subseteq f^{-1}(A)$.
III)$\Rightarrow$ II)
Let $U\subseteq Y$ be open. Then is $U^c=Y\setminus U$ closed. So we have $f^{-1}(Y\setminus U)=X\setminus f^{-1}(U)=f^{-1}(U)^c$ is closed. Since $f^{-1}(U)^c$ is closed, it is $f^{-1}(U)$ open.
II) $\Rightarrow$ I)
Let $x\in X$ be arbitrary. Let $f(x)\in V$ open. Then exists $\varepsilon >0$ such that, $B_\varepsilon (f(x))\subseteq V$. Since $f^{-1}(B_\varepsilon(f(x))\ni x$ is open, it exists $\delta > 0$ with $B_\delta(x)\subseteq f^{-1}(B_\varepsilon f(x))$, hence $f(B_\delta(x))\subseteq (B_\varepsilon f(x))$.
The problem here, is that I have not shown, that for every $\varepsilon$ we have such $\delta$, or am I mistaken? I have shown for a specific $\varepsilon$ we can find this $\delta$ which is to weak. How can I fix this?
Thanks in advance.
If you slightly retool your proof from II) to I), then you can fix this issue. Rather than supposing you have an open $V$ containing $f(x)$, then immediately discard it for a ball of some radius $\varepsilon$, start by supposing $\varepsilon > 0$, and considering $f^{-1}(B_\varepsilon(f(x)))$. Note that this is an open set, by assumption, which contains $x$. Thus, there is a $\delta$-ball $B_\delta(x)$ contained in $f^{-1}(B_\varepsilon(f(x)))$, as required.