Prove that $f(x)\to L$ as $x\to x_0$ iff $f(U\cap E)\subseteq V$ whenever $V\ni L$ and $U\cap E\ni x_0$ are open.

Question

Let $(X,d_{X})$ and $(Y,d_{Y})$ be metric spaces, let $E$ be a subset of $X$, and let $f:X\to Y$ be a function. Let $x_{0}\in X$ be an adherent point of $E$ and $L\in Y$. Then the following two statements are logically equivalent:

(a) $\displaystyle\lim_{x\to x_{0};x\in E}f(x) = L$

(b) For every open set $V\subset Y$ which contains $L$, there exists an open set $U\subset X$ containing $x_{0}$ such that $f(U\cap E)\subseteq V$.

MY ATTEMPT (EDIT)

Let us prove $(a)\Rightarrow(b)$ first.

Let $V\subset Y$ be an open set such that $L\in V$. Then there is an $\varepsilon > 0$ such that $B(L,\varepsilon)\subseteq V$.

Related to such $\varepsilon$ there is a $\delta$ such that $U = B(x_{0},\delta)\subset X$ satifying $f(U\cap E)\subseteq V$, and we are done.

Let us prove now that $(b)\Rightarrow(a)$

Let $\varepsilon > 0$.

If we choose $V = B(L,\varepsilon)$ there is an open set $U\subset X$ containing $x_{0}$ such that $f(U\cap E)\subseteq B(L,\varepsilon)$.

Let $\delta > 0$ such that $B(x_{0},\delta)\subseteq U$. Then we have \begin{align*} x\in E, d_{X}(x,x_{0}) < \delta \Rightarrow x\in U\cap E \Rightarrow f(x)\in B(L,\varepsilon) \Rightarrow d_{Y}(f(x),L) < \varepsilon \end{align*}

and we are done.

Answer 1

To be absolutely correct, I would suggest adding the following to your argument in $(a)\implies(b):$

Let $V\subset Y$ be open such that $L \in V.$ Then there exists $\epsilon >0$ such that $B(L,\epsilon)\subseteq V.$

Moreover for the argument in $(b) \implies (a),$ I would rephrase it as follows:

Let $\epsilon>0.$ In particular choosing $V=B(L,\epsilon),$ there exists an open set $U \subseteq X$ such that $x_0 \in U$ and $f(U \cap E)\subseteq B(L,\epsilon).$ Let $\delta>0$ such that $B(x_0,\delta) \subseteq U.$ Then $$x\in E,d_X(x,x_0)<\delta \implies x\in U\cap E\implies f(x)\in B(L,\epsilon)\implies d_Y(f(x),L)<\epsilon.$$