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Forgive me if this is a repeat post but it didn't have the solution I was looking for which is one by contradiction.

For the following proof, $\mathcal{O}(x)$ is the set of open neighborhoods of $x$.

So I suppose that if $(x_\alpha)_{\alpha \in A}$ converges to $x$, then $(f(x_\alpha))_{\alpha \in A}$ converges to $f(x)$. And assume $f$ is discontinuous. Then there exists $x \in X$ such that $V$ is an open neigborhood for $f(x)$ but $f^{-1}(V)$ is not an open neighborhood of $x$. Thus setting $A=$ the set of neighborhoods of some $y \in X$ ordered by reverse inclusion ($A \geq B \Rightarrow A \subseteq B$) we can construct a net $(y_U)_{U \in \mathcal{O}(y)}$ such that $$y_U \not\in f^{-1}(V)$$ which implies $$f(y_U) \not\in V.$$ For all $U_O \leq U$ some $U_0 \in \mathcal{O}(y)$. Choose any neighborhood of $y$, say $W$. Then for $T \geq W$, $T,W \in \mathcal{O}(y)$, one has that $$y_T \in W.$$ It follows then that there exists some $U_0$ such that whenever $U_0 \leq U$, $(y_U)_{U \in \mathcal{O}(y)}$ converges to $y$ thus $(f(y_U))_{U \in \mathcal{O}(y)}$ converges to $f(x)$ which implies $$f(y_U) \in V$$ a contradiction.

Basically, was the trick to let my directed indexing set be the set of open neighborhoods of a given point in the net? ordered by reverse inclusion.

homosapien
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1 Answers1

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Suppose $f$ is discontinuous. That is, there exists an $x \in X$ such that $f(x) \in V$ for some open $V \subset Y$ but

$$x \in (f^{-1}(V)^\circ)^c$$

This implies $x \in \overline{f^{-1}(V^c)}$, then there is a net $(x_\alpha)_{\alpha \in A}$ inside of $f^{-1}(V^c)$ converging to $x$. Then $(f(x_\alpha))$ converges to $f(x)$ which means $f(x_\alpha) \in V$ for all $\alpha$ but this is a contradiction as $f(x_\alpha) \in V^c$ and the proof is done.

homosapien
  • 4,157