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I'm trying to show this equivalence of different formulation of continuity. Could you have a check on my proof?

Let $(X, \tau_X)$ and $(Y, \tau_Y)$ be topological spaces and $f:X \to Y$ be continuous, i.e., $f^{-1} (O) \in \tau_X$ for all $O \in \tau_Y$. Then $f$ is continuous if and only if, $f(x_d) \to f(x)$ where $(x_d)_{d \in D}$ is a net such that $x_d \to x$.

Lemma: Let $A \subseteq X$. Then $\overline A$ is the set of limits of convergent nets with values in $A$. [A proof is provided here]

Proof: Assume that $f$ is continuous and $(x_d)_{d \in D}$ is a net in $X$ such that $x_d \to x$. Let $y := f(x)$ and $U$ is a neighborhood of $y$. Let $V := f^{-1} (U)$. Then $V$ is a neighborhood of $x$ due to the continuity of $f$. Then there is $d_V \in D$ such that $x_d \in V$ for all $d \ge d_V$. This implies $f(x_d) \in U$ for all $d \ge d_V$. Hence $f(x_d) \to y=f(x)$.

Conversely, let $U$ be closed in $Y$ and $V := f^{-1} (U)$. Let $(x_d)_{d\in D}$ be a net in $V$ such that $x_D \to x$. We want to prove $x \in V$. Let $y_d := f(x_d)$. Then $(y_d)$ is a net in $U$ such that $y_d \to y:=f(x) \in Y$. By our lemma, $y\in \overline U$. Because $U$ is closed, $U =\overline U$ and thus $y \in U$. Hence $x \in f^{-1}(y) \subseteq V$. As such, $V$ is closed.

Henno Brandsma
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Akira
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    It's at least original too use the closed set characterisation of continuity to do one half. Usually one sees the proof by contrdiction there. Nice find. – Henno Brandsma Jan 05 '22 at 01:38

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The very last part is better stated as: as $f(x) = y \in U$, by definition $x \in f^{-1}[U]= V$. So $V$ is closed under limits of nets and hence closed by the lemma.

"Psycho-notionally" I'd use $C$ or $F$ for a closed subset, personally I associate the letters $U$,$V$ and $O$ with open sets. It's minor, but proofs should minimise possible confusion, IMHO. And then I'd use, say, $D=f^{-1}[C]$ next for the inverse image we want to be closed (the next letter). It's easier on my memory.

Henno Brandsma
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