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Let $f:(X,\tau_X) \to (Y,\tau_Y)$ be a function between two topological spaces. Then $f$ is continuous $\iff$ $\forall Z \subset X: f(\overline Z) \subset \overline{f(Z)}$.

I want to prove this in terms of nets. $(\Rightarrow)$ is easy. Now for $(\Leftarrow)$ assume $x \in X$ and $x_\iota \to x$ where $\{x_\iota\}_{\iota \in (I,\leq)}$ is a net. I want to show that $f(x_\iota) \to f(x)$. I already know that $x \in \overline {\{x_\iota: \iota \in I \}}$. By assumption we thus have $f(x) \in \overline { \{f(x_\iota): \iota \in I\}}$. But now I only know that there is some $I' \subset I$ s.t. $\{f(x_\iota)\}_{\iota \in I'}$ converges to $f(x)$. Is this sufficient to show that the original net $\{f(x_\iota)\}_{\iota \in I}$ converges to $f(x)$ ?

I assume that we know that the closure of a set $Y$ is exactly the set of points which are limits of nets in $Y$.

Martin Sleziak
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1 Answers1

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I would do it with neighborhoods. Assume that $V$ is a neighborhood of $f(x)$, so there is a neighborhood $U$ of $x$ such that $f[U]\subseteq V$. Since $(x_i)_i\to x$ there is a $j\in I$ with $x_i\in U$ for all $i>j$. But then $f(x_i)\in V$ for all $i>j$.

To prove that $f$ is continuous at $x$, let $V$ be a neighborhood of $f(x)$ and assume that for each neighborhood $U$ of $x$ we have a non-empty $U-f^{-1}[V]$. The neighborhoods of $x$ ordered by reversed inclusion form a directed set $\cal U$. We can choose an $x_U$ in each $U-f^{-1}[V]$. It is easy to show that $(x_U)_\cal U$ converges to $x$, but $(f(x_U))$ does not converge to $f(x)$.

The property $x\in\overline{\{x_i\mid i\in I\}}$ is not really helpful. But you can make a stronger statement, namely $x\in\overline{T_j}$ for all $j\in I$, where $T_j$ denotes the tail $\{x_i\mid i\ge j\}$. Unfortunately, this still does not express that $x_i\to x$, but it rather means that $x$ is a cluster point of $(x_i)_i$. Then it is indeed possible to proof the $f(x)$ is a cluster point of $(f(x_i))_i$ using the characterization of continuity that you gave above.

By the way: The characterization of continuity of $f$ via the closure of subsets also has a local version: $$f\text{ is continuous at }x\iff x\in\overline A\text{ implies }f(x)\in\overline{f[A]}\text{ for all }A⊆X$$

Stefan Hamcke
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  • Your first paragraph proves $(\Rightarrow)$, but I still don't undertand how to prove $(\Leftarrow)$ by means of nets. Without nets it is quite easy since if $C \subset Y$ is closed then let $Z := f^{-1}(C)$. Using the assumption we get $f(\overline Z) \subset \overline{f(Z)} = \overline C = C$,s.t. $\overline Z \subset Z$. That is $Z = \overline Z$ and thus closed. –  Jan 03 '14 at 23:20
  • I've edited the proof for the other direction into my answer. It resembles the proof that a sequential continuous function at $x$ is continuous at $x$ if $x$ has a countable neighborhood base. – Stefan Hamcke Jan 04 '14 at 00:10