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We know that if $X,Y$ are topological spaces, then $f: X \to Y$ is continuous if and only if $f(\overline{E}) \subseteq \overline{f(E)}$, for all $E \subseteq X$.

I started studying nets by myself properly, and I think that the best way to get used to it is proving stuff using it. So I want to prove that assertion using nets. I am not interested in another proofs of the fact now.

I am having trouble in one of the implications.

  • $\implies$: Suppose that $f$ is continuous. Let $E \subseteq X$ be any set. If $\overline{E} = \varnothing$, there is nothing to do. If $\overline{E} \neq \varnothing$, take $y \in f(\overline{E})$. Write $y = f(x)$, with $x \in \overline{E}$. Then we can take a net $(x_\alpha)_{\alpha \in A} \subseteq E$ such that $x_\alpha \to x$. Since $f$ is continuous, $f(x_\alpha)\to f(x) = y$. But $f(x_\alpha) \in f(E)$ for all $\alpha \in A$ will give us $y \in \overline{f(E)}$, so $f(\overline{E})\subseteq \overline{f(E)}$ and we're done.

So far, so good.

  • $\impliedby$: Let $(x_\alpha)_{\alpha \in A} \subset X$ such that $x_\alpha \to x$. If we check that $f(x_\alpha)\to f(x)$, we're done. Let $V \subseteq Y$ be a neighbourhood of $f(x)$. Since $x_\alpha \to x$, we have $x \in \overline{\{x_\alpha \in X \mid \alpha \in A\}}$, so that the hypothesis gives: $$f(x) \in f(\overline{\{x_\alpha \in X \mid \alpha \in A\}})\subseteq \overline{f(\{x_\alpha \in X \mid \alpha \in A)}.$$Hence $V \cap f(\{x_\alpha \in X \mid \alpha \in A\})\neq \varnothing,$ and we obtain $\alpha_0 \in A$ such that $f(x_{\alpha_0}) \in V$. Since I just followed my nose so far, I'd expect that $\alpha \succeq \alpha_0$ would imply $f(x_\alpha) \in V$, ending the proof. But I couldn't show that.

Can someone help me? Thanks.

Ivo Terek
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2 Answers2

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Suppose that $f$ is not continuous at $x$. Then, there is a net $(x_{\alpha})_{\alpha\in A}$ such that $x_{\alpha}\to x$, but $f(x_{\alpha})\not\to f(x)$. This means that $f(x)$ has a neighborhood $V$ with the following property: $$\forall\beta\in A,\exists\alpha_{\beta}\in A:\alpha_{\beta}\succsim \beta,\,f(x_{\alpha_{\beta}})\notin V.$$ It is not difficult to check that $(x_{\alpha_{\beta}})_{\beta\in A}$ is a subnet of $(x_{\alpha})_{\alpha\in A}$ (in the definition given by Folland, 1999, p. 126, just take $B=A$ and, for each $\alpha_0\in A$, take $\beta_0=\alpha_0$; then, $\beta\succsim\beta_0$ will imply that $\alpha_{\beta}\succsim\beta\succsim\beta_0=\alpha_0$, as required). This implies that $x_{\alpha_{\beta}}\to x$, so $$f(x)\in f\left(\overline{\{x_{\alpha_{\beta}}\,|\,\beta\in A\}}\right)\subseteq \overline{f\left(\{x_{\alpha_{\beta}}\,|\,\beta\in A\}\right)}.$$ It follows that $V\cap\overline{f\left(\{x_{\alpha_{\beta}}\,|\,\beta\in A\}\right)}\neq\varnothing$, so $V\cap f\left(\{x_{\alpha_{\beta}}\,|\,\beta\in A\}\right)\neq\varnothing$. However, for any $\beta\in A$, $f(x_{\alpha_{\beta}})\notin V$, which is a contradiction.

P.S.: I told you nets would be useful. :-)

triple_sec
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Lets prove using the negative. Suppose $f$ do not is continuous at some point $y$, so for each $V$ open neighbourhood of $y$, there exists $V^\prime \subset V$ another neighborhood with $f^{-1}(V^\prime)$ not open, that means that there exists $x_{V^\prime} \in f^{-1}(V^\prime)\setminus \mathring{f^{-1}(V^\prime)} $. Consider now the net $x_{V^\prime}$ where the index form a directed set with the reverse inclusion order (A

Ricardo Correa da Silva
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