The Closed Set Definition of Continuous Maps on Topo. Spaces

# The Closed Set Definition of Continuous Maps on Topological Spaces

Recall from the Continuous Maps on Topological Spaces page that we say $f : X \to Y$ is continuous at $a \in X$ if there exists local bases $\mathcal B_a$ of $a$ and $\mathcal B_{f(a)}$ of $f(a)$ such that for every $B \in \mathcal B_{f(a)}$ there exists a $B' \in \mathcal B_{a}$ such that:

(1)
\begin{align} \quad f(B') \subseteq B \end{align}

On the The Open Neighbourhood Definition of Continuous Maps on Topological Spaces page we saw that $f : X \to Y$ is continuous at $a \in X$ if for every open neighbourhood $V$ of $f(a)$ there exists an open neighbourhood $U$ of $a$ such that:

(2)
\begin{align} \quad f(U) \subseteq V \end{align}

On the Equivalent Statements Regarding Continuous Maps on Topological Spaces page that we saw that that the following statements are equivalent:

• (1) $f : X \to Y$ is continuous on all of $X$.
• (2) For every open set $V$ in $Y$ we have that $f^{-1}(V)$ is open in $X$.
• (3) For every basis $\mathcal B_Y$ of $Y$ we have that for all $B \in \mathcal B_Y$ that $f^{-1}(B)$ is open in $X$.

We will now look at yet another equivalent definition of a map being continuous on all of $X$ with regards to closed sets. We must first prove the following lemma though.

 Lemma 1: Let $X$ and $Y$ be topological spaces and let $f : X \to Y$. Then for all $S \subseteq Y$ we have that $f^{-1}(S^c) = (f^{-1}(S))^c$.
• Proof: Let $x \in f^{-1}(S^c)$. Then $f(x) \in S^c$ so $f(x) \not \in S$ and $x \not \in f^{-1}(S)$ so $x \in (f^{-1}(S))^c$. Therefore $f^{-1}(S^c) \subseteq (f^{-1}(S))^c$.
• Now let $x \in (f^{-1}(S))^c$. Then $x \not \in f^{-1}(S)$ so $f(x) \not \in S$ so $f(x) \in S^c$ and $x \in f^{-1}(S^c)$. Therefore $f^{-1}(S^c) \supseteq (f^{-1}(S))^c$.
• Since $f^{-1}(S^c) \subseteq (f^{-1}(S))^c$ and $f^{-1}(S^c) \supseteq (f^{-1}(S))^c$ we have that then:
(3)
 Theorem 2: Let $X$ and $Y$ be topological spaces and let $f : X \to Y$. Then $f$ is continuous on all of $X$ if and only if for all closed sets $V$ in $Y$ we have that $f^{-1}(V)$ is closed in $X$.
• Proof: $\Rightarrow$ Suppose that $f$ is continuous on all of $X$. Then for all open sets $V$ in $Y$ we have that $f^{-1} (V)$ is an open set in $X$.
• Let $V$ now be a closed set in $Y$. Then $V^c$ is an open set in $Y$ and so $f^{-1}(V^c)$ is open in $X$. But by Lemma 1:
• So $(f^{-1}(V))^c$ is open in $X$ which implies that $f^{-1}(V)$ is closed in $X$. Since $V$ is an arbitrary closed set in $Y$ we see that for every closed set $V$ in $Y$ we have that $f^{-1}(V)$ is closed in $X$.
• $\Leftarrow$ Suppose that for every closed set $V$ in $Y$ we have that $f^{-1}(V)$ is closed in $X$. Let $V$ instead be an open set in $Y$. Then $V^c$ is closed in $Y$ so $f^{-1}(V^c)$ is closed in $X$ and by Lemma 1:
• So $(f^{-1}(V))^c$ is closed in $X$ which implies that $f^{-1}(V)$ is open in $X$. Since $V$ is an arbitrary open set in $Y$ we see that for every open set $V$ in $Y$ we have that $f^{-1}(V)$ is open in $X$. Therefore $f$ is continuous on all of $X$. $\blacksquare$