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)

\begin{align} \quad f^{-1}(S^c) = (f^{-1}(S))^c \quad \blacksquare \end{align}

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:

(4)

\begin{align} \quad f^{-1}(V^c) = (f^{-1}(V))^c \end{align}

• 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:

(5)

\begin{align} \quad f^{-1}(V^c) = (f^{-1}(V))^c \end{align}

• 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 $X$. $\blacksquare$