Equivalent Statements Regarding Cts. Maps on Topo. Spaces

Equivalent Statements Regarding Continuous Maps on Topological Spaces

Recall from the Continuous Maps on Topological Spaces page that if $X$ and $Y$ are topological spaces then a map $f : X \to Y$ is said to be continuous at the point $a \in X$ if there exists local bases $\mathcal B_a$ of $a$ and $\mathcal B_{f(a)}$ of $f(a)$ such that for all $B \in \mathcal B_{f(a)}$ there exists a $B' \in \mathcal B_a$ such that:

(1)

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

Furthermore, we said that $f$ is continuous on all of $X$ if $f$ is continuous at each point $a \in X$.

Additionally, on the The Open Neighbourhood Definition of Continuous Maps on Topological Spaces page we looked at an equivalent definition for the continuity of $f$ at $a \in X$. We saw that $f$ is continuous at $a \in X$ if and only 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}

We will now look at some equivalence statements regarding $f$ being continuous on ALL of $X$.

Theorem 1: Let $X$ and $Y$ be topological spaces and let $f : X \to Y$. Then the following are equivalent: a) $f$ is continuous on all of $X$ b) For every open set $V$ in $Y$ we have that $f^{-1}(V)$ is an open set in $X$. c) For every basis $\mathcal B_Y$ of $Y$ and for every $B \in \mathcal B_Y$ we have that $f^{-1}(B)$ is open in $X$.

• Proof: $a) \implies b)$ Suppose that $f$ is continuous on all of $X$, and let $V$ be an open set in $Y$. We want to show that $f^{-1}(V)$ is an open set in $X$. We will accomplish this by showing that $f^{-1}(V)$ is the union of a collection of open sets.

• Note that for all $a \in f^{-1}(V)$ we have that $f(a) \in V$. Since $V$ is an open set, we have that $\mathrm{int} (V) = V$. So, for all $f(a) \in V$ by definition of $V$ being an open set there exists an open neighbourhood $V_{f(a)}$ of $f(a)$ such that:

(3)

\begin{align} \quad f(a) \in V_{f(a)} \subseteq V \end{align}

• By the theorem on the The Open Neighbourhood Definition of Continuous Maps on Topological Spaces page, we have that for all $a \in f^{-1}(V)$ and for all open neighbourhoods $V_{f(a)} \subseteq V$ of $f(a)$ there exists an open neighbourhood $U_a$ of $a$ such that:

(4)

\begin{align} \quad f(U_a) \subseteq V_{f(a)} \subseteq V \end{align}

• Therefore $U_a \subseteq f^{-1} (V)$. Hence we see that:

(5)

\begin{align} \quad f^{-1} (V) = \bigcup_{a \in f^{-1} (V)} U_a \end{align}

• But then $f^{-1} (V)$ is the union of a arbitrary collection of open sets which shows that $f^{-1} (V)$ is open in $X$.

• $b) \implies c)$. Suppose that for all open sets $V$ in $Y$ we have that $f^{-1} (V)$ is open in $X$. Let $\mathcal B_Y$ be any basis of the topology on $Y$. Recall that $\mathcal B_Y$ is a subset of the topology on $Y$, and hence every $B \in \mathcal B_Y$ is an open set. Hence $f^{-1} (B)$ is open in $X$ for all $B \in \mathcal B_Y$.

• $c) \implies a)$. Suppose that for all bases $\mathcal B_Y$ of the topology on $Y$ we have that every $B \in \mathcal B_Y$ is such that $f^{-1}(B)$ is open. Consider the following local basis of $f(a)$:

(6)

\begin{align} \quad \mathcal B_{f(a)} = \{ B \in \mathcal B_Y : f(a) \in B \} \end{align}

• We know that $\mathcal B_{f(a)}$ is indeed a local basis of $f(a)$ by one of the theorems on the Basic Theorems Regarding Local Bases of a Point in a Topological Space page.

• Now let $V$ be any open neighbourhood of $f(a)$. Since $\mathcal B_{f(a)}$ is a local basis of $f(a)$ we have that for every open neighbourhood $V$ of $f(a)$ there exists a $B \in \mathcal B_a$ such that:

(7)

\begin{align} \quad f(a) \in B \subseteq V \end{align}

• Therefore $a \in f^{-1}(B)$ and since $B \in \mathcal B_Y$ we have by hypothesis that $f^{-1} (B)$ is open, and so $f^{-1}(B)$ is an open neighbourhood of $a$.

• Recall from the The Open Neighbourhood Definition of Continuous Maps on Topological Spaces page that $f$ is continuous at $a$ if and only if for all open neighbourhoods $V$ of $f(a)$ there exists an open neighbourhood $U$ of $a$ such that $f(U) \subseteq V$. In this case, take $U = f^{-1}(B)$. Then:

(8)

\begin{align} \quad f(U) = f(f^{-1}(B)) = B \subseteq V \end{align}

• Therefore $f$ is continuous at $a \in X$, but $a$ was arbitrary. Hence $f$ is continuous on all of $X$. $\blacksquare$