The Open Neighbourhood Definition of Cts. Maps on Topo. Spaces

The Open Neighbourhood Definition of Continuous Maps on Topological Spaces

Recall from the Continuous Maps on Topological Spaces page that if $X$ and $Y$ are topological spaces then a function $f : X \to Y$ is said to be Continuous at a 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 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}

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

We will now look an alternative but similar definition of a map $f : X \to Y$ being continuous at a point $a \in X$.

Theorem 1: Let $X$ and $Y$ be topological spaces and let $f : X \to Y$. Then $f$ 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 $f(U) \subseteq V$.

• Proof: $\Rightarrow$ Let $f : X \to Y$ and suppose that $f$ is continuous at $a \in X$. Then 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:

(2)

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

• 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 by definition for every open neighbourhood $V$ of $f(a)$ that there exists $B \in \mathcal B_{f(a)}$ such that:

(3)

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

• But $B$ itself is an open neighbourhood of $f(a)$, and so since $f$ is continuous we have that since $B \in \mathcal B_{f(a)}$ that there exists a $B' \in \mathcal B_a$ such that:

(4)

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

• But $B' \in \mathcal B_a$ is an open neighbourhood of $a$ $(*)$, so set $U = B'$. Therefore from $(*)$, $(**)$, and $(***)$ we have that:

(5)

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

• Hence, for every open neighbourhood $V$ of $f(a)$ there exists an open neighbourhood $U$ of $a$ such that $f(U) \subseteq V$.

• $\Leftarrow$ Suppose now that for every open neighbourhood $V$ of $f(a)$ there exists an open neighbourhood $U$ of $a$ such that $f(U) \subseteq V$.

• To show that $f$ is continuous we must show that there exists an local bases $\mathcal B_{f(a)}$ of $f(a)$ and $\mathcal B_a$ of $a$ such that for all $B \in \mathcal B_{f(a)}$ there exists a $B' \in \mathcal B_a$ such that $f(B') \subseteq B$.

• Consider the following (rather trivial) local bases of $a$ and $f(a)$:

(6)

\begin{align} \quad \mathcal B_{a} = \{ U : U \: \mathrm{is \: an \: open \: neighbourhood \: of \:} a \} \quad \quad \mathcal B_{f(a)} = \{ V : V \: \mathrm{is \: an \: open \: neighbourhood \: of \:} f(a) \} \end{align}

• Then for every $V = B \in \mathcal B_{f(a)}$ (open neighbourhood of $f(a)$) we are given that there exists an open neighbourhood $U = B' \in \mathcal B_a$ such that:

(7)

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

• Therefore $f$ is continuous at $a$. $\blacksquare$

From Theorem 1 above, we see that then $f : X \to Y$ is continuous on all of $X$ if and only if for all $a \in X$ we have that for all open neighbourhoods $V$ of $f(a)$ there exists an open neighbourhood $U$ of $a$ such that $f(U) \subseteq V$.