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)
• 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)
• 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 $f(U) \subseteq V$.