Generalizing Continuity to Maps on Topological Spaces

Generalizing Continuity to Maps on Topological Spaces

The reader should be somewhat familiar with the definition of continuity of a function $f$ mapping $\mathbb{R}$ into itself. Recall that $f$ is said to be continuous at $a \in \mathbb{R}$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $\mid x - a \mid < \delta$ then $\mid f(x) - f(a) \mid < \epsilon$.

Our aim is to generalize this concept of continuity of real-variable functions to functions/maps $f : X \to Y$ where $X$ and $Y$ are topological spaces.

With the definition of continuity of a function $f$ mapping $\mathbb{R}$ into itself, we see an equivalent definition for the continuity of $f$ at $a$ is that for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $x \in (a - \delta, a + \delta)$ then $f(x) \in (f(a) - \epsilon, f(a) + \epsilon)$.

The following theorem will make the notion of continuity at a point $a \in \mathbb{R}$ a little more abstract in preparing us for the definition of a function $f : X \to Y$ on topological spaces to be continuous at a point.

 Theorem 1: Consider the set $\mathbb{R}$ with the usual topology $\tau$ of open intervals on $\mathbb{R}$. Let $f$ be a function mapping $\mathbb{R}$ into itself. Then $f$ is continuous at a point $a \in \mathbb{R}$ if and only for each open set $V$ containing $f(a)$ there exists an open set $U$ containing $a$ such that $f(U) \subseteq V$.
• Proof: $\Rightarrow$ Suppose that $f$ is continuous at a point $a \in \mathbb{R}$. Then for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $x \in (a - \delta, a + \delta)$ then $f(x) \in (f(a) - \epsilon, f(a) + \epsilon)$.
• Now let $V \in \tau$ be such that $f(a) \in V$. Provided that $V$ is not the empty set, we have that $V$ will either be an open interval containing $f(a)$ or a union of open intervals - one of which contains $f(a)$. Hence let $(c, d)$ be the interval such that:
(1)
\begin{align} \quad f(a) \in (c, d) \subseteq V \end{align}
• Since $f(a) \in (c, d)$ we have that $c < f(a) < d$ so $f(a) - c, d - f(a) > 0$. Let $\epsilon$ be defined as:
(2)
\begin{align} \quad \epsilon = \min \{ f(a) - c, d - f(a) \} \end{align}
• Then we have that:
(3)
\begin{align} \quad (f(a) - \epsilon, f(a) + \epsilon) \subseteq (c, d) \subseteq V \end{align}
• Since $f$ is continuous we have that for this $\epsilon$ there exists a $\delta > 0$ such that for all $x \in (a - \delta, a + \delta)$ we have that $f(x) \in (f(a) - \epsilon, f(a) + \epsilon)$. Take $U = (a - \delta, a + \delta)$. Then:
(4)
\begin{align} \quad f(U) \subseteq (f(a) - \epsilon, f(a) + \epsilon) = V \end{align}
• $\Leftarrow$ Suppose that for $a \in \mathbb{R}$ we have that every every open set $V$ containing $f(a)$ is such that there exists an open set $U$ containing $a$ such that $f(U) \subseteq V$.
• For some $\epsilon > 0$ let $V = (f(a) - \epsilon, f(a) + \epsilon)$. Then there exists an open set $U$ containing $a$ such that $f(U) \subseteq V$. Since $U$ is an open set that is nonempty $U$ must contain an interval containing $a$, say:
(5)
\begin{align} \quad a \in (c, d) \subseteq U \end{align}
• So $a \in (c, d)$ tells us that $c < a < d$. Therefore $a - c, d - a > 0$ and so Define:
(6)
\begin{align} \quad \delta = \min \{ a - c, d - a \} \end{align}
• Hence we see that:
(7)
\begin{align} \quad a \in (a - \delta, a + \delta) \subseteq (c, d) \subseteq U \end{align}
• So, for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $x \in (a - \delta, a + \delta)$ then $f(x) \in (f(a) - \epsilon, f(a) + \epsilon)$, so $f$ is continuous at $a \in \mathbb{R}$. $\blacksquare$