Continuity of Functions on Metric Spaces

# Continuity of Functions on Metric Spaces

Recall from the Limits of Functions on Metric Spaces that if $(S, d_S)$ and $(T, d_T)$ are both metric spaces and $A \subseteq S$ then a function $f : A \to T$ is defined for all $x \in A$ by some rule $f(x) = y$ where $A = D(f)$ is called the domain of $f$ and $T = C(f)$ is called the codomain of $f$ (where the range of $f$ is defined to be the set $R(f) = \{ f(x) = y \in T : x \in A \}$).

We have already looked at the continuity of functions that map from the real numbers to the real numbers - but we can define a more general definition of continuity for a function $f$ on metric spaces.

 Definition: Let $(S, d_S)$ and $(T, d_T)$ be metric spaces and let $f : S \to T$. The function $f$ is said to be Continuous at the Point $p \in S$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $d_S(x, p) < \delta$ then $d_T(f(x), f(p)) < \epsilon$. If $A \subseteq S$ and $f$ is continuous at every point $a \in A$ then $f$ is said to be Continuous on $A$.

Equivalently, $f$ is said to be continuous at $p \in S$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that the image under $f$ of the ball centered at $p$ with radius $\delta$ ( $f(B_S(p, \delta))$) is fully contained in the ball centered at $f(p)$ with radius epsilon $(B_T(f(p), \epsilon)$, i.e.:

(1)
\begin{align} \quad f(B_S(p, \delta)) \subseteq B_T(f(p), \epsilon) \end{align}

Now note that if $p$ is an accumulation point of $S$ then if $f$ is continuous at $p$ then by the definition of continuity we have that for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $d_S(x, p) < \delta$ then $d_T(f(x), f(p)) < \epsilon$ - so for all $x \in S \setminus \{ p \}$ and $d_S(x, p) < \delta$ we have that $d_T(f(x), f(p)) < \epsilon$ and so we have that:

(2)
\begin{align} \quad \lim_{x \to p} f(x) = f(p) \end{align}

Furthermore if $p$ is instead an isolated point of $S$ then there exists a $\delta > 0$ such that $B(p, \delta)$ contains no points of $S$ other than $p$, i.e., $S \cap B(p, \delta) \setminus \{ p \} = \emptyset$. Therefore, we have that for all $\epsilon > 0$ if we choose $\delta > 0$ then $B(p, \delta)$ only contains $p$ and $f(B_S(p, \delta)) = \{ f(p) \} \subseteq B_T(f(p), \epsilon)$ so any $f$ where $f(p)$ is defined is therefore continuous at $p$.