The Continuity of Composite Functions on Metric Spaces

# The Continuity of Composite Functions on Metric Spaces

Recall from the Continuity of Functions on Metric Spaces page that if $(S, d_S)$ and $(T, d_T)$ are metric spaces and $f : S \to T$ then $f$ 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 $d_T(f(x), f(p)) < \epsilon$.

Equivalently, we said that $f$ is continuous at $p$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that:

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

On the Sequential Criterion for the Continuity of a Function on Metric Spaces page we also proved a very important theorem which said that $f$ is continuous at $p$ if and only if for all sequences $(x_n)_{n=1}^{\infty}$ from $S$ that converge to $p$ we have that the sequence $(f(x_n))_{n=1}^{\infty}$ converges to $f(p)$.

We will now continue in looking at what can be said about the continuity of functions. Let $(S, d_S)$, $(T, d_T)$, and $(U, d_U)$ all be metric spaces and let $f : S \to T$ and $g : T \to U$. We can obtain a composite function $g \circ f : S \to U$ defined for all $x \in S$ by:

(2)
\begin{align} \quad (g \circ f)(x) = g(f(x)) \end{align}

In the following theorem we will see how the continuity

 Theorem 1: Let $(S, d_S)$, $(T, d_T)$, and $(U, d_U)$ be metric spaces and let $f : S \to T$ and $g : T \to U$. If $f$ is continuous at $p$ and $g$ is continuous at $f(p)$ then $g \circ f$ is continuous at $p$.
• Proof: To show that $h$ is continuous at $p$ we must show that for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $d_S(x, p) < \delta$ that then:
(3)
\begin{align} \quad d_U(g(f(x)), g(f(p))) < \epsilon \end{align}
• Since $g$ is continuous at $f(p)$ we have that for all $\epsilon > 0$ there exists a $\hat{\delta} > 0$ such that if $d_T(f(x), f(p)) < \hat{\delta}$ $(***)$ then:
(4)
\begin{align} \quad d_U(g(f(x)), g(f(p))) < \epsilon (****) \end{align}
• Also, since $f$ is continuous at $p$ we have that for all $\delta$ ($> 0$) that there exists a $\bar{\delta} > 0$ such that if $d_S(x, p) < \bar{\delta}$ $(*)$ then:
(5)
\begin{align} \quad d_T(f(x), f(p)) < \hat{\delta} (**) \end{align}
• So then for any given $\epsilon > 0$, take $\delta = \bar{\delta}$. Then if $(*)$ holds then $(**)$ holds and by $(***)$ we then have that $(****)$ holds.
• Hence, for all $\epsilon > 0$ there exists a $\delta = \bar{\delta} > 0$ such that if $d_S(x, p) < \delta$ then $d_U(g(f(x)), g(f(p)) < \epsilon$, so $g \circ f$ is continuous at $p$. $\blacksquare$