Open and Closed Set Crit. for Continuity of Functions on Met. Spaces

Open and Closed Set Criteria for Continuity of 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 a point $p \in S$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $d_S(x, p) < \delta$ then $d_T(f(x), f(p)) < \epsilon$.

We also said that $f$ is continuous on all of $S$ if $f$ is continuous at each point $p \in S$.

We will now look at two very important theorems which tell us when a function $f$ is continuous on all of $S$ or not in terms of open and closed sets.

Theorem 1: Let $(S, d_S)$ and $(T, d_T)$ be metric spaces and let $f : S \to T$. Then $f$ is continuous on all of $S$ if and only if for every open set $V$ in $T$ we have that the inverse image $f^{-1}(V)$ is open in $S$.

In the proof below we will use the "ball-definition" of continuity of $f$. Recall that $f : S \to T$ is continuous at a point $p \in S$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that $f(B_S(p, \delta)) \subseteq B_T(f(p), \epsilon)$.

  • Proof: $\Rightarrow$ Suppose that $f : S \to T$ is continuous on all of $S$ and let $V$ be an open set in $T$. Let $y = f(p) \in V$. Since $V$ is open in $T$ we have that there exists an $\epsilon > 0$ such that the ball (in $T$) centered at $y = f(p)$ with radius $\epsilon$ is fully contained in $V$, that is:
(1)
\begin{align} \quad B_T(y, \epsilon) = B_T(f(p), \epsilon) \subseteq V \end{align}
  • Since $f$ is continuous on all of $S$ we have that there exists a $\delta > 0$ such that:
(2)
\begin{align} \quad f(B_S(p, \delta)) \subseteq B_T(f(p), \epsilon) \end{align}
  • Hence we see that:
(3)
\begin{align} \quad f(B_S(p, \delta)) \subseteq V \\ \quad f^{-1}(f(B_S(p, \delta))) \subseteq f^{-1}(V) \\ \quad B_S(p, \delta) \subseteq f^{-1}(V) \end{align}
  • So for each $y = f(p) \in V$ there exists a ball (in $S$) centered at $p$ with radius $\delta > 0$ that is fully contained in $f^{-1}(V)$. Hence $\mathrm{int}(f^{-1}(V)) = f^{-1}(V)$, so $f^{-1}(V)$ is open in $S$.
  • $\Leftarrow$ Now suppose that for all open sets $V$ in $T$ we have that the inverse images $f^{-1}(V)$ are open sets in $S$. To show that $f$ is continuous on all of $S$ we must show that $f$ is continuous for every $p \in S$.
  • Let $p \in S$ be such that $y = f(p)$. Then for every $\epsilon > 0$, the ball centered at $y = f(p)$ with radius $\epsilon > 0$ is open in $T$, i.e., $B_T(y, \epsilon) = B_T(f(p), \epsilon)$ is open in $T$.
  • Since $B_T(f(p), \epsilon)$ is open in $T$ we have that the inverse image $f^{-1}(B_T(f(p), \epsilon))$ is open in $S$. Now, since $f(p) \in B_T(f(p), \epsilon)$ we have that $p \in f^{-1}(B_T(f(p), \epsilon))$ and since $f^{-1}(B_T(f(p), \epsilon))$ is open in $S$ there exists a $\delta > 0$ such that:
(4)
\begin{align} \quad B_S(p, \delta) \subseteq f^{-1}(B_T(f(p), \epsilon)) \\ \quad f(B_S(p, \delta)) \subseteq B_T(f(p), \epsilon) \end{align}
  • Therefore $f$ is continuous at each $p \in S$, so $f$ is continuous on all of $S$. $\blacksquare$
Theorem 2: Let $(S, d_S)$ and $(T, d_T)$ be metric spaces and let $f : S \to T$. Then $f$ is continuous on all of $S$ if and only if for every closed set $V$ in $T$ we have that the inverse image $f^{-1}(V)$ is closed in $S$.

The only difference between Theorem 1 and Theorem 2 is the replacement of the word "open" with "closed".

  • Proof: $\Rightarrow$ Suppose that $f$ is continuous on all of $S$. Then by Theorem 1, for every open set in $T$ the inverse image is open in $S$.
  • Let $V$ be any closed set in $T$. Then $T \setminus V$ is open in $T$ so $f^{-1}(T \setminus V) = f^{-1}(T) \setminus f^{-1}(V) = S \setminus f^{-1}(V)$ is open in $S$. Hence $f^{-1}(V)$ is closed in $S$.
  • So for every closed set $V$ in $T$ we have that $f^{-1}(V)$ is closed in $S$.
  • $\Leftarrow$ Suppose that for every closed set $V$ in $T$ we have that $f^{-1}(V)$ is closed in $S$. Let $U$ be an open set in $T$. Then $T \setminus U$ is closed in $T$, and $f^{-1}(T \setminus U)$ is closed in $S$. Furthermore:
(5)
\begin{align} \quad f^{-1}(T \setminus U) = f^{-1}(T) \setminus f^{-1}(U) = S \setminus f^{-1}(U) \end{align}
  • So $S \setminus f^{-1}(U)$ is closed and $f^{-1}(U)$ is open. So for every open set $U$ in $T$ we have that $f^{-1}(U)$ is open in $S$, so by Theorem 1, $f$ is continuous on all of $S$. $\blacksquare$
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