Sequential Criterion for the Continuity of a Function on Metric Spaces

# Sequential Criterion for the Continuity of a Function 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 $d_T(f(x), f(p)) < \epsilon$.

Equivalently, we said that $f$ is continuous at the point $p \in S$ 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}

We will look at an important theorem which says that $f$ is continuous at a point $p \in S$ if and only if for every sequence $(x_n)_{n=1}^{\infty}$ in $S$ that converges to $p$ we have that the sequence $(f(x))_{n=1}^{\infty}$ in $T$ converges to $f(p)$, i.e.,

(2)
\begin{align} \quad \lim_{n \to \infty} f(x_n) = f \left ( \lim_{n \to \infty} x_n \right ) \end{align}
 Theorem 1 (The Sequential Criterion for the Continuity of a Function on Metric Spaces): Let $(S, d_S)$ and $(T, d_T)$ be metric spaces and let $f : S \to T$. Then $f$ is continuous at the point $p \in S$ if and only if for all sequences $(x_n)_{n=1}^{\infty}$ in $S$ that converge to $p$ we have that $(f(x_n))_{n=1}^{\infty}$ in $T$ converges to $f(p)$.
• Proof: $\Rightarrow$ Suppose that $f$ is continuous at the point $p \in S$. Then for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $d_S(x, p) < \delta$ $(*)$ then:
(3)
\begin{align} \quad d_T(f(x), f(p)) < \epsilon \quad (**) \end{align}
• Let $(x_n)_{n=1}^{\infty}$ be any sequence in $S$ that converges to $p$. Since $(x_n)_{n=1}^{\infty}$ converges to $p$ we have that $\lim_{n \to \infty} x_n = p$, so $\lim_{n \to \infty} d_S(x_n, p) = 0$. So, for $\delta$ ($> 0$) there exists an $N(\delta) \in \mathbb{N}$ such that if $n \geq N(\delta)$ then:
(4)
\begin{align} \quad d_S(x_n, p) < \delta \quad (***) \end{align}
• So, for all $n \geq N(\delta)$ we have that $(***)$ holds, and because of $(***)$ we have by the continuity of $f$ at $p$ that $(*)$ then implies that $d_T(f(x_n), f(p)) < \epsilon$.
• So, if we let $N = N(\delta)$, then for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $d_T(f(x_n), f(p)) < \epsilon$, and so $\lim_{n \to \infty} d_T(f(x_n), f(p)) = 0$ which implies that:
(5)
\begin{align} \quad \lim_{n \to \infty} f(x_n) = f(p) \end{align}
• Therefore the sequence $(f(x_n))_{n=1}^{\infty}$ converges to $f(p)$.
• $\Leftarrow$ Suppose that all sequences $(x_n)_{n=1}^{\infty}$ from $S$ that converge to $p$ are such that the corresponding sequence $(f(x_n))_{n=1}^{\infty}$ from $T$ converge to $f(p)$ and suppose that $f$ is not continuous at $p$.
• Since $f$ is not continuous at $p$ there exists an $\epsilon_0 > 0$ such that for all $\delta > 0$ we have that if $d_S(x, p) < \delta$ $(****)$ then:
(6)
\begin{align} \quad d_T(f(x), f(p)) \geq \epsilon_0 \end{align}
• For each $n \in \mathbb{N}$, construct a sequence $(x_n)_{n=1}^{\infty}$ by taking $x_n \in S$ such that $d_S(x_n, p) < \frac{1}{n}$. Then we see that:
(7)
\begin{align} \quad 0 \leq \lim_{n \to \infty} d_S(x_n, p) \leq \lim_{n \to \infty} \frac{1}{n} = 0 \end{align}
• Hence $\lim_{n \to \infty} d_S(x_n, p) = 0$ so $\lim_{n \to \infty} x_n = p$, i.e., the sequence $(x_n)_{n=1}^{\infty}$ converges to $p$. So, for any $\delta > 0$ there exists an $N(\delta) \in \mathbb{N}$ such that $\frac{1}{N(\delta)} < \delta$. So, for all $n \geq N(\delta)$ we have that $(****)$ holds and so:
(8)
\begin{align} \quad d_T(f(x_n), f(p)) \geq \epsilon_0 > 0 \end{align}
• But then $\lim_{n \to \infty} d_T(f(x_n), f(p)) \neq 0$ so $\lim_{n \to \infty} f(x_n) \neq f(p)$ and so the sequence $(f(x_n))_{n=1}^{\infty}$ does not converge to $f(p)$ which is a contradiction. Therefore our assumption that $f$ was not continuous at $p$ was false. Hence, if every sequence $(x_n)_{n=1}^{\infty}$ that converges to $p$ is such that $(f(x_n))_{n=1}^{\infty}$ converges to $f(p)$ then $f$ is continuous at $p$. $\blacksquare$