Sequential Criterion for the Limit of a Function on Metric Spaces

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Recall from the Limits of Functions on Metric Spaces page that if $$(S, d_S)$$ and $$(T, d_T)$$ are both metric spaces, $A \subseteq S$ , $f : A \to T$ , and $p \in S$ is an accumulation point of $$A$$ . Then the following notation:

(1)

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

means that for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $x \in D(f) \setminus \{ p \}$ and $d_S(x, p) < \delta$ then $d_T(f(x), b) < \epsilon$ .

Equivalently we said that $\lim_{x \to p} f(x) = b$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $x \in (D(f) \setminus \{ p \}) \cap B_S(p, \delta)$ then $f(x) \in B_T(b, \epsilon)$ .

We will now look at an important theorem which says that $\lim_{x \to p} f(x) = b$ if and only if for every sequence $(x_n)_{n=1}^{\infty}$ in $D(f) \setminus \{ p \} = A \setminus \{ p \}$ that converge to $$p$$ we have that the sequences $(f(x_n))_{n=1}^{\infty}$ converge to $$b$$ .

Theorem 1: Let

$(S, d_S)$

and

$(T, d_T)$

be metric spaces,

A \subseteq S

f : A \to T

p \in S

be an accumulation point of

$A$

, and

b \in T

. Then

\lim_{x \to p} f(x) = b

if and only if for all sequences

(x_n)_{n=1}^{\infty}

A \setminus \{ p \}

(sequences of elements from

$A$

not containing

$p$

) that converge to

$p$

$S$

we have that the sequences

(f(x_n))_{n=1}^{\infty}

converge to

$b$

$T$

Proof: $\Rightarrow$ Suppose that $\lim_{x \to p} f(x) = b$ and let $(x_n)_{n=1}^{\infty}$ be a sequence of elements from $A \setminus \{ p \}$ such that $(x_n)_{n=1}^{\infty}$ converges to $$p$$ , i.e., $\lim_{n \to \infty} x_n = p$ .

Now since $\lim_{x \to p} f(x) = b$ then for all $\epsilon > 0$ there exists a $\delta > 0$ such that for all $x \in A \setminus \{ p \}$ such that $d_S(x, p) < \delta$ $$(*)$$ we have that $d_T(f(x), b) < \epsilon$ .

Since $\lim_{n \to \infty} x_n = p$ we have that for $\delta > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $d_S(x_n, p) < \delta$ . So whenever $n \geq N$ we have by $$(*)$$ that then $d_T(f(x_n), b) < \epsilon$ . Therefore $\lim_{n \to \infty} f(x_n) = b$ .

$\Leftarrow$ Now suppose that for all sequences $(x_n)_{n=1}^{\infty}$ contained in $A \setminus \{ p \}$ that converge to $$p$$ we have that $\lim_{n \to \infty} f(x_n) = b$ .

Assume instead that $\lim_{x \to p} f(x) \neq b$ . Then there exists an $\epsilon_0 > 0$ such that there exists an $x_n \in D(f) \cap B \left (p, \frac{1}{n} \right ) \setminus \{ p \}$ where $d_T(f(x_n),b) \geq \epsilon_0$ . Consider the sequence $(x_n)_{n=1}^{\infty}$ . Then $d_S(x_n, p) < \frac{1}{n}$ for all $n \in \mathbb{N}$ and so $\lim_{n \to \infty} d_S(x_n, p) \leq \lim_{n \to \infty} \frac{1}{n} = 0$ , so $\lim_{n \to \infty} x_n = p$ , i.e., $(x_n)_{n=1}^{\infty}$ converges to $$p$$ . However $d_T(f(x_n), b) \geq \epsilon_0 > 0$ for all $n \in \mathbb{N}$ so $\lim_{n \to \infty} f(x_n) \neq b$ which is a contradiction.

Hence, the assumption that $\lim_{x \to p} f(x) \neq b$ was false. Therefore $\lim_{x \to p} f(x) = b$ . $\blacksquare$

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Sequential Criterion for the Limit of a Function on Metric Spaces