Sequential Criterion for the Continuity of a Function

*This page is intended to be a part of the Real Analysis section of Math Online. Similar topics can also be found in the Calculus section of the site.*

# Sequential Criterion for the Continuity of a Function

We will now look at a very important result known as the Sequential Criterion for the Continuity of a Function.

Theorem 1 (Sequential Criterion for Continuity): Let $f : A \to \mathbb{R}$ be a function. Then $f$ is continuous at the point $c \in A$ if and only if for all sequences $(a_n)$ from $A$ with $\lim_{n \to \infty} a_n = c$ then we have that $\lim_{n \to \infty} f(a_n) = f(c)$. |

**Proof:**$\Rightarrow$ Let $f$ be continuous at the point $c \in A$ and let $(a_n)$ be any arbitrary sequence from $A$ such that $\lim_{n \to \infty} a_n = c$. We want to show that $\lim_{n \to \infty} f(a_n) = f(c)$.

- Let $\epsilon > 0$ be given. Since $f$ is continuous at $c \in A$ then exists a $\delta > 0$ such that if $x \in A$ and $\mid x - c \mid < \delta$ then $\mid f(x) - f(c) \mid < \epsilon$.

- Now since $\lim_{n \to \infty} a_n = c$ then for this $\delta > 0$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - c \mid < \delta$.

- Therefore for $n ≥ N$ we have that since $a_n \in A$ and $\mid a_n - c \mid < \delta$ then $\mid f(a_n) - f(c) \mid < \epsilon$, in other words, for $n ≥ N$, $(f(a_n))$ converges to $f(c)$ so $\lim_{n \to \infty} f(a_n) = f(c)$.

- $\Leftarrow$ We will prove the other direction by contradiction. Suppose that for all sequences $(a_n)$ from $A$ with $\lim_{n \to \infty} a_n = c$ and $\lim_{n \to \infty} f(a_n) = f(c)$ that instead $f$ is NOT continuous at $c \in A$.

- Then there exists an $\epsilon_0 > 0$ such that $\forall \delta > 0$ with $x \in A$ and $\mid x - c \mid < \delta$ then $\mid f(x) - f(c) \mid ≥ \epsilon_0$.

- Now for each $n \in \mathbb{N}$ let $\delta_n = \frac{1}{n}$. Choose $a_n$ such that $\mid x - c \mid < \delta_n = \frac{1}{n}$ and $\mid f(a_n) - f(c) \mid ≥ \epsilon_0$. But then we have a contradiction since $(a_n)$ converges to $c$ however $(f(a_n))$ does not converge to $f(c)$. Thus our assumption that $f$ was not continuous at $c$ was false. $\blacksquare$

# Sequential Criterion for the Discontinuity of a Function

Theorem 2 (Sequential Criterion for Discontinuity): Let $f : A \to \mathbb{R}$ be a function. Then $f$ is said to be discontinuous at $c \in A$ if and only if there exists a sequence $(a_n)$ from $A$ such that $\lim_{n \to \infty} a_n = c$ but the sequence $(f(a_n))$ does not converge to $f(c)$. |

We should note that the theorem above is simply the contrapositive of the sequential criterion for the continuity of a function. Thus if we can find a sequence $(a_n)$ from $A$ that converges to $c \in A$ but is such that the corresponding sequence $(f(a_n))$ does not converge to $f(c)$.