Sequential Continuity is Equivalent to Continuity in the Reals

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Theorem

Let $A \subseteq \R$ be a subset of the real numbers.

Let $c \in A$.

Let $f : A \to \R$ be a real function.


Then $f$ is continuous at $c$ if and only if:

for each sequence $\sequence {x_n}$ in $A$ that converges to $c$, the sequence $\sequence {\map f {x_n} }$ converges to $\map f c$.


Corollary

Let $I$ be a real interval.

Let $x \in I$.

Let $f : I \to \R$ be a real function.


Then $f$ is continuous at $x$ if and only if:

for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$ converging to $x$ we have:
$\map f {x_n} \to \map f x$


Proof

Sufficient Condition

It suffices to show that if $f$ is discontinuous at $c$:

there exists a real sequence $\sequence {x_n}$ in $A$ such that $\sequence {x_n}$ converges to $c$ but $\sequence {\map f {x_n} }$ does not converge to $\map f c$.

As $f$ is discontinuous, there exists some $\varepsilon > 0$ such that for all $\delta > 0$:

there exists $x \in A$ with $\size {x - c} < \delta$ such that $\size {\map f x - \map f c} \ge \varepsilon$.

Using this property, we can construct a sequence $\sequence {x_n}$ as follows:

for each $n \in \N$, pick $x_n \in A$ such that $\size {x_n - c} \le \dfrac 1 n$ and $\size {\map f {x_n} - \map f c} \ge \varepsilon$

Note that since:

$\ds \lim_{n \mathop \to \infty} \frac 1 n = 0$

We have by the Squeeze Theorem for Real Sequences:

$\ds \lim_{n \mathop \to \infty} \size {x_n - c} = 0$

so $\sequence {x_n}$ converges to $c$.

However $\sequence {\map f {x_n} }$ cannot converge to $\map f c$ since:

$\size {\map f {x_n} - \map f c} \ge \varepsilon > 0$

for all $n \in \N$.

Therefore, our $\sequence {x_n}$ satisfies our original demand.

$\Box$


Necessary Condition

Let $c \in \R$.

Let $\sequence {x_n}$ be a sequence in $A$ that converges to $c$.

Let $\varepsilon \in \R_{> 0}$.

Since $f$ is continuous at $c$, there exists $\delta > 0$ such that:

for all $x \in A$ with $\size {x - c} < \delta$, we have $\size {\map f x - \map f c} < \varepsilon$.

Additionally, since $\sequence {x_n}$ converges to $c$, there exists $N \in \N$ such that:

for all $n > N$ we have $\size {x_n - c} < \delta$.

Therefore, since $x_n \in A$ for all $n \in \N$:

for all $n > N$, we have $\size {\map f {x_n} - \map f c} < \varepsilon$.

Since $\varepsilon$ was arbitrary, we have:

$\sequence {\map f {x_n} }$ converges to $\map f c$.

$\blacksquare$


Sources

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