# Sequential Continuity is Equivalent to Continuity in the Reals/Sufficient Condition

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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 if $f$ is continuous at $c$:

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

## Proof

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.

$\blacksquare$