# Uniformly Convergent Sequence on Dense Subset

## Theorem

Let $X$ be a metric space.

Let $Y \subset X$ be dense.

Let $V$ be a Banach space.

Let $\sequence {f_n}$ be a sequence of continuous mappings $f_n : X\to V$.

Let $\sequence {f_n}$ be uniformly convergent on $Y$.

Then $\sequence {f_n}$ is uniformly convergent on $X$.

## Proof

Let $\epsilon>0$.

Let $N \in\N$ be such that $\norm {f_n - f_m} < \epsilon$ for $n, m > N$ on $Y$.

Let $x \in X$.

Then there exists a sequence $\sequence {y_n} \in Y$ with $y_n \to x$.

By continuity:

$\norm {\map {f_n} x - \map {f_m} x} \le \epsilon$

Because $V$ is complete, $\sequence {f_n}$ is uniformly convergent on $X$.

$\blacksquare$
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