Uniformly Convergent Sequence of Continuous Functions Converges to Continuous Function

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Theorem

Let $S \subseteq \R$.

Let $x \in S$.

Let $\sequence {f_n}$ be a sequence of real functions $S \to \R$ converging uniformly to $f: S \to \R$.

Let $f_n$ be continuous at $x$ for all $n \in \N$.


Then $f$ is continuous at $x$.


Corollary

Let $S \subseteq \R$.

Let $\sequence {f_n}$ be a sequence of real functions $S \to \R$ converging uniformly to $f : S \to \R$.

Let $f_n$ be continuous for all $n \in \N$.


Then $f$ is continuous.


Proof

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

Since $f_n \to f$ uniformly, there exists some $N \in \N$ such that:

$\size {\map {f_n} x - \map f x} < \dfrac \epsilon 3$

for all $x \in S$ and $n \ge N$.

Since $f_N$ is continuous at $x$, there exists some $\delta > 0$ such that:

for all $y$ with $\size {x - y} < \delta$, we have $\size {\map {f_N} x - \map {f_N} y} < \dfrac \epsilon 3$

Then for $y$ with $\size {x - y} < \delta$ we have:

\(\ds \size {\map f x - \map f y}\) \(=\) \(\ds \size {\map f x - \map {f_N} x + \map {f_N} x - \map {f_N} y + \map {f_N} y - \map f y}\)
\(\ds \) \(=\) \(\ds \size {\paren {\map f x - \map {f_N} x} + \paren {\map {f_N} x - \map {f_N} y} + \paren {\map {f_N} y - \map f y} }\)
\(\ds \) \(\le\) \(\ds \size {\map f x - \map {f_N} x} + \size {\map {f_N} x - \map {f_N} y} + \size {\map {f_N} y - \map f y}\) Triangle Inequality for Real Numbers
\(\ds \) \(<\) \(\ds 3 \times \frac \epsilon 3\)
\(\ds \) \(=\) \(\ds \epsilon\)

Since $\epsilon$ was arbitrary, $f$ is continuous at $x$.

$\blacksquare$


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