# Extension of uniformly continuous function

I want to prove this:

$f\in C((a,b))$ uniformly continuous. Then there exists $\tilde{f}\in C([a,b])$ extension of $f$.

I took $x_n\rightarrow a$ and defined $\tilde{f}(a)=\mathrm{lim}\;f(x_n)$. I saw that this is a good definition, the only thing that I'm not able to prove is that $\tilde{f}$ is continuous at $a$ (or $b$). Could you help me please?

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Note that $f$ continuous is equivalent to $f(lim x_n)=lim f(x_n)$ holds for every convergent sequence. – azarel Jan 19 '12 at 0:42

You need to justify that $\lim f(x_n)$ exists. It does exist, since uniformly continuous functions map Cauchy sequences to Cauchy sequences (you might need to prove this; but it follows directly from the definitions). Now show that for any sequence $y_n$ converging to $a$, that $f(y_n)$ converges to $\tilde f(a)$. This will show that $\tilde f$ is continuous at $a$. Of course, you have to deal with $b$ too...
You need to show the limit exists. Do this by verifying that $f(x_n)$ is a Cauchy sequence. Use the uniform continuity to do this.