Find a function that is defined and continuous on the closure of a set E.

If f is defined and uniformly continuous on E, can somebody show me that there's a function g defined and continuous on the closure of E such that g = f on E?

Since, for every limit point x of E, there is a sequence {x_{n}} in E such that limx_{n} = x, I define g as below.

g(x)=f(x) if x in E; g(x)=lim_{n}f(x_{n}) if x belongs to (the closure of E)\E.

Would it be a right start? If so, then how do I prove that g is defined and continuous on (the closure of E)\E by using the fact that f is uniformly continuous?

Re: Find a function that is defined and continuous on the closure of a set E.

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**Rita** If f is defined and uniformly continuous on E, can somebody show me that there's a function g defined and continuous on the closure of E such that g = f on E?

You should have a *lemma* that says that if $\displaystyle (x_n)$ is a Cauchy sequence in $\displaystyle E$ then $\displaystyle f(x_n)$ is a Cauchy sequence. And that last sequence is converges to some $\displaystyle b\in\overline{E}$.

More over if both $\displaystyle (x_n)~\&~(y_n)$ are Cauchy sequences in $\displaystyle E$ and $\displaystyle (x_n)\to b~\&~(y_n)\to b$ then $\displaystyle f(x_n)\to f(b)~\&~f(y_n)\to f(b)$. Now that gives you a *natural* way to define the extension of $\displaystyle f$ .