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

#### 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?

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

g(x)=f(x) if x in E; g(x)=limnf(xn) 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?

#### Plato

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 $\displaystyle f$ .

2 people