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

Nov 2012
41
0
Taiwan
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?
 
Aug 2006
22,432
8,612
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 \(\displaystyle f\) .
 
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