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- Feb 5, 2012

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I have a little trouble understanding a question. Hope you can figure this out for me.

**Question:**Let \((Y,\,d')\) be a complete metric space and \((X,d)\) be a metric space. Let \(A\subset X\) be a dense subset of \(X\) and \(f:A\rightarrow Y\) be a uniformly continuous mapping. Prove that there is a unique continuous extension \(g:X\rightarrow Y\), such that \(g|_{A}=f\) and \(g\) is uniformly continuous on \(X\).

So first of all I want to understand the precise meaning of what a continuous extension is. Is it that we extend the continuous mapping \(f\) to the whole space \(X\) such that \(g(s)=f(s)\) for all \(s\in A\) and call this new mapping \(g\) the continuous extension of \(f\)? Am I correct?