Theorem. Let $E$ be a dense subset of a metric space $X$, and let $f$ be a uniformly continuous real function defined on $E$. Prove that $f$ has a continuous extension from $E$ to $X$.

To prove this theorem we'll use this lemma whics was proved here :

Suppose $f$ is a uniformly continuous mapping of a metric space $X$ into a metric space $Y$ and prove that $\{f(x_n)\}$ is a Cauchy sequence in $Y$ for every Cauchy sequence $\{x_n\}$ in $X$.

Proof: Let $p\in X$ then exists Cauchy sequence $\{p_n\}$ in $E$ such that $p_n\to p$. Since $f$ is uniformly continuous then $f(p_n)$ is also Cauchy sequence in $\mathbb{R}^1$ which is complete. Hence $f(p_n)\to g(p)$. Thus to every $p\in X$ we associate $g(p)\in \mathbb{R}^1$.

We'll show that $g$ is uniformly continuous on $X$. It means that for any $\varepsilon>0$ $\exists \delta>0$ such that for any $x,y\in X$ with $d(x,y)<\delta$ implies that $|g(x)-g(y)|<\varepsilon.$

Also $g(x)=\lim\limits_{n\to \infty}f(x_n)$ and $g(y)=\lim\limits_{n\to \infty}f(y_n)$ where $x_n,y_n$ are Cauchy sequences on $E$ such that $x_n\to x$ and $y_n\to x$. It's easy to check that $d(x_n,y_n)\to d(x,y)$ as $n\to\infty$.

Since $f$ uniformly continuous on $E$ then for given $\varepsilon>0$ $\exists \delta>0$ such that for all $x_n,y_n\in X$ with $d(x_n,y_n)<\delta$ implies that $|f(x_n)-f(y_n)|<\frac{\varepsilon}{2}.$

Making limit conversion we get that $d(x,y) \color{blue}{\leqslant} \delta$. But symbol $\leqslant$ little worry me because statement holds if distance $<\delta$.

Can anyone show how to correct this moment.

  • $\begingroup$ @PhoemueX, Sorry but where I must to use it? $\endgroup$ – ZFR Sep 28 '15 at 17:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.