In my analysis and topology course, the following statement was brought up, with the professor stating there was a counterclaim.
For $E \subset \mathbb{R}$ such that every bounded continuous function from $E$ to ${\mathbb{R}}$ can be extended to a continuous function from ${\mathbb{R}}$ to ${\mathbb{R}}$, then $E$ is a closed subset of ${\mathbb{R}}.$
I spent a few hours and feel there are no counterclaims to this statement, but in lecture my professor said there was one. What is a counterexample to such a statement? The professor is infamous for giving us 'trick' questions however, so I just want to know if this was one of them.
Best Answer
There is no counterexample: the statement is true.
Let $E$ be a non-closed subset of $\Bbb R$; then there is a $p\in(\operatorname{cl}E)\setminus E$, and there is a sequence $\langle x_n:n\in\Bbb N\rangle$ of distinct points of $E$ that converges to $p$ in $\Bbb R$. Let $D=\{x_n:n\in\Bbb N\}$; $D$ is a relatively closed, discrete subset of $E$, so there are real numbers $r_n>0$ such that $(\operatorname{cl}B(x_m,r_m))\cap\operatorname{cl}B(x_n,r_n)=\varnothing$ whenever $m,n\in\Bbb N$ and $m\ne n$. For each $n\in\Bbb N$ there is a continuous function $f_n:E\to[0,1]$ such that $f_n(x_n)=1$, and $f_n(x)=0$ whenever $|x-x_n|\ge r_n$. Let
$$f:E\to[-1,1]:x\mapsto\sum_{n\in\Bbb N}(-1)^nf_n(x)\,;$$
then $f$ is continuous and bounded, but it cannot be extended continuously to $p$.