Counter Example: Subset of $\mathbb{R}$ such that every bounded continuous function extended to continuous function, then the subset is closed

Question

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.

Answer 1

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 $p$.

Answer 2

If $F$ were not closed, we have at least a sequence ${x_n}$ in $F$ converging to some $x\in\mathbb{R}-F$. Then $f(t)=\sin(1/(t-x))$ is bounded and continuous on $F$, but it cannot be extended continuously at $t=x$.