Assume $f:C \rightarrow \mathbb{R}$ is continuous is also uniformly continuous. Show $C$ is closed.

Question

Let $C\subset \mathbb{R}^n$ with the property that any function $f:C \rightarrow \mathbb{R}$ that is continuous has to be uniformly continuous. Show $C$ must be closed.

Here is my thought process: Suppose, by way of contradiction, C is not closed. Then there exists a sequence $(x_k)\subset C $ such that $(x_k)\rightarrow x*\notin C$.

At this point I am stuck because I need to find a function $f$ that is continuous but not uniformly continuous

Answer 1

Hint: Suppose that there exists a sequence $(x_n)$ such that $x_n\in C, lim_nx_n=x$ and $x$ is not an element of $C$. Consider $f(x)={1\over\|y-x\|}$ defined on $C$, it is continuous but not uniformly continuous.