Every open set is a continuous image of closed set

Question

I'm trying to prove that every open set is a continuous image of a closed set, after failing to provide a counter example. That is:

For every $U\subset\mathbb{R}^m$, there exists $C\subset\mathbb{R}^n$ and $f:C\to U $ a continuous function such that $f(C)=U$.

I tried solving this with $n=m$ to get a rough idea for a more general case. The best I could come up with was taking $C=U \cup \partial U$, closed set, and $f(x)=x$ if $x\in U$, however, I couldn't find a way to define $f:C \to U$ that is continuous.

I'd be glad for a hint in the right direction.

Answer 1

It's known that every open set arises as a union of countably many closed balls:

$$U = \bigcup_{k=1}^\infty B_k$$

Let $C$ also be a union of countably many closed balls, but make them disjoint:

$$C = \bigcup_{k=1}^\infty C_k \text{, where } C_k = \{x \in \mathbb{R}^m : \|x - (3k,0,\dots,0)\| \le 1\}$$

And let $f$ map $C_k$ to $B_k$ in an obvious way. Then $f$ is continuous and $f(C) = U$.