Let f : R → R be a (not necessarily continuous) function and suppose that f(U) is closed for every open set U ⊂ R. Then the image f(R) is discrete

Question

The question is to prove or disprove this statement but I cannot make any headway on it. I assume that the statement is true but I have tried proving by showing there are no limit points in $F(\mathbb{R})$ but cannot seem to do so. I think the proof may have something to do singletons in $F(\mathbb{R})$. I am also not sure if I can presume that $\mid f(U)\mid \le \mid{U\mid}$ in this case

Note: In this context the discrete set $S$ is one where $\forall x \in S, \;\exists\epsilon \; [(x-\epsilon),(x+\epsilon)]\cap S=\{x\}$

https://math.stackexchange.com/questions/186427/function-whose-image-of-every-open-interval-is-infty-infty and various duplicates... — Moishe Kohan, Oct 07 '24 at 17:13
It's much easier than that one though. For example, it's easy to construct $f$ such that $f(\mathbb R) = {0} \cup {1/n: n \in \mathbb N}$ and $0 \in f(U)$ for every nonempty open $U$. — Robert Israel, Oct 07 '24 at 17:21

Malady · Answer 1 · 2024-10-07T18:49:17.243

This is not the case. Consider $f:\mathbb{R}\rightarrow \mathbb{R}$ defined piecewise by: $$f(x)=\left\{ \begin{array}{ll} 0 && x\in \mathbb{Q} \\ \frac{1}{\lceil |x|\rceil} && x\notin \mathbb{Q}\end{array} \right.$$

This is in agreement with @Robert Israel’s comment.

The image of $f$ is exactly $K=\{\frac{1}{n}\mid n\in \mathbb{N}\}\cup \{0\}$ which is not discrete (since $0$ is a limit point).

If a subset $A \subset K$ contains $0$, then $A$ is closed in $\mathbb{R}$, since $0$ is the only limit point of $K$.

Since the preimage of $0$ is dense, the image of each open set is a subset of $K$ which contains $0$, and is thus closed.

Let f : R → R be a (not necessarily continuous) function and suppose that f(U) is closed for every open set U ⊂ R. Then the image f(R) is discrete

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