Suppose for a metric space every infinite subset has a limit point. What should be my strategy to construct a countable dense subset there? Additionally, how do I intuitively guess that with such a property the metric space has a countable dense subset.
Answer
HINT: Let $\langle X,d\rangle$ be a metric space in which every infinite subset has a limit point. For each $n\in\Bbb N$ let $D_n$ be a maximal subset of $X$ such that $d(x,y)\ge 2^{-n}$ whenever $x,y\in D_n$ with $x\ne y$. (You can use Zorn’s lemma to show that $D_n$ exists.)
- Show that each $D_n$ is finite.
- Show that $\bigcup_{n\in\Bbb N}D_n$ is dense in $X$.
I’m not sure how you’d guess this result. The hypothesis on $X$ does tell you that $X$ does not contain an infinite closed discrete subset, which in some sense says that the points of $X$ aren’t spread out too much, but that property alone isn’t enough to ensure that $X$ is separable: the result really does use the fact that $X$ is a metric space as well.
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