Given an infinite set $X$ with no limit points, is $X$ unbounded? (In an arbitrary metric space)
I only know how to do this in $\mathbb{R}^k$.
Since $X$ has no limit points, $X$ is closed. An infinite set with no limit point also cannot be compact, because we can choose a ball around each point of $X$ with no other points in it, and such cover has no finite subcover.
Then in $\mathbb{R}^k$ we have $\text{closed} \land \text{bounded} \implies \text{compact}$ (Heine-Borel theorem), so by simple logic knowing it is not compact we have: $$\neg\text{compact} \implies \left( \neg\text{closed} \lor \neg\text{bounded} \right),$$ and since it is closed by definition, it must be unbounded.
The question is, how can I show this in an arbitrary metric space, and not just in $\mathbb{R}^k$? Just to clarify, I'm asking this because of my own curiosity, I only stumbled upon this in relation to trying to solve something else (in $\mathbb{R}^k$) and was wondering if it works in general.
Best Answer
Consider any infinite set $X$ equipped with the discrete metric $$d(x,y) = \begin{cases} 0 & \text{if }x = y\\ 1 &\text{otherwise}.\end{cases}$$ Then $X$ is infinite, topologically discrete, and bounded.
Here's another example that you might find less pathological. Let $(a_n)_{n\in \mathbb{N}}$ be an increasing sequence of rational numbers in the interval $(0,\sqrt{2})$, converging to $\sqrt{2}$ from below. Then $X = \{a_n\mid n\in \mathbb{N}\}$ is an infinite, topologically discrete, bounded subset of $\mathbb{Q}$ with its usual metric.