# Is an infinite set with no limit point unbounded in an arbitrary metric space?

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.

## 3 Answers

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.

• Thank you. Topological discreteness implies it has no limit points, right? In that case, what is the mistake in OP's "proof" ? – evaristegd Jul 18 at 2:25
• @evaristegd Yes, topological discreteness is equivalent to having no limit points. The "mistake" (it's not a mistake, because the OP never claimed to have a proof that works in a general metric space) is the use of the Heine-Borel characterization of compact sets, which is only valid in special metric spaces like $\mathbb{R}^k$. – Alex Kruckman Jul 18 at 2:27
• You can certainly equip that set with the discrete metric, but then the Heine-Borel theorem doesn't apply to it. When I said "is only valid in special metric spaces like $\mathbb{R}^k$", I meant $\mathbb{R}^k$ with its standard metric. @evaristegd – Alex Kruckman Jul 18 at 2:38
• It's worth noting that the discrete metric isn't induced by a norm, so the notion of "Bounded set" may have a slightly different meaning than what the OP had in mind. – Bar Alon Jul 18 at 12:00
• @BarAlon Hmm, possibly. But the OP only mentions metric spaces in the question (no norms), and "bounded" has a standard meaning in metric spaces (contained in some ball of finite radius), so I think it's pretty likely that this is the notion they're asking about. – Alex Kruckman Jul 18 at 13:08

Simple example: $$l_p$$ space with $$1\le p\lt \infty$$. Sequence $$a_k$$ with the $$k^{th}$$ coordinate $$=1$$ and all others $$=0$$. $$||a_k||_p=1$$, but {$$a_k$$} has no limit point.

• Is $\{a_k \}$ an infinite set? – evaristegd Jul 18 at 3:48
• Yes: $l_p$ spaces are infinite dimensional. – herb steinberg Jul 18 at 3:57
• I see. I started reading again about it, and, indeed, they are spaces of infinite sequences. Thank you – evaristegd Jul 18 at 5:03

No, see $$X=(0,1)$$ and $$A = \{\frac{1}{n}: n \in \mathbb{N}, n \ge2 \}$$ e.g. All subsets of $$X$$ are bounded.

• @evaristegd that’s the only exception. I edited. – Henno Brandsma Jul 18 at 5:08