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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.

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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.

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  • $\begingroup$ Thank you. Topological discreteness implies it has no limit points, right? In that case, what is the mistake in OP's "proof" ? $\endgroup$ – evaristegd Jul 18 at 2:25
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    $\begingroup$ @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 $\mathbb{R}^k$. $\endgroup$ – Alex Kruckman Jul 18 at 2:27
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    $\begingroup$ 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 $\endgroup$ – Alex Kruckman Jul 18 at 2:38
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    $\begingroup$ 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. $\endgroup$ – Bar Alon Jul 18 at 12:00
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    $\begingroup$ @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. $\endgroup$ – Alex Kruckman Jul 18 at 13:08
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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.

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  • $\begingroup$ Is $\{a_k \}$ an infinite set? $\endgroup$ – evaristegd Jul 18 at 3:48
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    $\begingroup$ Yes: $l_p$ spaces are infinite dimensional. $\endgroup$ – herb steinberg Jul 18 at 3:57
  • $\begingroup$ I see. I started reading again about it, and, indeed, they are spaces of infinite sequences. Thank you $\endgroup$ – evaristegd Jul 18 at 5:03
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No, see $X=(0,1)$ and $A = \{\frac{1}{n}: n \in \mathbb{N}, n \ge2 \}$ e.g. All subsets of $X$ are bounded.

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

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