# Is every compact discrete metric space singleton?

So I was working on a proof that every compact discrete metric space is finite, and my result was that not only is every one finite, but every one is just a singleton set. I just want to confirm whether this is true or not.

In my proof I showed that a set K in (M, discrete) is sequentially compact only if every sequence in K is a constant sequence in the form (x, x, x, . . .).

EDIT: The above is incorrect. I didn't consider sequences with constant tails in my proof, and if I did it would have led me to the conclusion that every compact discrete metric space is finite, but not necessarily singleton.

EDIT 2: It turns out that even when considering sequences with constant tails, singleton sets are still the only sets that only contain sequences with constant tails. So something else in my proof is wrong.

The only sets that contain only constant sequences are singleton sets, so therefore, every compact discrete metric space is singleton.

• If $M = \{x,y\}$ can't you form two constant sequences? Apr 17 '20 at 17:16
• The "In my proof I showed that" part is wrong. Apr 17 '20 at 17:17
• I believe what you should've shown is that such a sequence would be eventually constant, not just constant. Apr 17 '20 at 17:17
• @lisyarus It is still not the case thate every sequence is eventually constant. Only the convergent ones. Apr 17 '20 at 17:18
• @HagenvonEitzen Oh, I somehow missed that convergence is not in the prerequisites. You are right, of course. Apr 17 '20 at 17:20

No. Let $$X$$ be any finite set, and define

$$d(x,y)=\begin{cases} 1,&\text{if }x\ne y\\ 0,&\text{if }x=y\;; \end{cases}$$

then $$d$$ is a metric on $$X$$ that generates the discrete topology, and $$X$$ is trivially compact in that topology, since every finite space is compact.

A space $$X$$ is sequentially compact if every sequence $$s_i \in X$$ has a convergent subsequence, i.e., for some increasing sequence of natural numbers $$i_j$$, the sequence defined by $$r_j = s_{i_j}$$ is convergent. I think in your proof you are mistakenly requiring the sequence $$i_j$$ to contain all sufficiently large $$i$$, i.e., you are requiring the sequence $$r_i$$ to be obtained from $$s_i$$ by omitting a finite number of elements. That is too restrictive.

For example if $$X = \{-1, +1\}$$. The sequence defined by $$s_i = (-1)^{i}$$ has a convergent subsequence given by $$r_j = s_{2j}$$, which omits infinitely many elements of $$s_i$$.

In general, any sequence of elements $$s_i$$ of a finite set $$X$$ must contain infinitely many occurrences of some element $$x$$ of $$X$$. If you form $$r_j$$ by omitting all $$i$$ such that $$s_i \neq x$$, then you will get a convergent subsequence of $$s_j$$. So any finite metric space is sequentially compact.

Aside: this problem is a bit easier if you use the characterisation of compactness in terms of open covers (if you have come across that).