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Any hints on how I should go about this? Thanks.

r123454321
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2 Answers2

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The statement is false if $|X|=1$, so countable here is probably supposed to be understood as countably infinite.

HINT: Let $X$ be the space, and let $x,y\in X$ with $x\ne y$. Let $r=d(x,y)>0$. Use the fact that $X$ is countable to show that there must be an $s\in(0,r)$ such that

$$\{z\in X:d(x,z)=s\}=\varnothing\;.$$

If $B(x,s)=\{z\in X:d(x,z)<s\}$, show that $B(x,s)$ and $X\setminus B(x,s)$ are non-empty open sets in $X$, and conclude that $X$ is not connected.

Brian M. Scott
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  • Why not (finite and not $|X|=1$ and not empty)? – BCLC Oct 18 '18 at 00:57
  • @BCLC: (Apparently I never saw this one.) The result is true for every countable metric space of cardinality greater than $1$, but I was sticking to the statement of the result in the question. Unfortunately, countable has two meanings in common use: properly it means of cardinality at most $\omega$ (or $\aleph_0$, if you prefer), but some use it to mean or cardinality exactly $\omega$, and for the result to be true as stated, the word must have the narrower sense. – Brian M. Scott Dec 17 '23 at 22:01
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Do you mean totally disconnected? In that case, take two points $x, y$ with $d(x, y) = a$. Then, since there are only countably many points in the space, and there are uncountably many real numbers less than $a$, there must be a distance $b < a$ such that there are no points exactly at a distance $b$ from $x$.

That means that the ball around $x$ with radius $b$ is both open and closed, and so is its complement. Thus you have partitioned the space into two disjoint clopen sets, one containing each point. Since the points were arbitrary, this means that the space is totally disconnected.

Arthur
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  • Thank you! One question though: why exactly is the ball around x with radius b both open and closed? Singleton sets are always closed, I get that, but why is it open as well? – r123454321 Sep 25 '13 at 08:37
  • What I've shown is not so much that the singletons are open, but more that no two points are in the same connected component. When I say "No points are at a distance $b$ from $x$", I mean exactly $b$. There might be points $z$ such that $d(x, z) < b$. The main point, however, is that the $b$-ball centered at $x$ is both open and closed at the same time, since ${z | d(x, z) < b}$ and ${z | d(x, z) \leq b}$ consist of exactly the same points. Thus no point outside the ball is in the same connected component as $x$. Since $y$ is arbitrary, no point but $x$ is in $x$'s connected component. – Arthur Sep 25 '13 at 08:50
  • The points of $\mathbb Q$ are dense in the real numbers.So no, you can never find such a $b$. – alexis Sep 25 '13 at 11:36
  • @alexis Yes, I can find such a $b$. Given a rational point $p$, the balls of radius (for instance) $\frac{\pi}{n}$ centered on $p$ are clopen, since the boundary is empty. For any other rational number $q$, choose $n$ so that the ball around $x$ is small enough not to contain $q$ (then the corresponding $\frac{\pi}{n}$ is my $b$). Thus you have a clopen neighbourhood around $x$ not containing $y$, proving that they are not part of the same connected component. – Arthur Sep 25 '13 at 11:42
  • I of course mean $p$ and $q$ in the last sentence in my previous comment, not $x$ and $y$. – Arthur Sep 25 '13 at 11:51
  • Oh sorry, I (mis)read that to say there would be no points at distance $< b$, but you meant exactly $b$. Makes sense now. (To show it's closed, you need to point out that the ball from $q$ to the separating point is disjoint and also open). – alexis Sep 25 '13 at 11:53