Show that $\mathbb Z$ with the $2$-adic metric is not connected.

Question

I'm teaching myself metric spaces with Sutherland's book, and was struggling to get started with this question. I'm defining connected as not disconnected, where a metric space is disconnected if it can be written as a disjoint union of two nonempty open sets.

Any help would be highly appreciated, thanks!

score 1 · Answer 1 · answered Oct 03 '22 at 10:13

1

The even and the odds form a disjoint open cover.

You can let $A = 2 \mathbb{Z}$ and $B = 2 \mathbb{Z} + 1$. Note that they are both open since the $2$-adic distance between an even and an odd number is always at least $1$.

answered Oct 03 '22 at 10:13

Sofía Marlasca Aparicio

3,238

1

Indeed, $A$ and $B$ are both open balls in the $2$-adic metric. – GEdgar Oct 03 '22 at 10:52

Anne Bauval · Accepted Answer · 2022-10-03T10:55:42.223

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The set of values of the $2$-adic metric $d$ on $\mathbb Z$ is $D:=\{0\}\cup\{2^{-n}\mid n\in\mathbb N\}$. Choose any $r\in(0,1)\setminus D$ and write $\mathbb Z=\{x\in\mathbb Z\mid d(x,0)<r\}\cup\{x\in\mathbb Z\mid d(x,0)>r\}$.

edited Oct 03 '22 at 10:55

answered Oct 03 '22 at 10:22

Anne Bauval

34,650

@GEdgar Isn't there some ambiguity about $\mathbb N$, which (to english-speakers) might denote ${1,2,\dots}$? If so, how should I denote ${0,1,2,\dots}$? – Anne Bauval Oct 03 '22 at 11:53

score 1 · Answer 3 · answered Oct 03 '22 at 12:38

No countable metric space $M$ with more than one point is connected. Simply take $a,b\in M$, with $a\ne b$, and take some number $r\in(0,d(a,b))$ which is not of the form $d(a,c)$ for some $c\in M$ (there must be such a number, since $(0,d(a,b))$ is uncountable). Then $B_r(a)$ is both closed (since it is equal to the closed ball with center $a$ and radius $r$) and open, but it is not empty (it contains $a$) and it is not $M$ (it doesn't contain $b$). So, $M$ is disconnected.

Show that $\mathbb Z$ with the $2$-adic metric is not connected.

3 Answers3