2

If $X$ are a infinte countable set, define a metric in $X$ such that no point are isolated (i.e. for all $a\in X, \epsilon>0$ there exists $x\in B(a;\epsilon)\setminus\{a\})$.

I tried this but I don't know if it's correct: once $X$ is infinite countable then it can be put in 1-1 correspondence with $\mathbb{Q}$. Then we can write $X=(x_r)_{r\in\mathbb{Q}}$ and define distance in $X$ by $d(x_r,x_s) = |r-s|$. Once $\mathbb{Q}$ is dense in $\mathbb{R}$ no point will be isolated. Is this approach correct?

I just don't feel it, once the integers could fit in this with this euclidean norm, and still will not be a discrete metric space... Thanks in advance.

Bernard
  • 179,256
AnalyticHarmony
  • 2,085
  • 3
    It doesn't say that every infinite set will have no isolated points. Just that $X$ itself doesn't have any. You're right about the choice of metric, and to some extent, it's the only way to do this (or, any countable metric space without isolated points is homeomorphic to the rational numbers). – Asaf Karagila Sep 08 '18 at 18:32
  • Not every infinite, but every infinite countable, that's right? – AnalyticHarmony Sep 08 '18 at 18:34
  • 4
    Yes, every countably infinite metric space has an isolated points or it is homeomorphic to the rationals. – Asaf Karagila Sep 08 '18 at 18:35

1 Answers1

6

There is no "choice" but to go to $\mathbb{Q}$: every countable metric space without isolated points is homeomorphic to $\mathbb{Q}$. You can find proofs here.

So your idea works: take a bijection $f$ between the countable set $X$ and $\mathbb{Q}$ (which can be done as their both in bijective correspondence with, say $\mathbb{N}$) and define $d(x,x') = |f(x) - f(x')|$ where the right hand side has the absolute value as defined on the reals. Then the countable set has a metric that makes it isometric to $\mathbb{Q}$ in the usual distance (by definition basically), so these spaces are now homeomorphic so $X$ has no isolated points just like $\mathbb{Q}$ has none.

Henno Brandsma
  • 250,824