1

Let $\langle X,d \rangle$ be a metric space such that there is an uncountable set $Y \subseteq X$ such that any two distinct points of $Y$ have distance greater than one. Show that $X$ is not separable.

I know this is a very similar problem, but that one involves a fixed distance between points in the subset, not a lower bound. Also, I can't make sense of any of the answers.

Edit: To clarify, I don't understand the answers to the linked problem. One claims that if $\alpha$ is this fixed distance, then for every element in the subset there is always an element in the metric space that is at most $\frac{\alpha}{4}$ away. I don't understand why this is forced, or how the solution proceeds from there. Another answer claims the subset is discrete, which I also don't see. The last one seems good, but I think it claims that the metric space being separable implies that this uncountable subset is also separable. Is that true?

tedthered
  • 11
  • 2
    The same approach works: in answer to question you linked we don't actually need distance between points of $Y$ to be exactly $\alpha$, it's enough that it's at least $\alpha$. – mihaild May 02 '23 at 22:45
  • Separability is not hereditary but second-countability is, and for metric spaces, these two properties are equivalent. This is why "the metric space being separable implies that this uncountable subset is also separable". – Anne Bauval May 03 '23 at 04:22
  • 1
    Thank you! That was the last piece of the puzzle I needed – tedthered May 03 '23 at 04:57
  • Does this answer your question? Every subspace of a separable metric space is separable – Anne Bauval May 03 '23 at 05:51
  • "One claims that if α is this fixed distance, then for every element in the subset there is always an element in the metric space that is at most $α/4$ away": not anywhere in the metric space $X$ (otherwise you could as well uselessly take $a(x)=x$) but in a dense subset $A.$ "why this is forced": by definition of "$A$ is dense in $X$". "or how the solution proceeds from there": $a:M\to A$ is injective because if $x\ne x',$ $d(a(x),a(x'))\ge d(x,x')-d(x,a(x))-d(a(x'),x)\geα-α/4-α/4>0.$ Hence every dense subset $A$ of $X$ is uncountable. – Anne Bauval May 03 '23 at 06:16
  • "Another answer claims the subset is discrete, which I also don't see": for every $x\in M,$ ${x}$ is an open subset of $M,$ since it is equal to $B(x,α/2)\cap M.$ In my opinion, the best answer was the one explained in my previous comment. The accepted one (explained by my first comment above or by the proposed duplicate) was flawed, but I corrected it in a comment. – Anne Bauval May 03 '23 at 06:28

0 Answers0