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Let $\sim$ be equivalence relation on $\mathbb{R}$ such that:

$x \sim y \iff (x = y$ or $x, y \in \mathbb{Z})$

Show that $X = \mathbb{R}/_{\sim}$ does not satisfy axiom $1A\mathbb{N}$.

Hint: consider $[0]$.

So neighborhoods of $[0]$ are neighborhoods of $p^{-1}([0]) = \mathbb{Z}$, so if I take

$B_1 = \bigcup_{k \in \mathbb{Z}}(k-1/1, k+1/1)$,

$B_2 = \bigcup_{k \in \mathbb{Z}}(k-1/2, k+1/2)$,

$B_3 = \bigcup_{k \in \mathbb{Z}}(k-1/3, k+1/3)$,

$B_n = \bigcup_{k \in \mathbb{Z}}(k-1/n, k+1/n)$,

I can't see why does this not work as a countable basis for $[0]$, because I think that if I take any neighborhood of $[0]$, I can find some $B_i$ contained in it.

So my question is: what am I missing and how to prove that the given space doesn't satisfy that axiom.

Pan Miroslav
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  • Your axiom $1A\mathbb{N}$ is having a countable local base? – Henno Brandsma Mar 21 '17 at 12:28
  • yes. I will try to rewrite my sets, because I've checked your link and it does not contradicts (at least I don't see it still) the sets I am thinking of.

    edit: actually, I will try to change it to a bit different system of sets.

     – Pan Miroslav Mar 21 '17 at 13:05
  • @PanMiroslav What's $1A\mathbb{N}$ axiom? And do I understand correctly that the $\mathbb{R}/\sim$ is simpy a bouquet of countably many spheres? – freakish Mar 21 '17 at 14:18
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    Take an open set that has a set of intervals around each $k \in \mathbb{Z}$ that gets smaller and smaller as $k$ increases. No $B_n$ can be a subset of it. – Henno Brandsma Mar 21 '17 at 17:33
  • Thank you! This example contradicting my sets actually really helped to clarify it. – Pan Miroslav Mar 21 '17 at 21:51

1 Answers1

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In my answer here ,I show that $\mathbb{N} \subset \mathbb{R}$ does not a countable base of neighbourhoods, so we cannot have a countable family of open sets $U_n, n \in \mathbb{N}$, such that

$\mathbb{N} \subset U_n$ for all $n$ and for any open $O$ with $\mathbb{N} \subset O$, we have $U_n \subset O$.

But a supposed local base $\{B_n: n \in \mathbb{N}\}$ at $[0]$ when pulled back under $q$, so $\{q^{-1}[B_n]: n \in \mathbb{N}\}$ would exactly be such a non-existent family of neighbourhoods:

This holds, because if $\mathbb{N} \subset O$ is open, $q^{-1}[q[O]] = O$ is open and so $q[O]$ is open and contains $[0]$. So then $[0] \in B_n \subset q[O]$ for some $n$, would exist and then $q^{-1}[[0]] = \mathbb{N} \subset q^{-1}[B_n] \subset q^{-1}[q[O]] = O$. So the $q^{-1}[B_n]$ would form a countable base of neighbourhods of $\mathbb{N}$. And we cannot have a countable one as proved.

Your supposed solution fails (you mean to take $q[B_n]$ of course, otherwise they're not in the quotient space), if you look at my proof:

Define $$O = q[\cup_{k \in \mathbb{Z}} (k-\frac{1}{2(|k|+1)} ,k+\frac{1}{2(|k|+1)})]$$ Then $B_n \subset O$ fails at the interval around $n$, at least. It doesn't really matter whether we identify $\mathbb{Z}$ or $\mathbb{N}$ to one class.

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