Let $\mathcal{R}$ be the equivalence relation on $\mathbb{R}$ consisting of those pairs $(r, s)$ of real numbers with the property that there exist two integers $m$ and $n$ such that $r − s = m + n\sqrt{2}$. Show that the resulting quotient space is not Hausdorff.
Show that the quotient space is not Hausdorff.
general-topologyquotient-spacesseparation-axioms
Related Solutions
Hints: It actually shouldn't be too difficult to just go through the proof by hand without relying on other theorems.
- ($\Rightarrow$) If $X$ is not Hausdorff, then there are distinct points $x , y \in X$ which cannot be separated. Show that $p ( x , 0 ) , p ( y , 0 )$ cannot be separated in $Y$. If $A \subseteq X$ is not closed, let $x \in \overline{A} \setminus A$. Show that the points $p ( x , 0 ) , p ( x , 1 )$ (which are distinct) cannot be separated in $Y$.
- ($\Leftarrow$) This should just be working through the cases. First note that if $x , y \in X$ are distinct, then they can be separated by open sets in $X$, and it is not too hard to separate $p ( x , i ) , p ( y , j )$ for any $i,j \in \{ 0 , 1 \}$. So the real difficulty will be in separating $p ( x , 0 ) , p ( x , 1 )$ for $x \in X$. But note that we only have to worry about this when $p ( x , 0 ) \neq p ( x , 1 )$. When does this happen? (Obvious choices for open sets should jump out at you.)
$\def\RR{\mathbb{R}}$There is no separation condition which will do the job. That's a vague statement, so here is a precise one: There is a subset of $\mathbb{R}^2$ which (equipped with the subspace topology) does not have condition $\dagger$.
Proof: Let $A$ and $B$ be two disjoint dense subsets of $\mathbb{R}$, neither of which contains $0$. (For example, $\mathbb{Q} +\sqrt{2}$ and $\mathbb{Q}+\sqrt{3}$.) Let $$X = (A \times \RR_{\geq 0}) \cup (B \times \RR_{\leq 0}) \cup (\{0\} \times \RR_{\neq 0}) \subset \RR^2.$$ Define $(x_1,y_1)$ and $(x_2, y_2)$ to be equivalent if $x_1=x_2$ and, in the case that $x_1=x_2=0$, that $y_1$ and $y_2$ have the same sign.
Verification that this is a closed equivalence relation: $X^2$ is a metric space, so we can check closure on sequence. Let suppose we have a sequence $(x_n, y_n) \sim (x'_n, y'_n)$ with $\lim_{n \to \infty} x_n=x$, $\lim_{n \to \infty} y_n=y$, $\lim_{n \to \infty} x'_n=x'$ and $\lim_{n \to \infty} y'_n=y'$. We must verify that $(x,y) \sim (x',y')$. First of all, we have $x_n = x'_n$, so $x=x'$ and, if $x=x' \neq 0$, we are done. If $x=x'=0$, we must verify that $y$ and $y'$ have the same sign. But $y_n$ and $y'_n$ weakly have the same sign for all $n$, so they can't approach limits with different signs.
Verification that $X/{\sim}$ is not Hausdorff: We claim that no pairs of open sets in $X/{\sim}$ separates the images of $(0,1)$ and $(0,-1)$. Suppose such open sets exist, and let $U$ and $V$ be their preimages in $X$. Then there is some $\delta$ such that $(A \cap (-\delta, \delta) )\times \RR_{\geq 0} \subset U$ and $(B \cap (-\delta, \delta) )\times \RR_{\geq 0} \subset B$. Then $U \cap \RR \times \{ 0 \}$ is an open set which contains $(-\delta, \delta)$. By the density of $B$, there must be a point of $B \cap (- \delta, \delta)$ in $U \cap \RR$, and then this gives an intersection between $U$ and $V$.
Best Answer
SKETCH: Let $D=\left\{m+n\sqrt2:m,n\in\Bbb Z\right\}$; for each $r\in\Bbb R$, the $\mathcal{R}$-equivalence class $[r]$ of $r$ is $r+D=\{r+d:d\in D\}$. Use this result to show that $D$ is dense in $\Bbb R$ and hence that $[r]$ is dense in $\Bbb R$ for each $r\in\Bbb R$, and use that fact to show that if $U$ and $V$ are non-empty open sets in the quotient space, then $U\cap V\ne\varnothing$. Conclude that the quotient space is not Hausdorff.