Problem 2

[TODO]

Exercise 7.2 (An equivalence relation with closure)   Let $\sim$ be an equivalence relation on a topological space $X$. Assume each equivalence class is a closed set in $X$. Then a set of finitely many points in $X/{\sim}$ is closed in the quotient topology.

Proof. Let $S$ be a set of finitely many points in $X/{\sim}$:

$$S = \{[x_1], \dots, [x_n]\}$$ (7.3)

To show $S$ is closed, we show its inverse image (under the quotient map) is closed:

$$q^{-1}(S) = \{x \in X\;\vert\; q(x) \in S\} = \bigcup_{i=1}^n \{x\in X \;\vert\; x \sim x_i\}$$ (7.4)

We assumed each equivalence class is a closed set in $X$, so this is a finite union of closed sets, which is closed.

For an example of $X$ a Hausdorff where its quotient $X/{\sim}$ is not, consider $X=\mathbb{R}$ under the relation $a\sim b \iff a-b\in\mathbb{Q}$. Then the quotient space $X/{\sim}$ is not Hausdorff.

This can be seen by supposing distinct equivalence classes $[x]$ and $[y]$ lie in disjoint open sets $U$ and $V$. Select representatives $x<y$. Select an interior neighborhood $B_\delta(x)\subseteq q^{-1}(U)$ for some $\delta>0$. Approximate

$$\vert y-x-r\vert<\delta$$ (7.5)

for some rational $r$. Then $y-r\in B_\delta(x)$ implies $[y-r]$ in $U$. But the rationality of $r$ implies $[y]=[y-r]$. Therefore, $U$ and $V$ are not disjoint. $\qedsymbol$