Proving that $X/R$ is Hausdorff $\implies$ $R$ closed.

Question

As the title says, I'm trying to prove that if $X/R$ is a Hausdorff space then $R\subset X\times X$ is closed.

I have several questions about this:

$(1)$ What exactly is $R$?

I thought of $R$ as an equivalence relation, but I never thought of that relation (or any) as a set. What are the elements of this set?; my intuitive idea would be that $R$ contains elements of the equivalence classes (I would call them representants, but I don't know if that gives the correct idea), this is, an element $x_1$ such that because of it I can have the class $[x_1]\subset X/R$, an element $x_2\neq x_1$ such that $[x_2]\subset X/R$, etc; but I don't know if the way I'm trying to visualize this is right...

$(2)$ $X/R$ Hausdorff?.

I find less intuitive thinking a set of equivalence classes as a Hausdorff space than a more regular one (namely, $\mathbb{R}^n$ or similar). Then the hypothesis here is that given $[x],[y]\subset X/R\;\exists U,V\subset X\times X$ open such that $[x]\in U,[y]\in V, U\cap V=\emptyset$.

$(2.1)$ But why $R\subset X\times X$?

Following the idea of $(1)$ maybe would be correct think of $R$ as the set $\{(a,b)|\;r:X\to X,\, r(a)=b\}$?. Here the $b$ value would be one 'representant' of the class, then if $a_1,a_2\subset[b]$ would be $r(a_1)=r(a_2)=b$. But again, I don't know if my intuition is ok...

$(2.2)$ Finally, why is $R$ closed if $X/R$ Hausdorff.

I've considered several ways, none of them seems to work for me:

$(a)$ Showing that accumulations points of $R$ belongs to $R$. Problem: How do I relate accumulation points and the Hausdorff condition?.

$(b)$ $R$ closed in $X\times X$ if $R=R_1\times R_2$, each of them included in $X$. Problem: How to show that $R_i$ is closed in X?.

Also, is there an intuitive way of picturing quotient spaces?. Recommendation of books with a clear treatment about this topic are highly appreciated.

Answer 1

Firstly, by definition a relation on $X$ is a subset of $X \times X$. This is how relations are defined in set theory! Any subset of $X \times X$ is called a (binary) relation on $X$. Instead of $(x,y) \in R$ one also writes $xRy$. An equivalence relation of course has to satisfy some extra conditions (which I assume you know). These imply that the sets of the form $[x] = \{y \in X: xRy \}$, the classes of $X$ under $R$, partition $X$.

Then $X/R = \{[x]: x \in X \}$, by definition, and we have the so-called quotient map $q: X \rightarrow X/R$, defined by $q(x) = [x]$, and the topology on $X/R$ is defined by $O \subset X/R$ open iff $q^{-1}[O] ( = \{x: q(x) = [x] \in O \})$ open in $X$.

Now, assume $X/R$ is Hausdorff, we can show $R \subset X \times X$ is closed: take $(x,y) \in (X \times X) \setminus R$, and we have to show the latter set is open, so we need to find $O_1 \times O_2 \subset X \times X$ (basic) open such that $(x,y) \in O_1 \times O_2 \subset (X \times X)\setminus R$, or equivalently $(O_1 \cap O_2) \cap R = \emptyset$. But $(x,y) \notin R$ by definition means that $xRy$ does not hold, and this means (check it!) that $[x] \neq [y]$.

Now apply the Hausdorff condition on $X/R$ (where the points are exactly these classes!) to find open disjoint sets $O,O'$ such that $[x] \in O \subset X/R$ and $[y] \in O' \subset X/R$ with $O \cap O' = \emptyset$.

By definition, $x \in O_1 := q^{-1}[O] \subset X$ is open, and also $y \in O_2:= q^{-1}[O'] \subset X$ is also open. And these sets are disjoint, as $O$ and $O'$ are. Finally check that $(O_1 \times O_2) \cap R = \emptyset$, as required.

Answer 2

Since you say $a R b$ iff $(a,b) \in X\times X$ and $a$ together with $b$ satisfy the equivalence relation, then $R$ can be viewed as some subset of $X\times X$ satisfying the relation.

Again, you can view the relation as all $(a,b)$ such that $a R b$, so clearly it can be viewed as a subset of $X\times X$.

You could make the above more rigorous showing that you can go back and forth bijectively between equivalence relation subsets of $X\times X$ and equivalence relations such that the two corresponding ideas are the same equivalence relation! But why?

Now $X / R$ is the quotient space of the topological space $X$, there exists a topology on it induced by the quotient map $p: X \to X/R$. See here.

Now if $X/R$ is Hausdorff then take two distinct points $x, y \in X/R$. Then there are open sets $U,V \ni x,y$, resp. such that $U \cap V = \varnothing$.

Let's look at what we want to prove as I'm stuck too! We want to show that $R \subset X\times X$ is closed in the product topology.

Google for things about "Hausdorff and closed", I found the wiki article: Hausdorff space to have a few hints.

It says that $f: X \to Y$ a quotient map and $Y$ Hausdorff implies $\ker(f) = \{(x,x') \in X\times X : f(x) = f(x')\}$ is closed.

Wow! That seems really related. Let's see how we can use it. $\ker(p) = \{(x,x') : p(x) = p(x')\}$ but notice that $p(x) = p(x') \iff (x,x') \in R$, thus the kernel is the equivalence relation! Thus your conjecture proven! QED

Now all you have to do is prove the result that you used. Granted it looks a lot like our original problem, but gave us some ideas.

Consider $X\times X - R$ (set diff). We need to show that it's open. Let $(x,x')$ be in the set, then $p(x) \neq p(x')$ by definition (ie. $(x,x') \notin R$), so take the disjoint open sets $U, U'$ around $p(x), p(x')$ resp. by Hausdorfness of $X/R$. Let $V = p^{-1}(U), \ V' = p^{-1}(U')$. If $(V\times V') \cap R \neq \varnothing$, then there's $(v,v') \subset V\times V'$ such that $p(v) = p(v')$, $p(v) \in U$ and $p(v') \in U'$ which are disjoint! It was messy, but contradiction we have! QED

Note: $V\times V'$ is clearly an open set in the product topology, containing $(x,x')$ and disjoint from $R$, thus $R$ is closed. That completes the proof.

The thing that enabled us to solve this problem was the observation that $xRy$ iff $p(x) = p(y)$ for the quotient map $p$ associated with $R$. From that we was able to abstract the mathematics and use dat techmology to solve a problem. What's dat all about?

Answer 3

Since $\pi:X \to X/R$ is a quotient map, it is continuous and surjective $ \implies \pi^*: X \times X \to X/R \times X/R $, defined as $ \pi^*(x,y) \to (\pi(x),\pi(y))$, is continuous and surjective. Moreover, the product of Hausdorff spaces is Hausdorff, so $ \Delta \subset X/R \times X/R$ is closed, and since $\pi^*$ is continuous, $\pi^{*-1}(\Delta) = R$ is closed.