Prove that this set is compact?

Question

We define $\mathcal{E}:=\{E \subseteq \mathbb{N^2} : \text{$E$ is a reflexive and symmetric relation}\} \subseteq \{0,1\}^\mathbb{N^2}$.

I have to prove that $\mathcal{E}$ is compact for the product topology.

My idea is to prove that $\{0,1\}^\mathbb{N^2}$ is compact and $\mathcal{E}$ is closed in this compact.

By Tychonoff's theorem the set $\{0,1\}^\mathbb{N^2}$ is compact (I think that $\{0,1\}$ has the Borel-Lebesgue property so it's compact).

Now, I have no idea to prove that $\mathcal{E}$ is closed (which argument could I use ?).

Thanks in advance !

Answer 1

$\{0,1\}$ is compact simply because it’s finite. The space $\{0,1\}^{\Bbb N^2}$ is homeomorphic to the middle-thirds Cantor set, by the way.

HINT: Suppose first that $A\subseteq\Bbb N^2$ is not reflexive; then there is an $n\in\Bbb N$ such that $\langle n,n\rangle\notin A$.

• Show that $\{U\subseteq\Bbb N^2:\langle n,n\rangle\notin U\}$ is an open nbhd of $A$ disjoint from $\mathscr{E}$. (This follows immediately from the definition of the product topology.)

Now suppose that $A$ is not symmetric; then there are $m,n\in\Bbb N$ such that $m\ne n$, $\langle m,n\rangle\in A$, and $\langle n,m\rangle\notin A$.

• Show that $\{U\subseteq\Bbb N^2:\langle m,n\rangle\in U\text{ and }\langle n,m\rangle\notin U\}$ is an open nbhd of $A$ disjoint from $\mathscr{E}$. (This also follows immediately from the definition of the product topology.)

Answer 2

You can prove more, both all reflexive and all symmetric relations are closed separately, and so is their intersection. Hint: Note that $(x, y) ∈ E \iff π_{x, y}(E) = 1$.