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 !


$\{0,1\}$ 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.)
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  • $\begingroup$ $A=E$ it's the same thing ? $\endgroup$ – Maman Nov 22 '16 at 20:38
  • $\begingroup$ @Maman: There is no $E$. There is the set $\mathscr{E}$ of reflexive, symmetric relations on $\Bbb N$. Its definition used the symbol $E$ as a dummy variable, but neither the question nor my answer talks about any specific relation $E$. We show that $\mathscr{E}$ is closed by showing that its complement is open. To do that we look at an arbitrary $A\notin\mathscr{E}$. Such a relation $A$ is either not reflexive or not symmetric; the answer considers both cases. (And I specifically did not want to use the variable $E$ for a relation that is not in the set $\mathscr{E}$.) $\endgroup$ – Brian M. Scott Nov 22 '16 at 20:43
  • $\begingroup$ For the product topology the open sets are the product of $\{x,y\}$ ? $\endgroup$ – Maman Nov 22 '16 at 20:46
  • $\begingroup$ @Maman: No. For each $\langle m,n\rangle\in\Bbb N^2$ you have a space $X_{\langle m,n\rangle}=\{0,1\}$, and you’re looking at the infinite product $\prod_{\langle m,n\rangle\in\Bbb N^2}X_{\langle m,n\rangle}$. Basic open sets in this product are sets of the form $\prod_{\langle m,n\rangle\in\Bbb N^2}U_{\langle m,n\rangle}$ such that each $U_{\langle m,n\rangle}$ is open in $X_{\langle m,n\rangle}$, and $\{\langle m,n\rangle\in\Bbb N^2:U_{\langle m,n\rangle}\ne X_{\langle m,n\rangle}\}$ is finite. $\endgroup$ – Brian M. Scott Nov 22 '16 at 20:50
  • $\begingroup$ So all the pairs are elementary open sets (a special case ) ? $\endgroup$ – Maman Nov 22 '16 at 20:54

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

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