Equivalence relation closed in product, but not closed?

Question

Exercise 2.4.C in Engelking's Topology book asks for an example of an equivalence relation $E$ on a space $X$ such that:

• $E$ is a closed subset of $X^2$, but
• $E$ is not a closed equivalence relation.

His definition of closed equivalence relation is that the natural quotient mapping $\varphi:X\to X/E$ maps closed sets to closed sets.

Question. Is there an example like this where $X$ is separable metrizable, and the equivalence classes are compact?

Answer 1

Yes.

Let $\mathbb{R}_+ = [0, \infty)$ and $X = [0, 1] \times \mathbb{R}_+$. Define an equivalence relation ${\sim}$ on $X$ so that for each $x > 0$ the pair $\{ (x, 0), (x, \frac{1}{x}) \}$ is an equivalence class and all other classes consist of one point.

• Of course, $X$ is separable, metrizable since it's a subset of $\mathbb{R}^2$.
• Every class is finite (one or two points), hence compact.
• It is routine to check that ${\sim}$ is closed in $X^2$.
• The set $F = \left\{ \left( x, \frac{1}{x} \right) : x > 0 \right\}$ is closed, but $\varphi[F]$ is not.