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Let $A\subseteq \mathbb R^2$ and let $\pi_1(A)$ be the projection of $A$ onto $\mathbb R$. What sets can be expressed as $\pi_1(A)$ for $A$ closed?

It is well known that $\pi_1(A)$ need not be closed. In fact, you can get any open set via $\pi_1(A)$. Because open sets in $\mathbb R$ are disjoint unions of open intervals, it suffices to show that for any $a,b$, there is $A\subseteq \mathbb R^2$ closed with $\pi_1(A) = (a,b)$. You can take $$A = \{ (x,y): x\in(a,b), y \geq \frac1{x-a} - \frac1{x-b} \}$$ You can additionally generate any $F_\sigma$ set in this way. Take $F = \bigcup_{i=1}^\infty C_i$ for $C_i\subseteq \mathbb R$ closed. Then $$F = \pi_1(\{(x,n):n\in\mathbb N,\, x\in C_n\times n\})$$ What other sets can be generated in this way?

The sets must (easily) be $\sigma$-compact (Is the projection of a closed set always a Borel set?) so this means that you can't get all $G_\delta$ sets (the irrationals are $G_\delta$ and aren't $\sigma$-compact). This is quite different from projections of sets in $\mathbb R\times \mathbb Q^c$ where you can get all analytic sets and in particular all Borel sets (proposition 4.1.1 of A Course in Borel Sets by Srivasatav).

daRoyalCacti
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For clarity: $\sigma$-compact here means "is a countable union of compact sets".

A subset of $\Bbb R$ is $F_\sigma$ if and only if it it $\sigma$-compact. (The forward direction is because closed subsets are $\sigma$-compact, the backward direction is because compact sets are closed.)

So the projections of closed sets in $\Bbb R^2$ are exactly the $F_\sigma$ sets.

Izaak van Dongen
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