# Is union of intersections of horizontal lines and a closed set in the plane a borel set?

Let $$F$$ be a closed set in the plane $$\mathbb{R}^2$$, define $$F_y=\{x\in\mathbb{R}|(x,y)\in F\}$$, is $$\bigcup_{y\in \mathbb{R}}F_y$$ a borel set in $$\mathbb{R}$$? Intuitively, it is just like compressing a closed set in the plane into a set in the real line. I become interested in this problem for this is the last step that I need to finish an exercise, but I struggle to figure out whether it is true or not. Any help will be appreciated.

• The set $\bigcup_y F_y$ is in fact just the image of $F$ under the projection onto the first factor - AKA, "the projection of $F$". This lead me to the answer here (just the bit at the end), which is the same as the answer below - although perhaps a bit easier to motivate/easier to generalise/cleaner if you know facts like "$\Bbb R^2$ is $\sigma$-compact", and "the image of a compact set is compact", and "a compact set in $\Bbb R$ is closed". Often in measure theory you argue by "approximating" something complicated by some simpler things. Mar 19 at 10:31

It is actually an $$F_{\sigma}$$ set (i.e. a countbale union of closed sets). In fact, $$\{x: x \in F_y \,\,\text {for some } y \in [-N,N]\}$$ is a closed set for each $$N$$.
Indeed, if $$(x_n,y_n) \in F$$ with $$|y_n|\leq N$$ and $$x_n \to x$$ then there is a subsequence $$y_{n_i}$$ converging to some $$y \in [-N,N]$$ and $$(x,y) \in F$$.