This should be easy to prove but I have no idea how to do it:


If $X \subseteq \mathbb{R}^2$ is borel then $f(X)$ is borel where $f(x,y) = x$


Thanks
Tobias



Projection of Borel set from $R^2$ to $R^1$
This is false; take a look at <https://en.wikipedia.org/wiki/Analytic_set> for a quick introduction. For details, look at Kechris's book on **Classical Descriptive Set Theory**. There you will find also some information on the history of this result, how it was originally thought to be true, and how the discovery of counterexamples led to the creation of descriptive set theory. Another good reference is **A second course on real functions**, by van Rooij and Schikhof.


Here are some additional details (that I added to [another thread](https://mathoverflow.net/a/48590/6085)). Recall that the *analytic* sets are the empty set, and the sets that are images of Borel subsets of $\mathbb R$ by Borel measurable functions $f:\mathbb R\to\mathbb R$. This notion makes sense in arbitrary topological spaces, and has been particularly studied in [Polish spaces](https://en.wikipedia.org/wiki/Polish_space) (separable completely metrizable spaces).


* The analytic sets are closed under countable unions and intersections, and contain the Borel sets. In all *uncountable* Polish spaces, there are analytic sets whose complement is not analytic. Suslin proved that the Borel sets of uncountable Polish spaces are precisely those analytic sets whose complement is analytic.
* In Baire space $\mathcal N=\mathbb N^{\mathbb N}$ (the countable product of discrete $\mathbb N$), analytic sets are the projections of *closed* subsets of ${\mathcal N}^2$.
This is false with $\mathbb R$ in place of $\mathcal N$. Note that in that case, closed sets are countable union of compact sets, so their projections are $F\_\sigma$. Baire space plays a special role in the theory, since a nonempty set in a Polish space is analytic iff it is the continuous image of $\mathcal N$.


The actual results in ${\mathbb R}$ are as follows:


* A set is analytic iff it is the projection of the complement of the projection of a closed subset of ${\mathbb R}^3$.
* A set is analytic iff it is the projection of a $G\_\delta$ subset of $\mathbb R^2$.
* There is a continuous $g:\mathbb R\to\mathbb R$ such that a set is analytic iff it is $g(A)$ for some $G\_\delta$ set $A$.
* A set if analytic iff it is $f(\mathbb R\setminus\mathbb Q)$ for some continuous $f:\mathbb R\setminus\mathbb Q\to\mathbb R$. (Note that if $f$ is actually continuous on $\mathbb R$, then $f(\mathbb R\setminus\mathbb Q)$ is Borel.)


