Here is a cute application of Baire category theorem that is in the spirit of your examples. > Assume that $f: {\mathbb{R}}^{\mathbb{N}} \rightarrow \mathbb{R}$ is a Borel measurable function with the property that if $x \sim y$, then $f(x)=f(y)$, where $x \sim y$ if and only if $\{x_n: n \in \mathbb{N}\}=\{y_n: n \in \mathbb{N}\}$ for $x,y \in {\mathbb{R}}^{\mathbb{N}}$. Then, there exists $x \in {\mathbb{R}}^{\mathbb{N}}$ such that $f(x)=x_k$ for some $k \in \mathbb{N}$. We can consider Cantor's diagonalization argument as a Borel map taking a sequence of reals and producing a real different than any element in the list. This theorem tells you that there is no Borel way to do diagonalization in a "uniform" way. "Uniform" here means that the sequences consisting of the same elements are diagonalized with the same element. This is the most basic version of Friedman's Borel diagonalization theorem. In *On the necessary use of abstract set theory*, Advances in Mathematics, 41 (1981), 209-280, Harvey Friedman proves this result (Proposition C, p. 229) using a forcing argument. Though, in the appendix of the same paper, he gives another proof based on Baire category theorem.