Conditions for a topological space to contain a non-Borel subset Clearly, some spaces such as countable Hausdorff spaces have trivial Borel sigma-algebra.
Are there some ways to check if a space contains a non-Borel subset?
For example, does $\mathbb{Q}^{\mathbb{R}}$ equipped with compact-open topology contain a non-Borel subset?
 A: One way to establish existence of non-Borel sets is by counting how many subsets there are vs how many Borel sets there are. You can apply this strategy to $\mathbb Q^{\mathbb R}$: first note this space has $|\mathbb Q^{\mathbb R}|=\aleph_0^{2^{\aleph_0}}=2^{2^{\aleph_0}}$ elements, and hence $2^{2^{2^{\aleph_0}}}$ subsets.
To count Borel sets, first count open subsets. The topology is generated by open subsets of the form $V(K,U)$, where $K\subseteq\mathbb Q$ is compact and $U\subseteq\mathbb R$ is open. There are $2^{\aleph_0}$ possibilities for each, so there are $2^{\aleph_0}$ of the form $V(K,U)$, and hence $2^{\aleph_0}$ basic opens (which are finite intersections of such sets). Any open subset in $\mathbb Q^{\mathbb R}$ is a union of some family of basic opens, so there are at most $2^{2^{\aleph_0}}$ open subsets.
Now we show by induction that any level of the Borel hierarchy contains at most $2^{2^{\aleph_0}}$ elements. This follows at once from the fact $(2^{2^{\aleph_0}})^{\aleph_0}=2^{2^{\aleph_0}}$. Finally, taking the union over $\aleph_1$ levels of the Borel hierarchy, we get there are at most $\aleph_1\cdot 2^{2^{\aleph_0}}=2^{2^{\aleph_0}}$ Borel sets.
Hence there are more subsets of $\mathbb Q^{\mathbb R}$ than there are Borel subsets, so there must be some non-Borel subsets.
