Suslin's theorem

Remarks

One direction is easy: every Borel set is analytic, and the complement of a Borel set is Borel. For the other direction, the Lusin separation theorem says that two disjoint analytic sets $A$ and $B$ can be separated by a Borel set $C$, that is, $A \subseteq C$ and $C \cap B = \emptyset$. Applied to $A$ and $X \setminus A$, it gives a Borel set $C$ with $A \subseteq C \subseteq A$, so $C = A$.

The hypothesis on the complement cannot be dropped. Every uncountable Polish space contains an analytic set that is not Borel, and the complement of such a set is not analytic.

In the notation of the projective hierarchy, the theorem says $\boldsymbol{\Delta}^1_1 = \boldsymbol{\Sigma}^1_1 \cap \boldsymbol{\Pi}^1_1$ is exactly the class of Borel sets.

Going deeper

For a student

The theorem is the standard way to prove that a set is Borel without writing it explicitly from open sets by countable unions and complements. One shows that the set and its complement are both projections of Borel sets, and the theorem does the rest. A common use is the result that a bijective Borel map between Polish spaces has a Borel inverse.