It is known that if there is a measurable cardinal then every $\Pi\_1^1$ set has the perfect set property (i.e it is either countable or contains a copy of $2^{\omega}$). Also if we have $\Pi\_1^1$-determinacy (or in other words $0^{\sharp}$) then we get that $\Sigma\_2^1$ has the perfect set property. Note the result that $\Sigma\_1^1$ has the perfect set property is a $ZFC$ result. 


Is there a reason as to why we need stronger infinity axioms to prove the perfect set property for $\Pi$ classes in comparison to what we need to prove the perfect set property for $\Sigma$ classes? This is weird because we have $\Pi\_1^1 \subseteq \Sigma\_2^1$. Do we actually need less than a measurable to prove the PSP for $\Pi\_1^1$?



Large cardinal axioms and the perfect set property
Solovay showed that the following are equivalent:


1. $\boldsymbol{\Sigma}^1\_2$ sets have the perfect set property
2. $\boldsymbol{\Pi}^1\_1$ sets have the perfect set property
3. $\aleph\_1^{L[a]} < \aleph\_1$ for every real $a$


You only need an inaccessible to force (3).


