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 say $2^{\omega}$). 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 $\Pi_1^1$?

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 $\Pi_1^1$?

It is known that if there is a measurable cardinal then every $\Pi_1^1$ set has the prefect set property (i.e it is either countable or contains say $2^{\omega}$). Also if we have $\Pi_1^1$-determinacy (or in other words $0^{\sharp}$) then we get that $\Sigma_2^1$ has the prefect set property. Note the result that $\Sigma_1^1$ has the prefect set property is a $ZFC$ result.

Is there a reason as to why we need stronger infinity axioms to prove the prefect 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 $\Pi_1^1$?

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 say $2^{\omega}$). 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 $\Pi_1^1$?