1
$\begingroup$

I have been studying basics of descriptive set theory lately. In the lecture notes I follow (sadly, the notes are written in Czech), there is a lemma (which is used to prove that the functions of first borel class and the functions of first baire class coincide):

Let X be a metric space and $A_n\in\mathbf{\Sigma}_\alpha^0$ (finitely or countably many). Then there exist disjoint $D_n\subseteq A_n$ such that $D_n\in\mathbf{\Sigma}_\alpha^0$ and $\bigcup\limits_n D_n =\bigcup\limits_n A_n$. Moreover, if $\bigcup\limits_n A_n = X$, then $D_n\in\mathbf{\Delta}_\alpha^0$.

I could prove the first part of the lemma but I can't prove the moreover part. I'd like to show that $X\setminus D_n\in\mathbf{\Sigma}_\alpha^0$. Obviously, we have $X\setminus D_n = \bigcup\limits_n A_n\setminus\bigcup\limits_{k=1}^\infty D_n^k $, where $D_n^k\in\mathbf{\Delta}_\alpha^0$ as $D_n\in\mathbf{\Sigma}_\alpha^0$. However, I can't see how to continue.

Thank you for any help!

$\endgroup$
1
$\begingroup$

Well, in fact it's trivial. If $\bigcup\limits_n A_n=\bigcup\limits_n D_n= X$, then (as $D_n$ are disjoint) $D_n^C = \bigcup\limits_{k\neq n} D_k\in\mathbf{\Sigma}_\alpha^0$, hence $D_n\in\mathbf{\Delta}_\alpha^0$.

I'm sorry I'd asked such a trivial question but you certainly know it - sometimes the most obvious things are the hardest one.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.