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We know that <math>\mathbf{\Sigma}^1_1</math> is closed under countable union and intersection from the page [[analytic set]]. Since the image of union is the union of images, the implication "<math>\mathbf{\Sigma}^1_n</math> is closed under countable intersection <math>\Rightarrow</math> <math>\mathbf{\Sigma}^1_{n+1}</math> is closed under countable union" would be trivial, but to prove that <math>\mathbf{\Sigma}^1_n</math> is closed under countable intersection seems to be nontrivial for me. A proof would be well appreciated.
Another example: the diagram in the German page says that <math>\mathbf{\Sigma}^1_n, \mathbf{\Pi}^1_n\subset\mathbf{\Delta}^1_{n+1}</math>, which is equivalent to <math>\mathbf{\Sigma}^1_n, \mathbf{\Pi}^1_n\subset\mathbf{\Sigma}^1_{n+1}</math>. This means that every <math>\mathbf{\Sigma}^1_n</math> set and every <math>\mathbf{\Pi}^1_n</math> is the projection of a <math>\mathbf{\Pi}^1_n</math> set; while the latter implication is trivially true, I have no idea about the former even with ''n'' = 1. The inclusion gives us the inclusion of σ-algebras <math>\mathbf{\Delta}^1_1\subset\sigma(\mathbf{\Sigma}^1_1)\subset\mathbf{\Delta}^1_2\subset\cdots\subset\mathbf{P}</math>. Perhaps in fact each inclusion is strict in
What's more, the inclusion <math>\mathbf{\Pi}^1_n\subset\mathbf{\Delta}^1_{n+1}</math> tells us that <math>\mathbf{\Sigma}^1_n</math> sets are precisely the projections of <math>\mathbf{\Delta}^1_n</math> sets, so <math>\mathbf{\Delta}^1_n</math> sets are precisely those sets such that themselves as well as their projections are all <math>\mathbf{\Delta}^1_n</math> sets: by definition <math>\mathbf{\Delta}^1_n\subset\mathbf{\Sigma}^1_n</math>, and the projections of <math>\mathbf{\Sigma}^1_n</math> sets are also in <math>\mathbf{\Sigma}^1_n</math>
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