Tagged Questions

43 views

Is there a subset of R such that their Cantor-Bendixson rank is the first limit ordinal?

I'm looking for a set $A \subset \mathbb{R}$ such that $\bigcap^\infty_{n=0} A^{(n)}$ is a perfect set (i.e $X'=X$) but $\forall n \in \mathbb{N}$ the set $A^{(n)}$ isn't perfect (where ...
92 views

Is any subset of the open ordinal space $[0,\Omega)$ $G_\delta$?

Consider the open ordinal space $[0,\Omega)$, where $\Omega$ is the first uncountable ordinal. Can I say that every subset of $[0,\Omega)$ is $G_\delta$? If yes, does this imply that $[0,\Omega)$ is ...
253 views

Definable order types without infinity axiom.

Denote by $ZF^\times$ the theory of $ZF$ without the axiom of infinity. We know that $V_\omega$, the set of all hereditarily finite sets in a model of $ZF$, is a model of $ZF^\times$. We further know ...
190 views

Complexity of the set of computable ordinals

According to http://en.wikipedia.org/wiki/Analytical_hierarchy The set of all natural numbers which are indices of computable ordinals is a $\Pi^1_1$ set which is not $\Sigma^1_1$. However, "the ...
421 views

Derived sets and ordinals

Given a set $P$ of real numbers, its derived set is the set of all accumulation points - $a\in P^\prime$ if every open set containing $a$ also contains an infinite number of points from $P$ ...
I heard at some point (without seeing a proof) that every countable, compact space $X$ is homeomorphic to a countable successor ordinal with the usual order topology. Is this true? Perhaps someone can ...