Testwiki:Reference desk/Archives/Mathematics/2013 February 5

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February 5

A quetion in the real line topology

Hi! Here is my question: Given a countable union of countable sets $A_i$ such that $\cup A_i=R$. Does it implies that at list one of this sets is dense in some open segment of R? and if yes, how to prove it? Thanks! Topologia clalit (talk) 12:32, 5 February 2013 (UTC)

Are you sure you mean a countable union of countable sets? Because that requires rejecting the axiom of choice.--190.18.159.129 (talk) 12:49, 5 February 2013 (UTC)

Nevermind, our Baire category theorem article claims it can be proven for the reals without using choice. So the answer is yes, it follows from the Baire category theorem.--190.18.159.129 (talk) 13:08, 5 February 2013 (UTC)

Hi, As a matther of fact I have looked over this theorem. But I don't se how it's connected. The written states that the reals are a Baire space while: "A Baire space is a topological space with the following property: for each countable collection of open dense sets Un, their intersection ∩ Un is dense." But in my question there is a countable union of sets (not all countable) where $\cup A_i=R$. But it is not given that the sets are dense in R. and I want to show that at least one of them is dense in some open interval of R. I don't need to show that their intersection is dense.. I don't understand the connection.. Thanks Topologia clalit (talk) 13:58, 10 February 2013 (UTC)

The connection is via the complement. Let $B_i$ denote the closure of $A_i$. You wish to show that one of the $B_i$ contains an interval. Since $\bigcup _i{B}_i\supseteq \bigcup _i{A}_i=ℝ$, it follows that $\bigcap _i(ℝ-B_i)=\mathrm{\varnothing }$. But each $(ℝ-B_i)$ is an open set, so by the contrapositive of the Baire category theorem, at least one of the $(ℝ-B_i)$ is not dense. That means there is some open interval it does not intersect, and thus $B_i$ contains that open interval.--190.18.159.129 (talk) 03:49, 11 February 2013 (UTC)