Cantor-schroder-bernstein use in proof

[jaqueh]

Homework Statement

Use the berstein theorem to show: if A\subseteqℝ and there exists an open interval (a,b) such that (a,b)\subseteqA, then the cardinal number of A=ℂ

Homework Equations

The theorem states that if card(A)≤card(B) and card(B)≤card(A), then card(A)=card(B)

The Attempt at a Solution

I don't know what is relevant and what the open interval (a,b) implies. I know that (a,b) is a subset of ℝ and its equivalent to ℂ, but i don't know if I am supposed to prove that or what I am supposed to do at all

 

[scurty]

I never used this theorem before, but I think I understand why it is used. Do you have any theorems that relate subsets to cardinalities? If A \subset B, then card(A) ? card(B).

 

[jaqueh]

yes if a subset b then card(a)≤card(b), aka there exists an F:A-(1-1)->B

 

[scurty]

Okay, good. I think this proof is pretty easy now, right?

You have (a,b) \subseteq A, so what is true about their cardinalities? You already told us that card((a,b)) is the continuum.

You also have A \subseteq \mathbb{R}, so what is true about their cardnalities?

I think at this point you can apply the theorem to get your final line of the proof!

 

[jaqueh]

well i guess c=(a,b) ≤ A and A≤ R=c , but can i assume that (a,b) is equal to c?

 

[scurty]

I know that (a,b) is a subset of ℝ and its equivalent to ℂ, but i don't know if I am supposed to prove that or what I am supposed to do at all

I thought when you stated this line that you knew card((a,b)) is the continuum. Did you not prove this in a theorem yet? It turns out that card((a,b)) is the continuum, but if you didn't prove that, you might need to take a different approach.

In real analysis, that fact was one of the first thing we proved involving cardnalities, so I would assume you have that at your disposal?

 

[jaqueh]

i don't have that at my disposal because this is just a proof writing class, i don't know how i would prove that (a,b) is the continuum using what i know though. that is the main bit that is confusing me