Cantor-schroder-bernstein use in proof

[jaqueh]

Homework Statement

Use the berstein theorem to show: if A$\subseteq$ℝ and there exists an open interval (a,b) such that (a,b)$\subseteq$A, 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