Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**Kazy****Member**- Registered: 2006-01-24
- Posts: 37

I need to prove the following:

a) Let A, B be sets contained in a universal set U. Suppose that A ≈B. prove that P(A) ≈ P(B) (power sets)

b) Let A, B be sets contained in a universal set U such that A is a subset of B. Suppose that A is countable and B is uncountable. prove that B - A is uncountable.

c) Using the Schroeder-Bernstein Theorem, prove that any two intervals of real numbers are numerically equivalent.

Shroeder-Bernstein Theorem: Let A and B be sets, and suppose that A <= B and B <= A. Then A ≈ B.

Can anyone help with any of those?

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

a.

Since A ~ B, there exists a 1-1 and onto function θ. So consider all the power sets of A. Just take every a in there, and replace it with θ(a). All these elements must be distinct, so as long as the power sets of A are distinct (which they are), then these will be the power sets of B.

b.

Taking a countable number of elements away from an uncountable set, it should be clear this will leave you with an uncountable set.

Let's assume the B - A is countable. Then B - A = {x : x∈B and x is not in A}. Consider B - A + (B intersect A) = B. B intersect A is countable since A is countable. Thus, we have B - A + (B intersect A) which must be countable since both B-A and B intersect A are countable. So B is countable. Contradiction.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

Pages: **1**