# Set Theory Problem

• September 5th 2012, 04:32 PM
Kiefer
Set Theory Problem
A is a convex, nonempty set. A is not bounded below and is bounded above. B* = sup A.
Prove:
Claim 1: (-inf, b*) is a subset of A
Prove:
Claim 2: A is a subset of (-inf, b*]

Case 1: b* is not an element of A
Prove:
Claim 2.1: A = (-inf, b*)

Case 2: b* is an element of A
Prove:
Claim 2.2: A = (-inf, b*)

(There are 4 proofs here)
• September 5th 2012, 05:36 PM
Plato
Re: Set Theory Problem
Quote:

Originally Posted by Kiefer
A is a convex, nonempty set. A is not bounded below and is bounded above. B* = sup A.
Prove:
Claim 1: (-inf, b*) is a subset of A
Prove: Claim 2: A is a subset of (-inf, b*]

Case 1: b* is not an element of A
Prove: Claim 2.1: A = (-inf, b*)

Case 2: b* is an element of A
Prove:Claim 2.2: A = (-inf, b*)

Don't you have a typo in that?
Surely you mean $A=(-\infty,b^*]~?$
• September 10th 2012, 07:39 PM
johnsomeone
Re: Set Theory Problem
(I'm assuming A is a subset of the reals.)

Claim: A is a subset of (-inf, b*]
Proof: b* = sup A.

Claim: (-inf, b*) is a subset of A
Proof: Because b* = sup A, there exists $a_{n} \in A s.t. a_{n}\nearrow b*$. Because A not bounded below, there exists $c_{n} \in A$ s.t. $c_{n}\searrow -\infty$.

Let $x \in (-\infty, b*)$. Then there exists $m_{1}$ s.t. $c_{m_{1}} < x$ and $m_{2}$ s.t. $x < a_{m_{2}} <= b*$.

Since $c_{m_{1}}, a_{m_{2}} \in A, c_{m_{1}} < x < a_{m_{2}}$, it follows by A convex that x in A. Thus $(-\infty, b*) \subset A$.

Claim: If b* in A, then $A = (-\infty, b*]$. If b* not in A, then $A = (-\infty, b*)$.

Proof: From previous two claims. Know that $(-\infty, b*) \subset A \subset (-\infty, b*]$.

If b* in A, then $(-\infty, b*) \subset A$, and b* in A, so that their union, $(-\infty, b*]$, is a subset of A.

But then $(-\infty, b*] \subset A \subset (-\infty, b*]$, thus $A = (-\infty, b*]$.

If b* not in A, then $A \subset (-\infty, b*]$ and $A \subset \{b*\}^{c}$, so A is contained in their intersection, $(-\infty, b*)$.

Thus $(-\infty, b*) \subset A \subset (-\infty, b*)$, thus $A = (-\infty, b*)$.