# Proof involving sets. NEED HELP

1. Feb 26, 2009

### arpitm08

Proof involving sets. NEED HELP!!!

1. The problem statement, all variables and given/known data

Prove directly "If A U B = A, then B is a subset of A." and also provide a proof by contrapositive of its converse.

2. The attempt at a solution

Here is what i did, but I don't know if it is right or not,

Direct Proof: Assume A U B = A, then x ∈ (A U B) and x ∈ A. So it follows that B ∈ A = B is a subset of A.
Contrapositive of Converse Proof: Assume that A U B ≠ A, then x ∈ (A U B) and x ∉ A. Then, B ∉ A and so B is not a subset of A.

I don't think this is right. Could someone help me out please??

2. Feb 27, 2009

### J$C Re: Proof involving sets. NEED HELP!!! Direct Proof: Assume A U B=A as you have. Then to show a set B is a subset of a set A the standard technique is to let x be in B, then show it is also in A. Notice if x is in B then it is clearly also in A U B and the conclusion follows from your initial assumption. Converse: If B is a subset of A then A U B=A Contrapositive of Converse: If A U B /neq A then B is not a subset of A Notice A U B /neq A but A is clearly a subset of A U B. So what's left to make that not equals is A U B is not a subset of A. That gives that there's an element in A U B that is not in A. Go from there. 3. Feb 27, 2009 ### arpitm08 Re: Proof involving sets. NEED HELP!!! How about now... Directly - Assume A U B = A, then x ∈ B. Then x ∈ (A U B), and since A U B = A, x ∈ A. So B is a subset of A. Contrapositive of Converse - Assume that A U B ≠ A. Since A is a subset of A U B, there must be an x ∈ (A U B), such that x ∉ A, since A U B ≠ A. This means that there is a y ∈ B, such that y ∉ A. So B is not a subset of A. Is that a complete proof?? 4. Feb 27, 2009 ### lanedance Re: Proof involving sets. NEED HELP!!! in contrapositive of converse though its implied, I think you should change it to A is a proper subset of AUB 5. Feb 27, 2009 ### arpitm08 Re: Proof involving sets. NEED HELP!!! Allright. Thanks!! =D 6. Feb 27, 2009 ### J$C

Re: Proof involving sets. NEED HELP!!!

Looks good.