Can you check this proof please: sets?

[workerant]

Homework Statement

A,B,C,D are sets.

Prove that if C is contained in A and D is contained in B, then C∩ D is contained in A∩ B.

Homework Equations

The Attempt at a Solution

Let x be any element.

Then There exists (x that belongs to C∩D) and (x does not belong to A∩ B)

So x belongs to C and x belongs to D

If x belongs to C, since C is contained in A, then x belongs to A.
If x belongs to D, since D is contained in B, then x belongs to B.
So x belongs to A intersect B, a contradiction.

Then the original statement is true.

Is it okay?

 

[quantumdude]

The Attempt at a Solution

Let x be any element.

Then There exists (x that belongs to C∩D) and (x does not belong to A∩ B)

That needn't be the case at all. Let $A=B=C=D=\mathbb{R}$.

 

[workerant]

Well, then can I say to assume that because I am trying to do a proof by contradiction.

I'm confused by what you are saying...I should say A=B=C=D=R and then what? I'm not sure I follow...is the rest okay?

 

[Office_Shredder]

Don't do proof by contradiction, it just muddles things. Everything else was fine.

So x belongs to C and x belongs to D

If x belongs to C, since C is contained in A, then x belongs to A.
If x belongs to D, since D is contained in B, then x belongs to B.
So x belongs to A intersect B,

Hence all x in C intersect D are in A intersect B, and C intersect D is a subset of A intersect B. It's much cleaner that way

 

[quantumdude]

Well, then can I say to assume that because I am trying to do a proof by contradiction.

I'm confused by what you are saying...I should say A=B=C=D=R and then what?

I was giving you a counterexample to demonstrate that your opening statement is false. If you let all 4 sets equal the real numbers then it is not the case that that there exists an $x\in C\cap D$ with $x\notin A\cap B$.

 

[quantumdude]

If x belongs to C, since C is contained in A, then x belongs to A.
If x belongs to D, since D is contained in B, then x belongs to B.
So x belongs to A intersect B, a contradiction.

I agree with Office Shredder that you should forget about proof by contradiction here, but I don't agree that this all by itself is fine. You should include a line that says that $x\in C\cap D$. After all, you're supposed to show that $(C \cap D) \subset (A \cap B)$. Those two sets should be connected by your argument.

 

[workerant]

thanks guys...I actually I did include such a line in my formal write-up so thanks