Proving Set Subsets and the Cauchy-Schwarz Inequality: Insights and Techniques

[courtrigrad]

Lets say you are given a bunch of statements and you need to ask some questions to prove them:

(a) How do you show that a set is a subset of another set.
I said to show that $x\in A$ and [itex]x\in B [/tex]. What else can you do to show what $A\subset B$? Could you assume from the following: If $A\cup B = B\cup A$ then $A\subset B$? (sorry, not experienced in set theory).<br /> <br /> (b) If $a$ and $b$ are real nonnegative real numbers, then $a^{2}+b^{2} \leq (a+b)^{2}$. Is this the Cauchy-Schwarz inequality? Basically, the questions that I ask in this case, is how can I prove that $a^{2}+b^{2} \leq (a+b)^{2}$ or $(a+b)^{2}\geq a^{2}+b^{2}$ and work from this (forward or backward)?<br /> <br /> Thanks[/itex]

 

[BSMSMSTMSPHD]

(a) Start by assuming x is a member of set A. Then show that it must be in set B. This will prove that A is a subset of B.

Also, if you can show $$A \cap B = A$$ then that works too.

(b) Multiply out the $$(a + b)^{2}$$ and notice the extra term. What can you say about the sign of this term given what you've assumed about a and b?

 

[0rthodontist]

a. What you show is that if x is in A, THEN x is in B. Just showing that there is some x that is in A and also in B is not enough. You should convince yourself of this.

b. What is (a + b)^2 also equal to?

By the way, you should be careful about the $$\subset$$ symbol. Depending on the context $$A\subset B$$ can mean that A is a subset of B and not equal to B. It may be better to say $$A \subseteq B$$

 

[courtrigrad]

Not actually trying to prove statements. Just trying to ask the right questions to develop the proof.

 

[MathematicalPhysicist]

for the first question if you don't mind some quantifiers to clear up what you need to prove:
$$\forall x(x\in A \rightarrow x\in B)$$

 

[nocturnal]

for the first question if you don't mind some quantifiers to clear up what you need to prove:
$$\forall x(x\in A \rightarrow x\in B)$$

I think the quantifier is superfluous in this statement, although I'm no logician.

 

[ircdan]

A is a subset B means every element x in A is also in B. So to show A is a subset of B, you have to show every element in A is also in B. You start by assuming there is some x(it's arbitrary) in A, then show that x is also in B. Since the x was arbitrary, it holds for all the x's in A, so that's why it works. More precisely, loop gravity's post sums up what it means for A to be a subset of B.