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

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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 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
 
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(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?
 
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
 
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Not actually trying to prove statements. Just trying to ask the right questions to develop the proof.
 
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)
 
loop quantum gravity said:
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.
 
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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.
 
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