Philosophy of basic set theory proofs involving or .

[scorpion990]

Philosophy of basic set theory proofs involving "or".

Hey!

I'm working through an Introduction to Analysis text, and I'm currently on the first chapter, which covers set theory. In one of the end-of-chapter problems, I'm asked to prove a basic theorem which leads to the following statement: x is an element of A, and (x is an element of B or x is an element of C).

My text (Maxwell Rosenlicht's Introduction to Analysis) lacks in the "example" department, and so for a little while, I wasn't sure how to handle this statement. I've been pondering this for a few days, and realized that "a or b" is true if:
1. a is true. b is false.
2. b is true. a is false.
3. both a and b are true.
So, I realized that the only thing I could do was split the argument into these three possibilities. I then showed that for my particular proof, all three possibilities lead to the same statement.

I have a quick question: Is it common practice in set theory to split "or" statements into the three possible statements which can make it true, and proceed in this way? It makes a lot of sense intuitively, but I've never seen it done in any professional papers. Is there any other way to handle such statements? Thanks.

 

[CRGreathouse]

One other way would be to break it into two cases:

a is true, b is unknown
a is false, b is true

 

[scorpion990]

Hmm... Yes. I noticed that the case in which both a and b are true is a repeat of another proof.

So is this an "acceptable" practice in analysis?

 

[evagelos]

Hey!

I'm working through an Introduction to Analysis text, and I'm currently on the first chapter, which covers set theory. In one of the end-of-chapter problems, I'm asked to prove a basic theorem which leads to the following statement: x is an element of A, and (x is an element of B or x is an element of C).

My text (Maxwell Rosenlicht's Introduction to Analysis) lacks in the "example" department, and so for a little while, I wasn't sure how to handle this statement. I've been pondering this for a few days, and realized that "a or b" is true if:
1. a is true. b is false.
2. b is true. a is false.
3. both a and b are true.
So, I realized that the only thing I could do was split the argument into these three possibilities. I then showed that for my particular proof, all three possibilities lead to the same statement.

I have a quick question: Is it common practice in set theory to split "or" statements into the three possible statements which can make it true, and proceed in this way? It makes a lot of sense intuitively, but I've never seen it done in any professional papers. Is there any other way to handle such statements? Thanks.

in a mathematical proof we use logical identities called laws of logic which are always true whether your a's or b's are true or false.

So the use of true or false have no significance at all.

try to prove a theorem by using tables of true or false values,i mean a long theorem not a short one it will get you into trouble.
lets say you want to prove AU(B&C)=(AUB)&(AUC) by using your approach
furthermore mathematicians use the true false approach ,because it is too difficult for them to
grasp very simple but powerfully proofs.
For example in proving that the empty set is a set of every other set in staring the proof they have to assume , xεΦ and from that to prove xεA,where A is any set . Is very difficult for them while for a logician is an easy thing.
And what do they do?They use the F----->T trick.
Because the proof is short one and it Will not get them into trouble had they had to curry on like
in a long proof

 

[scorpion990]

I'm afraid I don't quite understand the last post =(

 

[evagelos]

I have a quick question: Is it common practice in set theory to split "or" statements into the three possible statements which can make it true, and proceed in this way? It makes a lot of sense intuitively, but I've never seen it done in any professional papers. Is there any other way to handle such statements? Thanks.

Try to do it in proving ...AU(B&C)=(AUB)&(AUC)