# Set TheoryWatch

#1
Hi guys,

I've been asked to show that , where and are binary relations on some set .

This is my proof:

Is this a valid proof?
0
4 years ago
#2
(Original post by r3l3ntl3ss)
Hi guys,

I've been asked to show that .

This is my proof:

x

Is this a valid proof?
0
#3
(Original post by sarcasmrules)
done, had a problem with the curly brackets
0
4 years ago
#4
(Original post by r3l3ntl3ss)
done, had a problem with the curly brackets
This is what I would have done:

Multiply both sides by the union of R and S:

sth operated with sth is e (identity):

Union operation is commutative and associative, so:

I think yours is valid too.
0
4 years ago
#5
What you've written is false

.

For these kinda of problems where you have a set and and you want to show that , you have to show inclusion on both sides so that and .

So assume and prove that also and vice versa.
1
4 years ago
#6
(Original post by sarcasmrules)
This is what I would have done:

Multiply both sides by the union of R and S:

sth operated with sth is e (identity):

Union operation is commutative and associative, so:

I think yours is valid too.
is not . In fact , where can be taken as the universe where .

Also as an identity makes no sense in set theory. There's no axiom of set theory that forces you to have an identity element in a set.
0
4 years ago
#7
(Original post by 0x2a)
is not . In fact , where can be taken as the universe where .

Also as an identity makes no sense in set theory. There's no axiom of set theory that forces you to have an identity element in a set.
Oh crap, I read the title as "group theory"

I see where you're coming from. would be any value in the Universe.
0
#8
Ah I also forgot to mention that and are binary relations on some set .

Is it still the same principle, though?
0
4 years ago
#9
(Original post by r3l3ntl3ss)
Ah I also forgot to mention that and are binary relations on some set .

Is it still the same principle, though?
The thing is now that you say it's a binary relation (which I assume is symmetric), and defined such that , then is correct.

Consider , and so . And you can show the converse showing that .
0
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