Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Dec 2nd 2011, 05:10 PM
Nombredor
If I have ((A ∩ B) ∪ Ω), is that equivalent to Ω ?
• Dec 2nd 2011, 05:19 PM
Plato
Quote:

Originally Posted by Nombredor
If I have ((A ∩ B) ∪ Ω), is that equivalent to Ω ?

Well of course, if \$\displaystyle X\subset Y\$ then \$\displaystyle X\cup Y=Y~.\$
All set are subsets of \$\displaystyle \Omega\$.
• Dec 2nd 2011, 05:29 PM
Nombredor
And I am right in saying that (A -B) ∪ (B - A) is equivalent to (A ∪ B) - (A ∩ B) ?

And is it accurate to say that (A ∩ Ω) ∪ ~B is the same as Ω - B ?
• Dec 2nd 2011, 05:42 PM
Plato
Quote:

Originally Posted by Nombredor
And I am right in saying that (A -B) ∪ (B - A) is equivalent to (A ∪ B) - (A ∩ B) ?

That is correct.

Quote:

Originally Posted by Nombredor
And And is it accurate to say that (A ∩ Ω) ∪ ~B is the same as Ω - B ?

If ~B means \$\displaystyle B^c\$, the complement of \$\displaystyle B\$ then no.
\$\displaystyle (A\cap\Omega )\cup B^c=A\cup B^c\$ which is not necessarily \$\displaystyle B^c~.\$
• Dec 2nd 2011, 05:45 PM
Nombredor
~B means ''non B'' in this case.
• Dec 2nd 2011, 05:49 PM
Plato
Quote:

Originally Posted by Nombredor
~B means ''non B'' in this case.

"non B" is not a set theory term.
B complement is the set of all elements not in B.
• Dec 2nd 2011, 05:52 PM
Nombredor
Quote:

Originally Posted by Plato
\$\displaystyle (A\cap\Omega )\cup B^c=A\cup B^c\$ which is not necessarily \$\displaystyle B^c~.\$

How do they look different on a Venn diagram ?
• Dec 2nd 2011, 06:14 PM
Nombredor
Also, I was wondering whether I was right in saying that (A ∪ B) ∩ (A ∪ ~B) is equivalent to (A - B) ?
• Dec 2nd 2011, 06:24 PM
Annatala
The only thing I've ever seen big-omega used for in set theory is as another name for omega-1 (first uncountable ordinal). Does it also mean "universal set" in the context of intro set theory, then?
• Dec 2nd 2011, 06:27 PM
Nombredor
Quote:

Originally Posted by Annatala
Does it also mean "universal set" in the context of intro set theory, then?

Yes, that's what I meant it as. I've also seen people use just ''u'', but I prefer omega because ''u'' also looks a lot like the reunion symbol.
• Dec 2nd 2011, 06:27 PM
Nombredor
double post
• Dec 2nd 2011, 06:58 PM
Amer
Quote:

Originally Posted by Nombredor
Also, I was wondering whether I was right in saying that (A ∪ B) ∩ (A ∪ ~B) is equivalent to (A - B) ?

it is correct
• Dec 3rd 2011, 05:55 AM
Plato
Quote:

Originally Posted by Nombredor
Also, I was wondering whether I was right in saying that (A ∪ B) ∩ (A ∪ ~B) is equivalent to (A - B) ?

No that is not correct.
Let \$\displaystyle \Omega =\{1,2,3,4,5,6,7,8,9\},~A=\{1,2,3,4\}~\&~B=\{2,4,6 ,8\}\$.

Find \$\displaystyle (A\cup B)~\&~(A\cup B^c)\$.

What is their intersection? What is \$\displaystyle A-B~?\$
• Dec 3rd 2011, 07:28 AM
Amer
it is equal to A
• Dec 3rd 2011, 07:37 AM
Plato
Quote:

Originally Posted by Amer
it is equal to A

Correct. So the OP
Quote:

Originally Posted by Nombredor
I was right in saying that
(A ∪ B) ∩ (A ∪ ~B) is equivalent to (A - B) ?

Is incorrect.
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last