Quick question about the empty set and the power set

• November 17th 2011, 05:18 PM
Robinator
Quick question about the empty set and the power set
Let's say we have a set A = {1, 2, 3, 4}

I understand that, for example, $\left\{{2, 3}\right\}\in P(A)$, but why is it that $\left\{{\varnothing}\right\}\subseteq P(A)$?
• November 17th 2011, 05:26 PM
TheChaz
Re: Quick question about the empty set and the power set
Take a subset S of A.
Then this subset is an element of P(A).
S is in P(A)
For a set to be a subset of P(A), each of its elements must be in P(A).
Ie each of its elements must be a subset of A.
The empty set is a subset of A.
The set containing the empty set is therefore a subset of P(A)
• November 17th 2011, 05:34 PM
Robinator
Re: Quick question about the empty set and the power set
So if I took the set {{2, 3}} it becomes a subset of P(A) rather than an element, correct? I think I understand, thank you.
• November 18th 2011, 08:37 AM
MoeBlee
Re: Quick question about the empty set and the power set
Quote:

Originally Posted by Robinator
Let's say we have a set A = {1, 2, 3, 4}

I understand that, for example, $\left\{{2, 3}\right\}\in P(A)$, but why is it that $\left\{{\varnothing}\right\}\subseteq P(A)$?

Is every member of {0} a member of PA? Yes, since 0 is the only member of {0} while 0 is in PA. So {0} is a subset of PA.
• November 18th 2011, 08:40 AM
MoeBlee
Re: Quick question about the empty set and the power set
Quote:

Originally Posted by Robinator
So if I took the set {{2, 3}} it becomes a subset of P(A) rather than an element, correct?

Yes, {{2 3}} is a subset of PA, and {{2 3}} is a member of PPA, but {{2 3}} is not a member of PA.