I' trying to construct the power set of $A = \{\emptyset, \{a\}\}$ and would appreciate some help.

Now the definition of a power set says that it's the set of all possible subsets of a given set. Normally, this means that no matter what the set, $\emptyset$ is always included in the power set. How should this be handled here? If I say that $\emptyset$ is in the power set $A$, then I observe that $\emptyset \in A$ and start wondering if I've made a mistake (as $\emptyset$ is not the subset of $A$ - or is it?)

  • $\begingroup$ By $\phi$, do you mean the empty set $\varnothing$? The empty set is a subset of every set. $\endgroup$ – Henning Makholm Jul 21 '14 at 14:59
  • $\begingroup$ {$\varnothing$} is also a subset of A. There two other subsets. $\endgroup$ – Mick Jul 21 '14 at 15:06
  • $\begingroup$ Actually, the third element of the power set is $\{\{a\}\}$. $\endgroup$ – A Bajaj Jul 21 '14 at 15:10

If $B=\{1,2\}$, then the set of all possible subsets of $B$ is $\{\varnothing,\;B, \;\{1\},\;\{2\}\}$.

Analogously, if $B$ is any set with two distinct elements, say $B=\{x,y\}$, then the set of all possible subsets of $B$ is $\{\varnothing,\;B, \;\{x\},\;\{y\}\}$.

Now, notice that your set $A$ is the set $B$ with $x=\varnothing$ and $y=\{a\}$. So, the power set of $A$ is $\mathcal{P}(A)=\{\varnothing,\;A,\;\{\varnothing\},\;\{\{a\}\}\}$.

As you have pointed out, always $\varnothing\in \mathcal{P}(A)$. But, in that specific case, we also have $\{\varnothing\}\in \mathcal{P}(A)$.

Remark 1: $\varnothing$ and $\{\varnothing\}$ are not the same thing.

Remark 2: If a set $S$ has $n$ elements, then the power set of $S$ has $2^n$ elements. That result can help you to construct power sets (particularly, for do not forget any subset).

  • $\begingroup$ Very nice and detailed answer. Thanks a lot! Looks like the empty set is not as simple as it seems. ^.^ $\endgroup$ – ankush981 Jul 22 '14 at 4:09

The power set should be $\{\phi , \{\phi \}, \{\{a\}\}, A\}$.

The definition of subset is $A \subseteq B$ iff every element of A is an element of B.

So yes $\phi$ is a subset as well as $\{\phi\}$. Also don't forget A itself.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.