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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?)

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  • $\begingroup$ 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
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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).

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  • $\begingroup$ 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
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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.

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