I am working on a problem with the following set:

$$S = \{\varnothing,\{\varnothing\}\}$$

My solution was: $P(S) = \{\varnothing, \{\varnothing\},\{ \varnothing , \{ \varnothing \}\}\}$, but the solution on the textbook shows: $P(S) = \{ \varnothing , \{ \varnothing \}, \{\{\varnothing\}\}, \{ \varnothing , \{ \varnothing \}\}$

Where does the $\{\{\varnothing\}\}$ come from?


  • 11
    $\begingroup$ Pretend that $S = \{ A, B \}$ and then substitute afterwards. $\endgroup$ – Qiaochu Yuan Sep 25 '11 at 18:48

From the fact:

$S=\{a,b\}$ where $a=\varnothing, b=\{\varnothing\}$.

Now $P(S)=\{\varnothing,\{a\},\{b\},\{a,b\}\}$. Plug in the values of $a,b$, and you have $\{b\} = \{\{\varnothing\}\}$.

The general rule of thumb (which you might find out later is actually a theorem) is that if you have $S$ with $k$ many elements then it has $2^k$ many subsets. If you ended up with a power set which has only three elements, you definitely missed one.

  • $\begingroup$ Thanks a lot, I got confused with the Ø at the beginning, thinking it was a from S = {a,b} $\endgroup$ – AlexBrand Sep 25 '11 at 19:07

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