4
$\begingroup$

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?

Thanks

$\endgroup$
1
  • 11
    $\begingroup$ Pretend that $S = \{ A, B \}$ and then substitute afterwards. $\endgroup$ – Qiaochu Yuan Sep 25 '11 at 18:48
9
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

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.