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I am working with Power Sets and I am stumped conceptually on a problem that looks as such: A = {a, {a}}.

Find P(P(A)).

I am under the impression the Power set of A would be: {∅ , a}

and then...

P(P(A)) would be: {{∅ }, {a}, {{∅ }} {a, ∅ }}

Which would resulting in a cardinality of 4.

anothercoderbro
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    Welcome to Mathematics Stack Exchange. Note that, if $A$ has $2$ elements, its power set should have $2^2=4$. (You're missing a couple elements) – J. W. Tanner Feb 03 '20 at 20:10
  • first one is 4, second one is 16. –  Feb 03 '20 at 20:11
  • Thank you for the welcome @J. W. Tanner. I thought they would be considered to be the same element, thus not counted again for Power Set. – anothercoderbro Feb 03 '20 at 20:13

1 Answers1

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Hint:

Try to instead look at finding $\mathcal{P}(\mathcal{P}(\{a,b\}))$ and after you are done, replace $b$ with $\{a\}$.

Additional hint:

So, you should be well familiar with that $\mathcal{P}(\{a,b\}) = \{\emptyset, \{a\},\{b\},\{a,b\}\}$. If you still struggle with finding the power set of that result, then try doing something similar as before... naming these something else. You should be able to find $\mathcal{P}(\{u,v,x,y\})$ and then after you are done, replace $u$ with $\emptyset$, replace $v$ with $\{a\}$, replace $x$ with $\{b\}$, and replace $y$ with $\{a,b\}$ (and then finally replace $b$ with $\{a\}$ at the end)

JMoravitz
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  • so {a} and {{a}} are different then in this context? I was under the impression they were the same element, thus the confusion. – anothercoderbro Feb 03 '20 at 20:11
  • they are different – J. W. Tanner Feb 03 '20 at 20:12
  • @anothercoderbro Absolutely. ${a}$ is not equal to ${{a}}$. Just like how $\emptyset$, a zero element set, is not equal to ${\emptyset}$ which is a set with one element (that one element being the empty set, but no less deserving of being considered an element than anything else) (which is a strong reminder of another mistake that you seem to have been making) – JMoravitz Feb 03 '20 at 20:13
  • a is an element, {a} is the set with 1 element which is a –  Feb 03 '20 at 20:13
  • Okay, I see what you are saying. Thank you @JMoravitz. – anothercoderbro Feb 03 '20 at 20:17