Use Mathematical Induction to show that for any set $S$, If $\vert S\vert = n$, then $\vert\mathcal P(S)\vert= 2^n$.
How would I approach this question? What would the base case be when I apply it to the left hand side?
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1The base case is $n=0$. – Oct 30 '20 at 09:41
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I understand that the base case is n=0, but how would I apply that to the power set? How would you use the 0 to prove a case? – nugh Oct 30 '20 at 09:42
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3For $n=0$, $S=\emptyset$ and therefore $\mathcal P(S)={\emptyset}$, which has $1$ element. – Oct 30 '20 at 09:43
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Then use induction as you suggested. – Plussoyeur Oct 30 '20 at 09:45
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5Does this answer your question? Prove that the power set of an $n$-element set contains $2^n$ elements – Chubby Chef Oct 30 '20 at 10:00
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related: https://math.stackexchange.com/questions/546414/what-is-the-proof-that-the-total-number-of-subsets-of-a-set-is-2n – halrankard2 Oct 30 '20 at 10:00