Answer by user65203 for Prove that the power set of an $n$-element set...
Alternative proof:Represent a subset of $X$ as a binary number such that its $k^{th}$ bit is set if and only if the $k^{th}$ element is taken.E.g., $\{a,b,c\}\to111$, $\{a,c\}\to101$, $\{c\}\to001$,...
View ArticleAnswer by Little Green Man for Prove that the power set of an $n$-element set...
Here is another one that uses the Binomial Theorem.Take, for example, the set $A = \{1, 2, 3, 4\}$Then, $P(A) = \{∅, \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{2, 4\}, \{3, 4\}, \{4,...
View ArticleAnswer by Luke Paluso for Prove that the power set of an $n$-element set...
I will prove the claim by induction. First suppose $|A|=1$. That is, $A=\{a_1\}$. Thus, $P(A) =\{ \{a\}, \{ \emptyset \} \}$. So, it can be concluded that $|A|=2^1=2$. My inductive hypothesis is that...
View ArticleAnswer by Amey Joshi for Prove that the power set of an $n$-element set...
Here is an another, combinatorial proof. Let $X$ be a set with $n$ elements. To form a subset of $X$, we go over each element of $X$ and exercise a choice of whether or not to include in the subset....
View ArticleAnswer by Frunobulax for Prove that the power set of an $n$-element set...
Suppose you've already shown that $X=\{1,2\}$ has $2^2=4$ subsets, namely ${\cal P}(X)=\{\emptyset,\{1\},\{2\},X\}$. Now you add a new element $a=3$ to get $Y=X\cup \{a\}=\{1,2,3\}$. The four subsets...
View ArticleAnswer by N. F. Taussig for Prove that the power set of an $n$-element set...
You misread the proof. Since set $X$ has $n$ elements, the induction hypothesis tells us that $|\mathcal{P}(X)| = 2^n$. The set $Y = X \cup \{a\}$ has $n + 1$ elements. It subsets are either subsets of...
View ArticleProve that the power set of an $n$-element set contains $2^n$ elements
Theorem. Let $X$ denote an arbitrary set such that $|X|=n$. Then $|\mathcal P(X)|=2^n$.Proof. The proof is by induction on the numbers of elements of $X$.For the base case, suppose $|X|=0$. Clearly,...
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