1 thought on “prove that |x| {}^{2} = x {}^{2}∣x∣ <br />2<br /> =x <br />2for any number x”
Answer:
Theorem. Let X denote an arbitrary set such that |X|=n. Then |P(X)|=2n.
Proof. The proof is by induction on the numbers of elements of X.
For the base case, suppose |X|=0. Clearly, X=∅. But the empty set is the only subset of itself, so |P(X)|=1=20.
Now the induction step. Suppose |X|=n; by the induction hypotesis, we know that |P(X)|=2n. Let Y be a set with n+1 elements, namely Y=X∪{a}. There are two kinds of subsets of Y: those that include a and those that don’t. The first are exactly the subsets of X, and there are 2n of them. The latter are sets of the form Z∪{a}, where Z∈P(X); since there are 2n possible choices for Z, there must be exactly 2n subsets of Y of which a is an element. Therefore |P(Y)|=2n+2n=2n+1. □
Answer:
Theorem. Let X denote an arbitrary set such that |X|=n. Then |P(X)|=2n.
Proof. The proof is by induction on the numbers of elements of X.
For the base case, suppose |X|=0. Clearly, X=∅. But the empty set is the only subset of itself, so |P(X)|=1=20.
Now the induction step. Suppose |X|=n; by the induction hypotesis, we know that |P(X)|=2n. Let Y be a set with n+1 elements, namely Y=X∪{a}. There are two kinds of subsets of Y: those that include a and those that don’t. The first are exactly the subsets of X, and there are 2n of them. The latter are sets of the form Z∪{a}, where Z∈P(X); since there are 2n possible choices for Z, there must be exactly 2n subsets of Y of which a is an element. Therefore |P(Y)|=2n+2n=2n+1. □