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First show that $|A| \leq |P(A)|$. Define the function $g: A\to P(A), g(a)=\{a\}$. Since inequality $a\ne b$ implies inequality $\{a\}\ne \{b\}$, that is $g(a)\neq g(b)$$g$ is one-to-one. Therefore, $|A| \leq |P(A)|$.

Show that we can't have $|A|=|P(A)|.$ Assume not, i.e., suppose that in fact $|A|=|P(A)|.$ Then there exists a bijection $f: |A|\to |P(A)|.$ Let $B=\{a\in A|a\notin f(a)\}\in P(A)$.

Since f is onto, there exists an $a_0\in A$ such that $f(a_0) =B$. Then we have the following.

If $a_0\in B$, then by definition of $B$$a_0\notin f(a_0)$, i.e. $a_0\notin B$. If $a_0\notin B$, then $a_0\in f(a_0)$, i.e. $a_0\in B$.

This contradiction prove that $|A|<|P(A)|.$

First show that $|A| \leq |P(A)|$. Define the function $g: A\to P(A), g(a)=\{a\}$. Since inequality $a\ne b$ implies inequality $\{a\}\ne \{b\}$, that is $g(a)\neq g(b)$$g$ is one-to-one. Therefore, $|A| \leq |P(A)|$.

Show that we can't have $|A|=|P(A)|.$ Assume not, i.e., suppose that in fact $|A|=|P(A)|.$ Then there exists a bijection $f: |A|\to |P(A)|.$ Let $B=\{a\in A|a\notin f(a)\}\in P(A)$.

Since f is onto, there exists an $a_0\in A$ such that $f(a_0) =B$. Then we have the following.

If $a_0\in B$, then by definition of $B$$a_0\notin f(a_0)$, i.e. $a_0\notin B$. If $a_0\notin B$, then $a_0\in f(a_0)$, i.e. $a_0\in B$.

This contradiction prove that $|A|<|P(A)|.$

First show that $|A| \leq |P(A)|$. Define the function $g: A\to P(A), g(a)=\{a\}$. Since inequality $a\ne b$ implies inequality $\{a\}\ne \{b\}$, that is $g(a)\neq g(b)$$g$ is one-to-one. Therefore, $|A| \leq |P(A)|$.

Show that we can't have $|A|=|P(A)|.$ Assume not, i.e., suppose that in fact $|A|=|P(A)|.$ Then there exists a bijection $f: |A|\to |P(A)|.$ Let $B=\{a\in A|a\notin f(a)\}\in P(A)$.

Since f is onto, there exists an $a_0\in A$ such that $f(a_0) =B$. Then we have the following.

If $a_0\in B$, then by definition of $B$$a_0\notin f(a_0)$, i.e. $a_0\notin B$. If $a_0\notin B$, then $a_0\in f(a_0)$, i.e. $a_0\in B$.

This contradiction prove that $|A|<|P(A)|.$