Clarification on theorem: no function to map A onto its power set P(A) In the proof of the theorem "For any set A, there does not exist a function mapping A onto its power set P(A)", there's a sentence (highlighted) that I couldn't follow. Contrary to what the illustration says, clearly {1, 3} comes from elements of A. Was it a typo or I missing something?

 A: You are confusing what the author means by "come from".
The author means  $f(1) = \{2\}$ so the output of the function, $\{2\}$ "comes from" the input of $1$.
$f(2) = \{2,3\}$ so the output of the function, $\{2,3\}$ "comes from" the input of $2$.
And $f(3) = \{1,2\}$ so the output of the function, $\{1,2\}$ "comes from" the input of $3$.
.... So what input, $x \subset A$ will give you $f(x) = \{1,3\}$?  
The answer is:  None!  There are only three possible inputs: 1,2,3; but there are eight possible outputs:  $\emptyset, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\},\{2,3\}, \{1,2,3\}$.  
$5$ of them do not "come from" any element of $A$: $\emptyset, \{1\},  \{3\}, , \{1,3\}, \{1,2,3\}$.
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This picture is if we define $B = \{$ then elements $a\in A$ so that $a \not \in f(a)\}$ then $1\not \in \{2\}= f(1)$ and $2\in \{2,3\}=f(2)$ and $3\not \in \{1,2\}$ so $B = \{1,3\}$.
Notice that $f(a) = B$ is impossible.   If $a \in B = f(a)$ then by definition $a \not \in B$.  But if $a \not \in f(a) = B$ then by definition $a \in B$.
So it is not possible to have have a function from $A\to P(A)$ where there is an $a \in A$ and $f(a) = B= \{$ then elements $a\in A$ so that $a \not \in f(a)\}$.
So $f$ can never be onto.
.....
You may think what if $B$ is empty and all elements are members of the sets they are mapped to.
Say $f(1) = \{1,2\};  f(2) = \{2,3\}, f(3)=\{1,2,3\}$.
Then $1 \in f(1); 2\in f(2); 3\in f(3)$ and $B= \{$ then elements $a\in A$ so that $a \not \in f(a)\}=\emptyset$
But .... what is $a$ where $f(a) = \emptyset$.  
Nothing.  There is not $f(a) = \emptyset$
And if you did have $f(a) = \emptyset$ then $a\in B$ because $a\not \in f(a) = \emptyset$.  So $B\ne \emptyset$ because $a\in B$.
[Example:  $f(1) = \emptyset$;  $f(2)=\{2,3\}; f(3) =\{1,3\}$ then $1 \not \in \emptyset= f(1)$ but $2 \in \{2,3\}$ so $2 \in f(2)$ and $3 \in f(3) = \{1,3\}$ so $B = \{1\}\ne \emptyset=f(1)$. 
And there $\{1\}$ doesn't "come from" $1$.
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No matter what set $A$ is, even if it is infinite, having an $a\in A$ so that $f(a) = B$ is impossible by logic.  If $f(a) = B$ and $a\in B$ then $a \in f(a)$ and $a\not \in B$.  ANd if $a\not \in B$ then $a\not \in f(a)$ and $a \in B$.
