Why do we show in this way that it is not the image of the function?

[evinda]

Hello! (Wave)

Theorem:
No set is equinumerous with its power set.

Proof:

Let $A$ be a set. We want to show that if $f: A \to \mathcal{P}A$ (a random function) then $f$ is not surjective.We define the set $D=\{ x \in A: x \notin f(x)\}$ and obviously $D \in \mathcal{P}A$.

We assume that there is a $a \in A$ such that $f(a)=D$.Then we have:

$$a \notin D \leftrightarrow a \notin f(a) \leftrightarrow a \in D, \text{ contradiction}$$

Therefore for each $x \in A, f(x) \notin D$, i.e. $f$ is not surjective.Could you explain me the proof from the point where we assume that there is an $a \in A$ such that $f(a)=D$?We have show that $D \in \mathcal{P}A$ and we want to show that $D \notin f(A)$ in order to show that $\mathcal{P}A$ isn't the image of $f$.
Why do we do it like that? (Worried)

 

[evinda]

I got it.. We want to show that $D \notin f(A)$.
So we suppose that $D \in f(A)$. That means that there is an $a \in A$ such that $f(a)=D$.

$$a \in D \leftrightarrow a \notin f(a) \leftrightarrow a \notin D$$

So we have found a contradiction..