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

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The discussion centers on the proof that no set is equinumerous with its power set, focusing on the function f from a set A to its power set P(A). The key argument involves defining a set D that contains elements of A not included in their corresponding f images. By assuming there exists an element a in A such that f(a) equals D, a contradiction arises when analyzing the membership of a in D. This contradiction demonstrates that D cannot be an image of f, proving that f is not surjective and thus establishing that the power set cannot be equinumerous with the original set. The proof effectively illustrates the fundamental concept of cardinality in set theory.
evinda
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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)
 
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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..
 

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