I am looking at some beginner set theory proofs in this online text.
One question I have is on exercise 6 (1.3.6) on page 2):
Prove that $\mathcal P(X) \subset X$ is false for any $X$.
The given answer is basically as follows:
Proof. Let $X$ be an arbitrary set; then there exists a set
$$Y = \{ u \in X : u \notin u \}.$$
Obviously, $Y\subset X$, so $Y\in\mathcal P(X)$, by Axiom of Power Set. If $Y\in X$, then we have $Y\in Y \iff Y \notin Y$ (a contradiction). This
proves that $\mathcal P(X) \not\subset X$.
I understand how one arrives at a contradiction if one assumes
that $Y\in X$. But does $Y$ necessarily have to be an element of $X$?
What about the case in which
$$Y \notin X?$$
Don't we have to show a contradiction for that too?
Best Answer
Hint Consider that you have defined $Y $ for a reason. Notably to show there is a subset of $\mathcal {P}(X) $ which is not an element of $X$.