There's no set $X$ such that $P(X)\subset X$.

Question

I want to show the above assertion, I guess I am not sure what's the definition of a set, I thought it's the basic building block.

Anyway, if the above titled assertion weren't true then for every $Y \subset X$ we will also have $Y \in X$, which should contradict $X$ being a set, but why is that?

Answer 1

If $\mathcal P(X)\subset X$ then $\forall Y:(Y\subseteq X\to Y\in X)$.   That is, all subsets of $X$ would be elements of $X$.

Now consider that $X$ is an element of the powerset of $X$ (ie $X\subseteq X$).   So...

Answer 2

If you know that no set can be an element of itself, then you can show that $P(X)$ is not a subset of $X$ by simply observing that $X\in P(X)$ while $X\notin X.$

If you don't know that, then you can still prove that $P(X)$ is not a subset of $X$ by using Bertrand Russell's trick: define the set $R=\{x\in X:x\notin x\},$ the set of all elements of $X$ which are not elements of themselves. (If no set is an element of itself, then $R=X$.) Then $R\in P(X)$ but $R\notin X,$ since assuming $R\in X$ leads to the contradiction $R\in R\leftrightarrow R\notin R.$

Answer 3

So forgive my english and my informal definition of a set

A set is a group of elements. So for example A = {1,2,3} is a set with three elements, which are 1, 2 and 3

Now, P(A) is called the set of parts of A. This set is the set of all possible subsets of A. In this case, P(A) = {Ø,{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}

So the important thing to note here is that the elements in P(A) are subsets themselves, and not the original elements in A. It is important to understand what the elements of a set are. I'd love to give a detailed explanation but I lack in that department

Assuming you get what I stated above, to prove what you want to prove is really easy. Suppose that P(A) is a subset of A. Then EVERY element in P(A) is also an element of A. This quickly leads to a contradiction, since while A had numbers as its elements, P(A) has subsets of those numbers as its elements.

A quick way of proving this as absurd is noting that $A \in P(A)$. So if P(A) was a subset of A then $A \in A$ which is absurd