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?

Can you please specify what definition of a set are you using? — Prasun Biswas, Oct 18 '17 at 03:11
This question is exercise 1.2 from Jech's third edition of Set Theory. — MathematicalPhysicist, Oct 18 '17 at 03:12
Also, $Y\subset X$ doesn't imply $Y\in X$. What it does imply is that $Y\in P(X)$. — Prasun Biswas, Oct 18 '17 at 03:12
I think it would be better if you precisely write out the definition of a set you are using rather than mentioning the book because I don't have the book you mention currently. — Prasun Biswas, Oct 18 '17 at 03:14
@PrasunBiswas I meant if the above assertion were not true then if $Y\in P(X)$ then $Y\in X$ where $X$ is a set. — MathematicalPhysicist, Oct 18 '17 at 03:14
If $P(X)$ stands for the power set of $X$, then element in $P(X)$ are not even points in $X$... — 3x89g2, Oct 18 '17 at 03:14

score 6 · Answer 1 · answered Oct 18 '17 at 03:38

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 $R\in R\leftrightarrow R\notin R.$

score 3 · Accepted Answer · answered Oct 18 '17 at 03:27

3

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...

answered Oct 18 '17 at 03:27

Graham Kemp

129,094

2

Note that this contradiction makes use of the axiom of foundation: https://en.wikipedia.org/wiki/Axiom_of_regularity – Christian Sykes Oct 18 '17 at 03:31
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@ChristianSykes that's what was missing for me, in Jech's book he doesn't say explcitly in chapter 1 what this axiom means. – MathematicalPhysicist Oct 18 '17 at 03:38

score 0 · Answer 3 · answered Oct 18 '17 at 03:43

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

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

3 Answers3

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