Prove that the relation $\subseteq$ over $℘(\mathbb{N})$ is not well-founded.

Question

I want to prove the following theorem:

Prove that the relation $\subseteq$ over $℘(\mathbb{N})$ is not well-founded.

I understand that to prove this we need to find a nonempty set $S \subseteq p(\mathbb{N})$, where for each $x \in S$, we can find $y \in S$, where $y \subseteq x$. I think that the main thing why I have difficulties with proving the theorem is that I don't have an intuition of why this is true. Like my brain doesn't believe the theorem is true, therefore I can't even know how to proceed. These are my thoughts which lead me to doubt about the theorem:

1. Let $S \subseteq p(\mathbb{N})$ be set, which has a minimal element $x$.
2. Then we can find a set, where $x$ is not a minimal any more by creating $S_n$ = $S \cup \{ x_n \}$, where $x_n$ is the set $x$ without any one element.
3. So like it seems that $x_n$ is a minimal in $S_n$.
4. And I feel that we can do this forever until $x_n$ is empty set, which is definitely minimal.

I know this can't be proof that the original theorem is false and I don't try to do this. I just want to build an intuition around the theorem, just to allow my brain to be comfortable with it and don't think that this is impossible. Could you please provide any hints, any thoughts so that I understand why this can be possible and will be able to go to the right direction in my proof.

Answer 1

The point is that if you do this "forever" starting from an $x$ that is infinite, you don't actually reach the empty set! The empty set might be what you would reach if you did it "forever and then one more time", taking the intersection of all the sets you got. But if you just iterate your process infinitely many times but don't do anything afterwards, there will be no last element of your set and thus no minimal element.

Explicitly, starting with $$S_0=\{\mathbb{N}\},$$ you could then take $$S_1=\{\mathbb{N},\mathbb{N}\setminus\{0\}\},$$ $$S_2=\{\mathbb{N},\mathbb{N}\setminus\{0\},\mathbb{N}\setminus\{0,1\}\},$$ and so on. If you then let $S$ be the union of all these sets $S_n$, then $$S=\{\mathbb{N},\mathbb{N}\setminus\{0\},\mathbb{N}\setminus\{0,1\},\dots\}$$ will not actually contain the empty set as an element (or any other minimal element), even though the intersection of all of its elements is empty.

(Incidentally, even if you did this process "forever and one more time", you might still not reach the empty set! For instance, in the example above, if instead of removing all of the natural numbers one by one, you only remove the even numbers, then the intersection of all the sets in $S$ would be the set of odd numbers, not the empty set.)