P(N) as a topological space

[SlipEternal]

So, given onemachine's question, I began thinking, what would the order topology on \(\displaystyle \mathcal{P}(\mathbb{N})\) look like? The order topology is the topology generated by the subbasis: \(\displaystyle S = \left\{\left\{A\in \mathcal{P}(\mathbb{N})\mid A \subseteq B\right\} \mid B \in \mathcal{P}(\mathbb{N})\right\} \cup\) \(\displaystyle \left\{\left\{A\in \mathcal{P}(\mathbb{N})\mid B \subseteq A\right\} \mid B \in \mathcal{P}(\mathbb{N})\right\}\).

After all, based on onemachine's problem, \(\displaystyle f:\mathbb{R} \to \mathcal{P}(\mathbb{N})\) with the property that for all \(\displaystyle x<y, f(x) \subsetneq f(y)\) is a homeomorphism from \(\displaystyle \mathbb{R}\) onto its image (where the image has the order topology induced by \(\displaystyle \subsetneq\)). This implies that the image is homeomorphic to a line. So, what if we view all of \(\displaystyle \mathcal{P}(\mathbb{N})\) as lines?

Let's redefine everything.
Let \(\displaystyle g: \mathbb{R} \to \mathcal{P}(\mathbb{Q})\) define the function \(\displaystyle g(x) = (-\infty,x)\cap\mathbb{Q}\).
Let \(\displaystyle h: \mathbb{Q} \to \mathbb{N}\) define some bijection (pick your favorite one).
Let \(\displaystyle f: \mathbb{R} \to \mathcal{P}(\mathbb{N})\) define the function \(\displaystyle f(x) = \left\{h(r) \mid r \in g(x)\right\}\).

This is essentially the same formulation that onemachine had. I just wanted it repeated here for clarity.

Next, let \(\displaystyle P = S_\mathbb{N}\). For each \(\displaystyle \sigma \in P\), define the following notations: \(\displaystyle \sigma f(x) := \left\{\sigma(n)\mid n\in f(x)\right\}\), \(\displaystyle \sigma f(\mathbb{R}) := \left\{\sigma f(x)\mid x\in\mathbb{R}\right\}\).

Does there exist \(\displaystyle P' \subseteq P\) with the following two properties:
1. \(\displaystyle \bigcup_{\sigma \in P'}{\sigma f(\mathbb{R})} = \mathcal{P}(\mathbb{N})\)
2. For each \(\displaystyle A \in \mathcal{P}(\mathbb{N})\), there exists \(\displaystyle x \in \mathbb{R}\) such that for each \(\displaystyle \sigma \in \left\{\sigma \in P' \mid A \in \sigma f(\mathbb{R})\right\}, (\sigma f)(x) = A\).

 

[daon2]

You could write \(\displaystyle P=\text{Aut}(\mathbb{N})\) or \(\displaystyle P = S_{\mathbb{N}}\) (as the infinite symmetric group) Maybe I am confused, but did you want to write \(\displaystyle P=\text{Aut}(\mathcal{P}(\mathbb{N}))\)?

 

[SlipEternal]

You could write \(\displaystyle P=\text{Aut}(\mathbb{N})\) or \(\displaystyle P = S_{\mathbb{N}}\) (as the infinite symmetric group) Maybe I am confused, but did you want to write \(\displaystyle P=\text{Aut}(\mathcal{P}(\mathbb{N}))\)?

The general idea is: \(\displaystyle \lim_{x \to -\infty}f(x) = \emptyset, \lim_{x \to \infty}f(x) = \mathcal{P}(\mathbb{N})\). So, the sets are essentially "growing" as \(\displaystyle x\) moves from negative to positive infinity. Automorphisms on \(\displaystyle \mathcal{P}(\mathbb{N})\) could in theory permute the full set with a finite set. They are more dangerous than the automorphisms I am currently looking at. I am hoping I can use the smaller group of automorphisms, but I have no clue how I would prove one way or the other whether it is possible.

Edit: I see the ambiguity. I am fixing it now.

 

[SlipEternal]

I fixed the notation up a little. I decided that I don't know nearly enough group theory to even begin making a dent on this problem. My initial intuition is that those two conditions will be satisfied by \(\displaystyle P' = P\), but I have no idea how to prove that, nor would I be surprised if that were not true.