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So I want to know if Hilbert's hotel "story" holds for this statement: $\wp (\mathbb{N}) \sim \wp (\mathbb{N})\smallsetminus \left \lbrace\emptyset\right\rbrace$

So, If the statement wasn't talking about the sets inside the power set, I could say it's exactly like Hilbert's hotel, so i'll just "move" one element and make room for the extra element that is not on the other (the empty set) But what I don't understand is how do I "move" a set down, because there is no order between the sets that are in the power set.

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    $\begingroup$ You don't have to enumerate the whole set. It suffices to choose a countable infinite subset and an enumeration thereof. $\endgroup$ – Zhen Lin Nov 30 '14 at 18:14
  • $\begingroup$ To be more specific about Zhen's comment: in this case since you have a convenient 'favored' set of subsets - the set $\{i\}: i\in\mathbb{N}$ - then you can run the 'hotel argument' on this set and say that your isomorphism just fixes all other elements of the power set. But note that some choice is inherent in the abstract argument, because without choice you can have things like amorphous sets (en.wikipedia.org/wiki/Amorphous_set) that can't be split into two infinite subsets. $\endgroup$ – Steven Stadnicki Nov 30 '14 at 18:20
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This time the hotel has rooms that are labeled not with single numbers, but with sets of numbers, possibly infinite sets. One room corresponds to the empty set, and has the label $\{\}$ on the door. One room has the label $\{5,23,119\}$ on the door. The Prime Presidential Penthouse Suite has every prime number on its door.

The hotel is full, with exactly one guest in each room.

Then the air conditioner breaks in the Empty Set Room $\{\}$. M. and Mme. Hilbert must find a new room for the guest in that room.

They move that guest to the room marked $\{1\}$.

They move the guest in the room $\{1\}$ to the room $\{1,2\}$.

They move the guest in the room $\{1,2\}$ to the room $\{1,2,3\}$.

They move the guest in the room $\{1,2,3\}$ to the room $\{1,2,3,4\}$.

(and so on)

Guests in the other rooms, including the $\{5,23,119\}$ room and the Prime Presidential Penthouse Suite, do not move. They keep their old rooms. No problem!


Note that there is nothing special about the proposed solution. Let $\def\S{\mathscr S}\S$ be any countable subfamily of $\wp(\Bbb N)\setminus\{\emptyset\}$, so we have $\S = \{S_1, S_2, S_3, \ldots\}$. Then move the guest from room $\{\}$ into room $S_1$, the guests in each $S_i$ into $S_{i+1}$, and leave the other guests where they are. The solution I described above has $S_i = \{1, \ldots i\}$, but any choice of distinct nonempty $S_i$ suffices.

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  • $\begingroup$ Thanks, Is it possible to define (foramlly) a bijection $f:\wp (\mathbb{N}) \rightarrow \wp (\mathbb{N})\smallsetminus \left \lbrace\emptyset\right\rbrace$ the will suffice the rules you stated? how? $\endgroup$ – wannabe programmer Dec 1 '14 at 16:53
  • $\begingroup$ The part below the horizonal line does exactly that. $\endgroup$ – MJD Dec 1 '14 at 16:57
  • $\begingroup$ I,m not sure about the part where i need to prove onto and 1:1 $\endgroup$ – wannabe programmer Dec 1 '14 at 16:59
  • $\begingroup$ Let $R$ be a nonempty subset of $\Bbb N$. We want to show that there is exactly one $D$ such that $f(D) = R$. If $R\in\S$, then $R= S_i$ for exactly one $i$. If $i>1$ then $R= f(S_{i-1})$, and if $i=1$ then $R=f(\emptyset)$. On the other hand, if $R\notin\S$ then $R=f(S)$. $\endgroup$ – MJD Dec 1 '14 at 17:03
  • $\begingroup$ Sorry, that last formula should have been $R=f(R)$. $\endgroup$ – MJD Dec 1 '14 at 17:12

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