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My question reads:

Let $M (\mathbb{N})= \{g\in {}^\mathbb{N}\mathbb{N} : g\text{ is a bijection}\}$. Prove that $M (\mathbb{N}) \sim \mathcal{P} (\mathbb{N})$.

Now, I was thinking of using the Cantor-Berstein Theorem, so I would need to define an injection in both directions. The one direction is straightforward, so I am okay with that part.

Then I wanted to say for the other direction we can define $f: \mathcal{P}$ ($\mathbb{N}$) $\to M (\mathbb{N}$) by for $X\in \mathcal{P}$ ($\mathbb{N}$), if $x\notin X$ then $f(2x)=2x$ and $f(2x+1)=2x+1$; if $x\in X$ then $f(2x)=2x+1$ and $f(2x+1)=2x$. Does this seem okay for this direction? Does this injection work?

Sam_U
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  • Convention question: are you assuming $0 \in \mathbb{N}$? – Theo Bendit Nov 01 '17 at 05:30
  • @TheoBendit Yes, sorry forgot to mention this. – Sam_U Nov 01 '17 at 05:36
  • Your $f$ is not a bijection for $X={3}$ – user340297 Nov 01 '17 at 05:38
  • @user340297 Why would it not work for 3? How could I fix my injection? I am using C-B thus I only want an injection. – Sam_U Nov 01 '17 at 05:39
  • @Sam_U $f(2)=f(3)=2$ so it's not injective and not surjective because $3$ has empty preimage. – user340297 Nov 01 '17 at 05:41
  • @Sam_U Not really, $f$ as you've defined in your post fixes every element that is not in $X$ and swaps odds and even numbers that are in $X$, but the largest and smallest odd/evens have problems not being mapped back into the set. $f(6)=6$ and $f(7)=7$ in your example – user340297 Nov 01 '17 at 05:46
  • Modifying your $f$ a bit works:

    Define $\phi:\mathcal{P}(\mathbb{N})\to M(\mathbb{N})$ by $\phi(X)=f_X$, as follows:

    $f_X(x)=x$ for all $x\notin X$,

    If $X$ is bounded above, order the element of $X$ in increasing order, $x_1<x_2<\cdots<x_n$ and $f_X(x_i)=x_{i+1}$ for all $i=1,2,\dots,n-1$, $f_X(x_n)=x_1$.

    If $X$ is unbounded, order elements in increasing order $x_1<x_2<\cdots$ and define $f(x_{2i+1})=f(x_{2i})$ and $f(x_{2i})=f(x_{2i+1})$ (swapping). You can check that this is an injection

     – user340297 Nov 01 '17 at 06:03
  • @user340297 Why do we need to consider boundedness in this case? We have not actually gone over bounds so I do not really understand this point. However, yes, I was trying to get at this idea of swapping the elements. – Sam_U Nov 01 '17 at 06:06
  • @Sam_U Problem with swapping is that when $X$ is a finite set (or equivalently bounded subset of naturals) with odd number of elements , there will always be one element left in $X$ that cannot be mapped to anywhere. This doesn't happen when $X$ is an infinite set – user340297 Nov 01 '17 at 06:09
  • I see that this is ([tag:proof-verification]) question, still it might be worth mentioning that there are a few questions around about this. For example, Cardinality of the set of bijective functions on $\mathbb{N}$? and other posts linked there. There is also this generalization posted on MO: Cardinality of the permutations of an infinite set. – Martin Sleziak Nov 02 '17 at 06:49
  • One problem with the notation in the way you wrote your proof is that you use $f$ in two different meanings. Once for a map from $\mathcal P(\mathbb N)$ to $M(\mathbb N)$ and then second time for what actually is the function $f(X)$. – Martin Sleziak Nov 02 '17 at 11:41
  • @AndrésE.Caicedo I thought that it might be useful to let you know that there is a question on meta related to tagging of this specific question: Is this elementary-set-theory or set-theory. – Martin Sleziak Nov 02 '17 at 12:07
  • @MartinSleziak Yes, I realized this was confusing and just editted the mistake. I wrote it down wrong. – Sam_U Nov 02 '17 at 15:52

1 Answers1

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Both directions will work, but in terms of working, I'd say that they're a little light (the second direction especially). There's nothing wrong with this if you're sparing the reader from tedious information that is obvious to both you and the reader, but if you're feeling less than certain about the veracity of your proofs, it helps to fill in the extra information.

For example, in your second proof, you could show that this proposed injection is an injection. You could suppose you have $X, Y \in \mathcal{P}(\mathbb{N})$, and show the resulting functions agree at each point.

Theo Bendit
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  • Is my first direction okay then? – Sam_U Nov 01 '17 at 05:57
  • I think both your directions are OK, user34029's concerns notwithstanding. I just think you should flesh them out if you feel under-confident. For the first direction, you could rely on the result from class to produce a bijection $\alpha : {^\mathbb{N}\mathbb{N}} \to \mathcal{P}(\mathbb{N})$ and let $\beta : M(\mathbb{N}) \to {^\mathbb{N}\mathbb{N}}$ be the inclusion map. Note why $\beta$ is injective, and hence why $\alpha \circ \beta$ is therefore injective. – Theo Bendit Nov 01 '17 at 06:35
  • I think for the first direction it would be enough to just use what we proved rather than give an explicit bijection, since I only want a injection because I am using the C-B Theorem. However, I am still not sure how to fix my second direction. I wrote it out and did a few examples of sets to see that it works and it seems to be fine. – Sam_U Nov 01 '17 at 06:45
  • Like I said, I think the second direction is fine. Essentially, you're grouping the natural numbers into consecutive pairs, and identifying elements of the subsets of $X$ by either swapping or not swapping the elements. – Theo Bendit Nov 01 '17 at 07:29
  • Well, what I had in mind was say $X$={$1$} then applying f we get that 0 goes to 0, 1 goes to 2, 2 goes to 1, 3 goes to 3, etc. – Sam_U Nov 02 '17 at 00:23
  • So, for the first direction what I have is okay, but if I wanted more detail I should give an explicit injection? – Sam_U Nov 02 '17 at 00:25
  • Yes, exactly. Prove all of your claims, and I think it'll all work out. – Theo Bendit Nov 02 '17 at 01:47
  • Okay, thank you. Then, to prove my second direction is an injection you suggested starting with two sets? – Sam_U Nov 02 '17 at 02:56
  • I am having some trouble showing that my second direction is an injection. – Sam_U Nov 02 '17 at 03:51
  • Let $f$ be the common map corresponding to sets $X$ and $Y$. We need to show $X \subseteq Y$ and $Y \subseteq X$. To prove the first, suppose $x \in X$, but for the sake of contradiction, $x \notin Y$. Since $x \in X$, we get $f(2x) = 2x$, but since $x \notin Y$, we also get $f(2x) = 2x + 1$, together implying $1 = 0$. – Theo Bendit Nov 02 '17 at 04:47
  • So, we begin our assumption by letting $X,Y\in P (\mathbb{N})$? – Sam_U Nov 02 '17 at 05:00
  • I am not following, don't we want to assume $f(X)=f(Y)$ and then show $X=Y$? – Sam_U Nov 02 '17 at 05:02
  • You might be getting yourself confused between the map from $\mathcal{P}(\mathbb{N})$ to ${^\mathbb{N}\mathbb{N}}$, and the map (which, in the question, you refer to as $f$), from $\mathbb{N}$ to $\mathbb{N}$, which an arbitrary set $X$ maps to. If we call the former map $\phi$, then I am starting by assuming $\phi(X) = \phi(Y)$, and calling this element of ${^\mathbb{N}\mathbb{N}}$ "$f$". So, $f$ maps from $\mathbb{N}$ to $\mathbb{N}$, and is the common result of using your construction to take two subsets $X$ and $Y$ of $\mathbb{N}$ and produce such a function in ${^\mathbb{N}\mathbb{N}}$. – Theo Bendit Nov 02 '17 at 05:10
  • Yes, okay, I think I am getting a bit confused. I should have given my map another name. Then yes, I sort of see that you are saying, but okay we assume that $\phi(X) = \phi(Y)$ and then this tells us that there is $f$=$g$ where these are bijections? It is what $\phi(X) = \phi(Y)$ equals that is confusing me sort of in your explanation. I am not seeing it too clearly. – Sam_U Nov 02 '17 at 05:15
  • $\phi(X)$ is an element of $M(\mathbb{N})$, which makes it a function bijective map from $\mathbb{N}$ to $\mathbb{N}$. – Theo Bendit Nov 02 '17 at 05:18
  • Right, because $\phi(X) = f$, then $\phi(X)$ is actually a bijective map. – Sam_U Nov 02 '17 at 05:20
  • So, then we say consider $x\in X$, then $f(2x)=2x$ – Sam_U Nov 02 '17 at 05:21
  • Yep. But if $x \notin Y$ and $f = \phi(Y)$ (as well as $f = \phi(X)$), then $f(2x) \neq 2x$. – Theo Bendit Nov 02 '17 at 05:22
  • Wait, so we have $\phi(X) = \phi(Y)$, then $\phi(X) = f= \phi(Y)$. Then, consider $x\in X$ and $x\notin Y$. Then, we have that $f(2x)=2x$ but also $f(2x) \neq 2x$, which is a contradiction. So, $x\in Y$? – Sam_U Nov 02 '17 at 05:25
  • But then how does this tell us that $X=Y$? – Sam_U Nov 02 '17 at 05:25
  • Well, it generates a contradiction when you assume that $\phi(X) = \phi(Y)$ and $X \not\subseteq Y$. What this means is $\phi(X) = \phi(Y) \implies X \subseteq Y$. Reversing the roles of $X$ and $Y$ also gives you $Y \subseteq X$, hence $X = Y$. – Theo Bendit Nov 02 '17 at 05:28
  • Okay, so what I have written in the previous line is fine for this one direction? This tells us that $X\subset Y$? – Sam_U Nov 02 '17 at 05:29
  • Yep. That's completely right. Hopefully it's convincing to you as well. – Theo Bendit Nov 02 '17 at 05:30
  • I understand this more clearly now, so thank you for all this explanation. Then for the case of $x\in Y$ we have $f(2x)=2x$ and $x\notin X$ we have again $f(2n)=2n+1$, which is a contradiction because $f(2n)=2n$? The only last minor detail I do not get is the 1 implies 0. – Sam_U Nov 02 '17 at 05:32
  • I said $f(2x) \neq 2x$ in the previous one for the first case so that might be a mistake. – Sam_U Nov 02 '17 at 05:33
  • The $1 = 0$ comment was about getting $2x = f(2x) = 2x + 1$ from the two equations, cancelling $2x$ from both sides, to get $0 = 1$. Aside from using $n$ instead of $x$, it seems like you're understanding it fine. – Theo Bendit Nov 02 '17 at 05:35
  • Come to think of it, the only thing I don't think we covered is showing that $\phi(X)$ is a bijection of $\mathbb{N}$. This is also something worth proving. I think it's pretty obvious, but if you're confused about how to go about doing it, then you should prove it for yourself. – Theo Bendit Nov 02 '17 at 05:37
  • Yes, I think I am fine now. Just when we are saying $x\in X$ we do not need to say $f(2x)=2x$ and $f(2x+1)=2x+1$ also. – Sam_U Nov 02 '17 at 05:40
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    Aside from this I think I am all good now. Thank you so much for explaining all of this it makes complete sense now! – Sam_U Nov 02 '17 at 05:41