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Im trying to prove that $\left|A\right|=\aleph$ for the following group: $A=\left\{f\in\mathbb{N}\rightarrow\left.\mathbb{N}\right|\forall n\le m\ .f\left(n\right)\le f\left(m\right)\right\}$ using Schröder–Bernstein theorem

To prove that $\left|A\right|\le\aleph$ I used the fact that $A\subseteq\mathbb{N}\rightarrow\mathbb{N}$ and therefore $\left|A\right|\le\left|\mathbb{N}\rightarrow\mathbb{N}\right|=\aleph_0^{\aleph_0}=\aleph$

For the other side I think I found an injective function $f\in P\left(\mathbb{N}\right)\rightarrow(\mathbb{N}\rightarrow\mathbb{N})$ which given a set $B\in P\left(\mathbb{N}\right)$ it will return the identity on $B$. In that case i will get $\aleph=2^{\aleph_0}=\left|P\left(\mathbb{N}\right)\right|\le\left|A\right|$

Is that correct or I missed something in the way?

Shiran Shaharabani
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  • What is $\aleph$? ${}{}{}{}{}{}{}{}$ – markvs Aug 17 '20 at 21:28
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    @JCAA I think this isn't so common anymore, but in older papers one often sees $\aleph$ used for $2^{\aleph_0}$ (i.e., $\vert \mathbb{R} \vert$). It looks like that's what OP is using it for here too. – Chris Eagle Aug 17 '20 at 21:29
  • Thank you Chris, I didn't know it is not being used anymore. – Shiran Shaharabani Aug 17 '20 at 21:34
  • I do not quite understand your "For the other side" Which injective function have you found? – Henry Aug 17 '20 at 21:35
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    If $f(B)$ is the identity on $B$, how does $f$ behave on $\Bbb{N} \setminus B$? (Also when you write "group", I think you should be writing "set".) – Rob Arthan Aug 17 '20 at 21:35
  • @Rob, thanks I fixed it in my post. For the question regards to f on N∖B , I assume it depend on what B is – Shiran Shaharabani Aug 17 '20 at 21:38
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    @Chris: It's quite common in introductory courses in Israel. – Asaf Karagila Aug 17 '20 at 21:44
  • @ShiranShaharabani: I think you still have "group" in the first sentence. I suspect you could come up with an injection such that $f(B)$ is the identity on $B$, but you'd have to do some more work to define the behaviour on $\Bbb{N} \setminus B$ and prove that you did have an injection, so that is what you have missed. I've posted an answer suggesting a simpler approach. – Rob Arthan Aug 17 '20 at 21:47
  • I think it is a bit harsh to close a question about non-decreasing sequences as a duplicate of a question about strictly increasing sequences, particularly as the work the OP has tried here is worthy of some comment. – Rob Arthan Aug 17 '20 at 21:52
  • @Rob: It's not about fair. It's about the fact that the proofs are very similar, and the end result is the same. Duplicates don't have to be word-for-word duplicates. In either case, I've added another duplicate. – Asaf Karagila Aug 17 '20 at 21:54
  • @AsafKaragila: the second duplicate you've identified doesn't have an accepted answer. At least one of the answers proposed for the first duplicate depends on the strictness and deson't use Schroeder-Bernstein, but the accepted one is OK on those scores So I'll leave it to you. – Rob Arthan Aug 17 '20 at 21:59
  • @RobArthan: Since when do we hold accepted answers as a benchmark for good answers? I also don't understand your argumentation here. Are bijections not injections? There were numerous instances in my experience where it was easier to come up with a second injection rather than proving surjectivity of a function. – Asaf Karagila Aug 17 '20 at 22:01
  • I thought the criteria for closing questions as duplicates required the original to have an accepted answer. – Rob Arthan Aug 17 '20 at 22:03

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Hint: to get an injection of $\Bbb{P}(\Bbb{N})$ into $A$, define $f(X)$ to the function such that $f(0) = 0$ and:

$$f(x + 1) = \left\{ \begin{array}{l@{\quad}l} f(x) & \mbox{$x \not\in X$}\\ f(x) + 1 & \mbox{$x \in X$} \end{array}\right.$$

Now show how you can recover $X$ from $f(X)$.

Rob Arthan
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