I recently came across a problem which required some knowledge about the self bijections of $\mathbb{N}$, and after looking up how to construct some different bijections I came across the result that the set of self bijections of $\mathbb{N}$ is uncountable.

And this got me wondering, what is the largest set for which its set of self bijections is countable? This obviously holds true for any finite set, but what is the last example of a set whose set of self bijections is countable?

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    $\begingroup$ An interesting side note: without the axiom of choice, it is consistent that there is an infinite set $A$ such that every self-bijection of $A$ fixes all but finitely many elements. Such an $A$ still has uncountably many self-bijections, but only countably many conjugacy classes of self-bijections - so essentially only countably many "really different" self-bijections. (Note that such a set clearly cannot have a countably infinite subset - without choice, $\aleph_0$ is no longer the minimal infinite cardinal!) $\endgroup$ – Noah Schweber Dec 24 '15 at 4:03
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    $\begingroup$ To slightly correct the note made by @Noah: $\aleph_0$ is always the unique minimal infinite cardinal, with or without choice. It just not always the minimum. $\endgroup$ – Asaf Karagila Dec 24 '15 at 15:05
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    $\begingroup$ Adding to the correction made by @AsafKaragila: to prove this, suppose $A$ is infinite. Pick $x\in A$ and let $B=A\setminus\{x\}$. Does $A\equiv B$? If not, then $A$ was not of minimal infinite cardinality. If so, then we have an injective map $f: A\rightarrow A$ whose range is $B$. But then $\{f^n(x): n\in\mathbb{N}\}$ is a countably infinite subset of $A$, so $\aleph_0\le\vert A\vert$. $\endgroup$ – Noah Schweber Dec 24 '15 at 16:24

There is no such maximal set, because $\aleph_0=|\mathbb{N}|$ is the smallest infinite cardinal.

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    $\begingroup$ Oh so this question is just equivalent to asking what is the largest finite number? $\endgroup$ – Brandon Thomas Van Over Dec 24 '15 at 2:10
  • $\begingroup$ Basically. I added an answer showing why. $\endgroup$ – Stella Biderman Dec 24 '15 at 2:19

Let $B(X)$ be the set of bijections of $X$ to itself.

To expand upon the already given answer, it's obvious that the number of bijections from a finite set to itself is finite because the number of functions from a finite set to itself is finite. Thus we are interested in infinite sets. It should also be obvious that if $|X|\geq|Y|$ then $|B(X)|\geq|B(Y)|$

Since there exists a bijection between two sets if and only if they have the same cardinality and the composition of two bijections is a bijection, the set exact you are looking at in this context doesn't matter, only its cardinality. To see this, let $|X|=|Y|$ and let $f:X\to Y$ be a bijection. Then $B(Y)=\{fgf^{-1}(x):g\in B(X)\}$ and so there are exactly the same number.

Now we have established:

1) No finite set has infinitely many bijections the set to itself.

2) $|B(S)|$ depends only on $|S|$.

3) $|X|>|Y|\Rightarrow |B(X)|\geq|B(Y)|$

It follows from the above that there is no set $X$ for which $|B(X)|$ is infinite and $B(X)<B(\mathbb{N})$ since $|\mathbb{N}|$ is the smallest infinite cardinal, which means that, by your observation, no set has countably infinite many bijections with itself! But then the largest set with countably many bijections to itself is finite, and every finite set exhibits this, so there is no largest set.

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    $\begingroup$ A big flaw: It is not true that $|X| < |Y|$ implies that there are more bijections $Y\to Y$ than there are $X\to X$, for infinite $X,Y$. This is independent of ZFC, so it's not likely to be "obvious";/ We have: $$2^{|X|} \le (\text{# of bijections } X\to X) \le |X|^{|X|} = 2^{|X|},$$ and similarly for $Y$. In some models of ZFC, there are infinite $X,Y$ with $|X| < |Y|$ but $2^{|X|} = 2^{|Y|}$. $\endgroup$ – BrianO Dec 24 '15 at 3:59
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    $\begingroup$ Hmmm. We do have $|X|<Y\Rightarrow |B(X)\leq B(Y)$ though, and that's actually all we need thankfully. Thanks for pointing this out though! $\endgroup$ – Stella Biderman Dec 24 '15 at 4:05
  • $\begingroup$ Yes, that's all that's needed. (I also answered, to fill in some details of my comment.) Accordingly you should change the first paragraph/last sentence too. Your argument doesn't need the 2nd paragraph, so that's just expository, yes? $\endgroup$ – BrianO Dec 24 '15 at 4:44
  • $\begingroup$ Correct. I just thought expanding upon that might be useful for the OP $\endgroup$ – Stella Biderman Dec 24 '15 at 5:47
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    $\begingroup$ nice..............+1@Stella $\endgroup$ – Bhaskara-III Dec 24 '15 at 17:21

Let's write $X!$ for the set of bijections $X\to X$. It is true that $$ |X| < |Y| \implies | X! | \le | Y! |. $$ However, it is not true that strict equality always holds. This is independent of set theory (ZFC). See below for details.

First, though, to address the question: There is no "largest [size of] set for which its set of self bijections is countable".

For finite $X,Y$, clearly the set of bijections $X\to Y$ are a subset of $Y^X$, the set of all functions $X\to Y$; so the set of bijections is finite.

The next largest size is $\aleph_0$, the cardinality of $X = \Bbb N$. The bijections $X!$ from this set to itself have cardinal $2^{\aleph_0}$, as shown below, so already the number of bijections is uncountable. If $Y$ is any larger set, then $|X!| = 2^{|X|} \le 2^{|Y|} = |Y!|$, so the cardinality of the bijections of $Y$ is uncountable too.

It is not true that $|X| < |Y|$ implies that there are more bijections $Y\to Y$ than there are $X\to X$, for infinite $X,Y$. This is independent of ZFC, so it's not likely to be "obvious";/ We have:

$$2^{|X|} \le (\text{# of bijections } X\to X) \le |X|^{|X|} = 2^{|X|},\tag{*} $$ To see the first inequality, consider the injection $$f\mapsto \big(x\mapsto (f(x), x)\big) \colon 2^X \to (\text{bijections } X \to 2\times X).$$

Similarly, (*) holds for $Y$.

However, in some models of ZFC, there are infinite $X,Y$ with $|X| < |Y|$ but $2^{|X|} = 2^{|Y|}$. In other models, there are no such $X,Y$ and the property is true. Assuming ZFC is consistent, neither is provable.

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  • $\begingroup$ Since I am not familiar with ZFC, could you give an example of your last statement? Or a specific model of ZFC where that assumption does not hold? $\endgroup$ – Brandon Thomas Van Over Dec 24 '15 at 4:44
  • $\begingroup$ Easton's Theorem shows that the continuum function $\kappa\to 2^{\kappa}$ is very underdetermined by ZFC. Subject to just a few constraints, for any of a wide variety of mappings there's a forcing extension (a model of ZFC) where that mapping is the continuum function. There are models where, $2^{\omega} = 2^{\omega_1}$, models where $2^{\omega_1} = 2^{\omega_2}$, others (like $L$) where all three are different; but the consequences for ordinary mathematics are probably few. <cont> $\endgroup$ – BrianO Dec 24 '15 at 5:31
  • $\begingroup$ <cont> A little more on this: math.stackexchange.com/questions/732285/… $\endgroup$ – BrianO Dec 24 '15 at 5:31
  • $\begingroup$ <and finally,> A lot more on this: mathoverflow.net/questions/67473/… $\endgroup$ – BrianO Dec 24 '15 at 5:39
  • $\begingroup$ Thank you very much for the references. $\endgroup$ – Brandon Thomas Van Over Dec 24 '15 at 5:49

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