I am studying a course on ZF Set Theory where I encountered the following question:
Does there exist a set admitting precisely four linear orders?
Since any set with $n$ elements exhibits $n!$ linear orders, it feels obvious to conclude that there does not exist such a set since $n! \neq 4$ for all $n \in \mathbb{N}$ (certainly not a finite set).
However, I was wondering if there is a proof (or counterexample) for the claim that is more rigorous than this observation. The question seemed to be posed as more than just a trivial consequence of this fact and so I am assuming that I have oversimplified the problem and that if I broaden my consideration to infinite sets, then it becomes potentially a lot more tricky to conclude that the statement is incorrect in such a simple manner.
I would be grateful for any clarity here.
Further Thoughts
Having now entertained the possibility of constructing infinite sets that exhibit precisely four linear orders, I have tentatively come up with an example that I think might work (edit: this does not work).
Perhaps if we define the set $$X := \mathbb{Z} \times \{ 1,2,3,4 \}$$
Then we can define the following four linear orders:
$$ \text{Order 1: $(a, i) <_1 (b, j)$ if and only if $a < b$ or ($a = b$ and $i < j$)}$$
$$ \text{Order 2: $(a, i) <_2 (b, j)$ if and only if $a > b$ or ($a = b$ and $i < j$)}$$
$$ \text{Order 3: $(a, i) <_3 (b, j)$ if and only if $i < j$ or ($i = j$ and $a < b$)}$$
$$ \text{Order 4: $(a, i) <_4 (b, j)$ if and only if $i < j$ or ($i = j$ and $a > b$)}$$
However, I am hesitant to accept this as an answer without first showing that these are indeed the only four possible linear orders here (since the question requires that there are precisely four). It's certainly the case that four is a lower bound here, but how do I show that there are no others (if indeed this is the case)?
Edit
The above example is not correct. Please refer to the answer I have produced below for the reason why that is the case. Thanks again to @spaceisdarkgreen and @EdwardH for their help in the comments.
The claim is False. As stated in the question, we can make use of the following lemma as part of the proof of the full claim. This helps us to deal with the case of finite sets (which our main argument will eventually reduce the problem down to):
We are now well equipped to make and prove the following claim with the use of the above lemma (thus showing that the answer to the question is that no such set exists):
Proof
We can prove this by contradiction. Assume that there exists a set $X$ that exhibits precisely four linear orders. Let us denote one of these by $(X, <)$. Then, we can immediately define new linear orders on $X$ by considering $\{ (X, <^b) : b \in X \}$ where $<^b$ is the same as $<$ except, we make the element $b$ our maximal element.
Therefore, the number of linear orders on $X$ is lower bounded by the cardinality of $X$. As a result, we require $|X| <4$, otherwise our assumption that $X$ has precisely $4$ linear orders would not hold.
Now that we know that $X$ is a finite set, we can use the fact that any finite set with cardinality $n$ has $n!$ linear orders to complete the proof (since $n! \neq 4$ for any choice of $n \in \mathbb{N}$). This makes direct use of the lemma above.
The set $X$ must satisfy the following $|X|! = 4$. However, this has no solutions for $|X| \in \mathbb{N}$ and so the cardinality of $X$ is undefined. This is a contradiction since cardinality is defined for all sets. Therefore, no such set $X$ can exist. $\quad \Box$
More explicitly, this means that if $|X| = 4,$ then we would have $24$ linear orders. If $|X| = 3,$ we would have $6$ linear orders. If $|X| = 2,$ we would have $2$ linear orders. If $|X| = 1$ or $|X| = 0$, then we would have $1$ linear order. And so it is clear that we cannot have a set that has precisely $4$ linear orders.