# Is $\emptyset : \emptyset \to \emptyset$ an isomorphism from $(\emptyset, \leq)$ to $(\emptyset, \leq)$?

I was asked to determine whether the following statement is true:

If every function $$F : P \to P$$ is a homomorphism from $$(P, \leq)$$ to $$(P, \leq)$$, with $$\leq$$ an arbitrary order, then $$|P| = 1$$.

It is straightforward to observe that $$|P| \not> 1$$. However, $$|P| = 0$$ seems to satisfy the first part of the predicate.

If $$P = \emptyset$$ there is one and only one function over $$\emptyset^2$$ that maps from the empty set to itself; namely, $$\emptyset$$. The definition of a homomorphism $$F$$ involves statements of the form: for all $$x, y$$ in $$P$$ occurs this and that involving $$F$$... So to refute that $$F$$ is a homomorphism one is to find a counter-example to these properties. Of course, such counter examples cannot be found in the empty set. So $$\emptyset : \emptyset \to \emptyset$$ is a homomorphism from $$(\emptyset, \leq)$$ to $$(\emptyset, \leq)$$.

The notation $$\emptyset : \emptyset \to \emptyset$$ is odd but seems formally correct, because $$\emptyset$$ is a function and it does have itself as domain and range. However, I'd incidentally like to know if this notation is correct indeed.

Now, my question is the following. Is $$\emptyset : \emptyset \to \emptyset$$ an isomorphism between $$(\emptyset, \leq)$$ and $$(\emptyset, \leq)$$? It seems to be the case that $$\emptyset$$ thus considered is not only a function, but that it is its own inverse ($$\emptyset^{-1} = \emptyset$$).

• It is an isomorphism for a silly reason: $\emptyset$ is one-to-one and onto (both vacuously) and preserves an order (which will be empty, of course). Mar 4 at 23:08
• If you are taking some sort of logic course including discussion of e.g. first-order logic, it's possible that in this course, models of a first-order theory are by definition always non-empty, and "$\le$ is an order on $P$" means "$(P, \le)$ is a model of the theory of partial/linear orders". If so, then the statement is true. I think pretty much any combination of "the statement was intended to be true/false, and it is true/false" is possible, depending on the definitions and whether whoever wrote the question was absent-mindedly thinking of a different definition or not :) Mar 5 at 3:18

As you surmise, the empty set, $$\emptyset$$, can be be viewed as a function $$\emptyset \to \emptyset$$ and it preserves the only possible ordering, $$\le$$, that you can put on its range $$\emptyset$$. So, yes, the statement in the title of your question is true, and the statement you that you have been asked to determine is false.

• Thanks for answering my question. Mar 5 at 19:16

Note that $$\mathrm{id}_{\varnothing}=\varnothing$$; that is, the empty set is also the identity map from $$\varnothing$$ to $$\varnothing$$. Therefore, you have $$\varnothing\circ\varnothing=\varnothing=\mathrm{id}_{\varnothing}$$, so indeed you have that $$\varnothing$$ is an invertible function $$\varnothing\to\varnothing$$ with $$\varnothing=\varnothing^{-1}$$.

You can also verify that if we view $$\varnothing$$ as a function from $$\varnothing$$ to $$\varnothing$$ we must conclude that it is one-to-one and onto.

I expect that the statement you were given was expected to include a non-empty clause,

If $$P\neq\varnothing$$ and every function $$F\colon P\to P$$ is a homomorphism from $$(P,\leq)$$ to $$(P,\leq)$$, with $$\leq$$ an arbitrary order, then $$|P|=1$$.

(I'm not wild about "an arbitrary order" there either... poor phrasing, IMHO...) Unless your definition of partially ordered set requires non-emptyness...

It is not uncommon to see people forget to include a nonemptiness clause when they should. For example, it is typical to see the statement:

A function $$f\colon X\to Y$$ is one-to-one if and only if it has a left inverse.

but the statement is false if we do not assume $$X\neq\varnothing$$.

• Thanks for the detailed answer. Mar 5 at 19:16