Could the empty set be complete partial order?

Question

According to the definition, $(S,\leq)$ is cpo when each totally ordered subset of $S$ has supremum. I'm sure $(\emptyset,\emptyset)$ is a total order, but I don't know whether $(\emptyset,\emptyset)$ is cpo.

Answer 1

Under the definition that you have given, the answer is no. It is not complete.

However, one has to ask whether or not the empty set is defined to be a partial order to begin with; if not then not only the question is moot, but the empty set itself is not a totally ordered subset of any partial order. Therefore we do not require that the empty set has a bound.

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In mathematics definitions are not absolute in most cases. And exactly these sort of end-cases are the common differences between two available definitions. If you define a partial order to be any set, including the empty set, which has a certain relation attached to it; and you define a complete partial order if every subset has a least upper bound, then indeed $\varnothing$ is not a complete linear order.

If we do not allow the empty set to be considered a partial order, then every complete partial order has a minimum, which is the least upper bound of the empty set; so we don't need to exclude the empty set from our definition of completeness. (Note that under this definition the real numbers are not complete, but if we add $\pm\infty$ they are complete)

Another alternative is to modify the definition that every bounded set has a least upper bound/greatest lower bound. In this definition the real numbers are complete again, but we still have to exclude the empty set, because it is bounded by every real number.

To sum up, mathematics is not context free. Everything depends on the definitions you are working with, which may vary slightly from context to context.