Is $(0,1)\cup[2,3)$ not Dedekind complete or does it have a gap somewhere?

Question

I learned that a linear order is connected if and only if it is Dedekind complete (any nonempty bounded subset has lub and glb) and has no gap ($\forall a < b \in L, (a, b) \neq \emptyset$).

I think $(0,1) \cup [2,3)$ as a subspace of $\mathbb{R}_{usual}$ is disconnected since $(0,1)$ is clopen, but then it means it is either not Dedekind complete, or it has a gap. I can not find any gap in between, and I don't know why it is not Dedekind complete. Can anyone tell me why?

Answer 1

You're right.

What you've discovered is that "Dedekind completeness" and topological connectedness are not necessarily the same thing. They are only such under the following considerations:

1. the topology in question is the order topology,
2. the order is dense (i.e. for any distinct elements $x$ and $y$ in the space, there is a third, $z$, such that $x < z < y$).

Under these assumptions, any Dedekind-complete ordered and topological set is topologically connected, and vice versa.

However, note that to make sense of them, we need both an order and a topology on the space in question. These two things can be chosen separately, and that is, in fact, what is going on in the case in question. Typically, if we are given an order, but no topology, we will assume the order topology to treat the space topologically, but we don't have to. If we are given both, and that topology is at variance with the order topology, the specified topology overrides the order topology "default", and topological and ordering results about the space in question need not be in agreement.

And that is what is happening in the case here. We are considering this set as inheriting both order and topology, individually, from $\mathbb{R}$. The inherited order on $(0, 1) \cup [2, 3)$ is equivalent to that on, say, $(0, 2)$ (or even $(0, 1)$ itself, but I choose the other for metrical consistency and hence intuitiveness): to see this, consider that $(0, 2) = (0, 1) \cup [1, 2)$, and then note that the original set has the same interval structure but that we've in effect simply relabeled the points in the second interval from $[1, 2)$ to $[2, 3)$.

But the inherited topology on $(0, 1) \cup [2, 3)$ is not the same as the order topology from the previous inherited order. This can be seen by noting that the order-equivalence to $(0, 2)$ would make the order topology connected (order-equivalence implies equivalence of order topologies), since that is a connected subset of $\mathbb{R}$, but as you've seen, this set, under topological inheritance, is disconnected.

Or in other words, your observation proves the inequivalence of the inherited order's order topology and the directly-inherited topology from $\mathbb{R}$, for this set.