Example of a complete unbounded dense linearly ordered set that isn't isomorphic to $\mathbb{R}$

Question

I know as a fact that $\mathbb{R}$ is the unique (upto isomorphism) complete linearly ordered field. But if we remove the "field" condition and replace it with "dense unbounded set", then I was wondering what is the situation then.

One example I came up was this: Just put $2^{2^\mathbb{N}}-$many copies of $[0,\infty)$ in front of $\mathbb{R}$, and put $2^{2^\mathbb{N}}-$many copies of $(-\infty,0]$ at the back of $\mathbb{R}$ like this:

$\cdots+(-\infty,0]+(-\infty,0]+\mathbb{R}+[0,\infty)+[0,\infty)+\cdots$.

Then this set is dense, unbounded, and complete linearly ordered set (am I right?), and by a cardinality argument, this set isn't isomorphic to $\mathbb{R}$. Please correct me if I am wrong!

I just realized that I wanted to ask some more:

If we instead put $2^\mathbb{N}-$many copies of $[0,\infty)$ and $(-\infty,0]$ at the front and back of $\mathbb{R}$ (instead of $2^{2^{\mathbb{N}}}-$many such copies), then how would the result change?

Also, do you have another example of a complete unbounded dense linearly ordered set that isn't isomorphic to $\mathbb{R}$ in your mind?

Answer 1

In fact, you don't even need $2^\mathbb{N}$-many: $\omega_1$-many additional "blocks" is enough to prevent isomorphism with $\mathbb{R}$ (and this is worth noting since as far as we know we might have $\omega_1<2^\mathbb{N}$). Specifically, $\omega_1$-many copies of $[0,1)$ glued together one after the other results in a complete dense linear order which is not separable (= does not have a countable dense subset). This is important enough to have a special name - it's called the Long Ray, and it and its variants show up a lot in general topology.

Answer 2

Another example is $[0,1] \times [0,1]$ lexicographically ordered and its minimum and maximum removed. The element $(1,0)$ is not contained in a separable open interval, hence it's not isomorphic to the long ray (and not to the reals).