Prove if they are linear continuum

Question

We have this two sets $\mathbb{N} \times [0,1)$ and $[0,1) \times \mathbb{N}$ and we have to see if they are linear continuum or not.

My doubt here is that I don't know where I have to work in order to prove the least upper bound property and the second condition to be linear continuum, if I have to work with subset of $\mathbb{R}\times\mathbb{R}$ or in $\mathbb{N}\times [0,1)$.

Well and I think that the first set is not a linear continuum because if you take a set like $$ S = \{ (x,y) \in\mathbb{N} \times [0,1) ; x > 0 \} $$ you don't have the first condition.

Answer 1

$\mathbb{N} \times [0,1)$ is order isomorphic to $[0,+\infty)$ via

$(n,t) \to n+t$. And the latter is a continuum (in Munkres' definition).

$[0,1) \times \mathbb{N}$ does not obey the denseness condition: no point lies between $(0,0)$ and $(0,1)$ e.g. or $(t, n)$ and $(t,n+1)$ more generally.

Answer 2

To get a feel for this, see if you can find any initial segment for the two ordered sets. When I do this, there is no doubt that $\mathbb{N} \times [0,1)$ is our candidate.

You asked where to work? How about directly finding the least upper bound?

Let $S \subset \mathbb{N} \times [0,1)$. If $r = (n, a)$ is an upper bound for $S$, then for every $(m,b) \in S$, $m \le n$. The set

$\tag 1 \{ m \mid (m,b) \in S\}$

must contain a maximum, call it $n_1$. So we must have

$\tag 2 S \, \bigcap \, [\{ n_1 \} \times [0,1)] \ne \emptyset$

Since $\{ n_1 \} \times [0,1)$ is a linear continuum, the nonempty set $\text{(2)}$ has a lub of the form $(n_1, a_1)$.

The point $(n_1, a_1) \in \mathbb{N} \times [0,1)$ is evidently the least upper bound for the set $S$.