In James Munkres's Topology I came across the definition of linear Continuum as follows:
Definition: A simply ordered set L having more than one element is called a linear continuum if the following hold:
- L has the least upper bound property.
- If $x<y$, there exists $z$ such that $x<z<y$.
We can see that standard examples of Linear Continuum are made from subsets of $\mathbb{R}$. My question:
Can we find a Linear continuum that is not made from $\mathbb{R} ?$
Can there be a countable Linear continuum?
Suppose $L=\{a_{i}:i\in\mathbb{N}\}$ is a countable linear continuum. Let $x_{1},y_{1}\in L$ such that $x_{1}<y_{1}$. For all $n$ we can pick $x_{n}$ such that $x_{n-1}<x_{n}<y_{n-1}$ and $a_{2n}\not\in (x_{n},y_{n-1})$ where $$(x_{n},y_{n-1})=\{a\in L:x_{n}<a<y_{n-1}\}.$$ Furthermore we can pick $y_{n}$ such that $x_{n}<y_{n}<y_{n-1}$ and $a_{2n+1}\not\in(x_{n},y_{n})$.
Note that $X=\{x_{n}:n\in\mathbb{N}\}$ is a non-empty set in $L$ with upper bound $y_{1}$. Since $L$ has the least upper bound property there is an $m\in\mathbb{N}$ such that $a_{m}$ is the supremum of $X$. However either $a_{m}<x_{m}$ or $a_{m}>y_{m}>x_{m}$, so either $a_{m}$ is not an upper bound or it is not the least upper bound.
So a linear continuum can not be countable.
I am not sure about your first question, I will think about it and expand this answer when I know how to solve that part.