Can there exists a countable continuum? 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?
 A: A linear continuum $L$ is a connected normal $T_1$ space (Munkres shows both in his book) and a connected normal $T_1$ space with at least two points has at least size continuum ($|\mathbb{R}|=\mathfrak{c}$), because we can find a continuous function $f:X \to \Bbb R$ (Urysohn's lemma) that is $0$ in one point and $1$ in the other so $f[L]$ will contain at least $[0,1]$ by connectedness (intermediate value theorem), and hence have size at least continuum.
Many continua exist that are not metrisable (so not embeddable in the reals or such spaces), the long line is one classical example (countably compact but not compact), the lexicographically ordered square (compact but not separable) (both covered in Munkres' book) are very well-known ones. In some models of ZFC we can find Suslin continua too (ccc but not separable); I quite like those myself. 
In ZFC we can find Aronszajn lines (more obscure) etc. So we certainly have a lot more (consistent) examples than just subsets of the reals!
A: 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.
