For a set $\mathbb{X}$ given order relation extended from that of $\mathbb{R}$, if $\mathbb{X} \supseteq \mathbb{R}$ then $\mathbb{X} = \mathbb{R}$? For a set $\mathbb{X}$ given order relation extended from that of $\mathbb{R}$, if $\mathbb{X} \supseteq \mathbb{R}$ then $\mathbb{X} = \mathbb{R}$ ?
Motivation for this question rose from an intuitive question that when I draw a virtual continuous straight line between $1$ and $2$ and virtually pick a point $x$ between them, then should $x$ be in $\mathbb{R}$?
*A note:
Extended real line is a trivial one. What I'm curious is more about the density (density in the sense of usual English) of, for example, virtual line between $0$ and $1$. 
Possibly, my question might not have been stated properly; I do not know how to put it rigorously.  But I guess you can get the point that I tried to talk about.
Isomorphic type of set inclusion in $\mathbb{X} \supseteq \mathbb{R}$ can be allowed.
 A: One can extend the order of $\mathbb{R}$ to proper supersets of $\mathbb{R}$.  
For example, consider the set of rational functions with coefficient in $\mathbb{R}$. Any such function $f(x)$ can be written as $\frac{P(x)}{Q(x)}$ where the lead coefficient of $Q(x)$ is positive. Define an order on the rational functions by declaring such an $f(x)$ to be positive if the lead coefficient of $P(x)$ is positive.
If we consider the reals as (constant) rational functions, this ordering extends the ordering on the reals.
For more dramatic examples, consider the non-standard models of analysis. 
A: No. You can have a set $\mathbb{X}$ with linear order such that $\mathbb{R} \subsetneq \mathbb{X}$ and the restriction of that order from $\mathbb{X}$ to $\mathbb{R}$ coincides with the standard order on $\mathbb{R}$.
Example 1. Strictly speaking, in order to have what you described it isn't even necessary to squeeze in new points "between" existing points of $\mathbb{R}$. You can also add new points "beyond the end". For instance, you can look at a new set $\mathbb{X} = \mathbb{R} \cup \{+\infty\}$, where $+\infty$ is defined to be strictly larger than any element of $\mathbb{R}$.
Example 2. If you want to squeeze things, you can also squeeze them without any harm. You can do this in a very straightforward manner. For instance, let us squeeze a new element, let's denote it by $\eta$, "immediately to the right of" 0. This can be done like this: let $\mathbb{X} = \mathbb{R} \cup \{\eta\}$, where $\eta$ is just a symbol. If you want set-theoretical strictness, you can take $\eta$ to be any set that doesn't belong to $\mathbb{R}$.
Now define an order on $\mathbb{X}$ like this: if $a,b \in \mathbb{X}$, let us say that $a \leq b$ if either one of these holds:


*

*Both $a$ and $b$ are in $\mathbb{R}$, and $a \leq b$ in the normal sense.

*$a = \eta$, $b \in \mathbb{R}$, and $b > 0$ in the normal sense.

*$b = \eta$, $a \in \mathbb{R}$, and $a \leq 0$ in the normal sense.

*$a = b = \eta$.


You can check that the definition of linear order is satisfied.
Example 3. If you don't like this very artificial and useless squeezing of a single element, you can find an example that looks more natural. For instance, take $\mathbb{X} = \mathbb{R} \times \mathbb{R}$ and define lexicographic order on $\mathbb{X}$: $(a, b) \leq (a', b')$ if either $a < a'$, or $a = a'$ and $b \leq b'$. You can think of $\mathbb{R}$ as a subset of $\mathbb{X}$ by means of an embedding $a \to (a, 0)$. This construction is kind of like squeezing a whole new line of real numbers around each original real number.
Example 4. Moving on to something more real, something that people do use and study, you can take a look at the field $\mathbb{R}^*$ of nonstandard real numbers. The set of standard real numbers $\mathbb{R}$ can be thought of as a subset of the nonstandard ones, i.e. $\mathbb{R} \subseteq \mathbb{R}^*$. This set $\mathbb{R}^*$ contains so called infinitely small numbers, so for instance if $\varepsilon \in \mathbb{R}^*$ is infinitely small and positive, then $0 < \varepsilon < \alpha$ for any positive standard number $\alpha$. So this $\varepsilon$ is kind of like the $\eta$ in example 2: it is squeezed between $0$ and all standard positive numbers. And you also have infinitely large numbers, i.e. there is an element $N \in \mathbb{R}^*$ such that $N > x$ for every $x \in \mathbb{R}$.
A: No. You can even order the whole plane $\mathbb R^2$ by $(x,y)<(x',y')$ if $x<x'$ or ($x=x'$ and $y<y'$).
Now every horizontal line or vertical line is a copy of $\mathbb R$.
