one to one correspondance between points on the number axis and real numbers My question is arose by the three statements: 


*

*An interval could be thought of as a line segment on the number axis
according to this book.

*I think it is true that every line segment has two end points.

*The Cantor-Dedekind axiom:

The points on a line can be put into a one-to-one correspondence with
  the real numbers.

so the unit interval [0,1] corresponds to a line segment AB on the number axis with its end points corresponding to real numbers 0 and 1 respectively , (0,1) is an interval different from [0,1], so I think it must correspond  to a line segment CD on the number axis different from AB, CD must have two end points different from the end points of AB, so what are the real numbers the two end points of CD respectively corresponding to ? Can we name them using some symbolic notations? Is there something wrong with my reasoning here ? 

The closed ray $[a,+\infty) = \{ x \ge a \}$ has one more additional point than the open ray $(a,+\infty) = \{ x \gt a \}$, but both are defined using the real number $a$ as an 'endpoint'.
Just for fun, let's define the subset $P \subset \mathbb R$ by
$\quad P = \{ y \in \mathbb R \,| \, y = x^2 \text{ for some } x \in \mathbb R 
 \text{ where } x \text{ has a multiplicative inverse} \}$
These are the positive real numbers. Do you feel compelled to say that this set has an endpoint?

The following is not helpful to the OP, but answers the 'question heading' so I leave it for what its worth.
Every open interval can be mapped bijectively to $\mathbb R$; see for example this link. Also,
Let $A = \{a_0,a_1,\dots,a_n\}$ be a finite set that is disjoint from $\mathbb R$. Then an explicit bijective function $f$ from $A \cup \mathbb R$ to $\mathbb R$ can be defined:
$$   
    f(x) = \left\{\begin{array}{lr}
        x, & \text{for } x \in \mathbb R \text{ and } x \notin \mathbb N\\
        x + (n+1), & \text{for } x \in \mathbb N\\
        k, &  \text{for } x \in A  \text{ and } x = a_k 
        \end{array}\right\} 
$$
