one to one correspondance between points on the number axis and real numbers

Question

My question is arose by the three statements:

1. An interval could be thought of as a line segment on the number axis according to this book.
2. I think it is true that every line segment has two end points.

3. 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 ?

Answer 1

Your reasoning is fine, and I see why you find it "strange". Clearly the real numbers have no end points, because no matter how large or small you go, you can always find a real number larger or smaller than it. So then where do $0,1$ map to if you set up a bijection between $[0,1]$ and $\mathbb{R}$? The answer is - any bijection between $[0,1]$ and $\mathbb{R}$ will be "strange" and you can map $0,1$ where ever you like.

Here's a well known example.

Let $f:[0,1]\rightarrow (0,1)$ be defined by: $$f(x) = \left\{ \begin{array}{1 1}\frac{1}{2} & \mbox{if } x = 0\\\frac{1}{2^{n+2}} & \mbox{if } x = \frac{1}{2^n}, n\in \mathbb{N}_0\\x & \mbox{otherwise}\end{array} \right.$$

It's easy to see that $f$ is a one-to-one correspondence, because it essentially takes care of $0$ and $1$ by mapping them to $1/2$ and $1/4$, then the rest of the points of the form $1/2^n$ in $(0,1)$ are "shifted" to $1/2^{n+2}$.

Now we construct two more functions:

Let $g:(-\pi/2, \pi/2)\rightarrow \mathbb{R}$ be defined by $g(x) = \tan(x)$

Let $h:(0,1)\rightarrow (-\pi/2, \pi/2)$ be given by $h(x) = -\pi/2 + \pi x$

It's easy to show that these are also bijections. But now: $$g\circ h\circ f:[0,1]\rightarrow\mathbb{R}$$ Is a one-to-one correspondence! Let's see where $0$ and $1$ go:

$(g\circ h\circ f)(0) = g(h(f(0))) = g(h(1/2)) = g(0) = 0$

$(g\circ h\circ f)(1) = g(h(f(1))) = g(h(1/4)) = g(-\pi/4) = -1$

Clearly $0$ and $-1$ are in no way "end points" of $\mathbb{R}$. The key is that any one-to-one correspondence between $[0,1]$ and $\mathbb{R}$ will be very "unnatural", because you need to put the end points somewhere artificially as I did above.

Answer 2

It is not true that every line segment has two end points. Don't forget that a point has no size, it represents a place, not a thing. In the case of the $[0, 1]$ interval you have the endpoints $0$ and $1$, but in the case of the $(0, 1)$ interval you have an infintiy of points that converge to $0$, respectively $1$. You can think that the left endpoint is $\lim_{x\to 0} x$ with $x>0$ and the right point is $\lim_{x\to 1} x$ with $x<1$.

Answer 3

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?

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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: $$ f(x) = \left\{\begin{array}{lr} x, & \text{for } x \in \mathbb R \text{ and } x \notin \mathbb N\\ x + (n+1), & \text{for } x \in \mathbb N\\ k, & \text{for } x \in A \text{ and } x = a_k \end{array}\right\} $$