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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 ?

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    $\begingroup$ An open interval and a closed interval have the same end points, 0 and 1 in this case. If one allows line segments with or without the end points, it means that different line segments can have the same end points. Where is the problem? $\endgroup$ Commented Nov 11, 2017 at 8:27
  • $\begingroup$ @Gribouillis Thanks, please have a look at the updated post $\endgroup$
    – iMath
    Commented Nov 11, 2017 at 8:46
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    $\begingroup$ Every interval has two endpoints. An interval is not determined by its endpoints; to determine an interval, besides specifying its endpoints, you also have to specify which endpoints are included in the interval. There are four intervals with endpoints $0$ and $1$: they are the closed interval $[0,1],$ the open interval $(0,1),$ and the two half-open intervals $[0,1)$ and $(0,1].$ So the answer to your question is that the endpoints of the open interval $(0,1)$ are named "$0$" and "$1$". $\endgroup$
    – bof
    Commented Nov 11, 2017 at 9:12
  • $\begingroup$ The problem is handled easily by making the convention that a line segment of positive length corresponds to a closed interval of same length. And an open interval corresponds a line segment minus its end points. Perhaps you might have seen some graphs of discontinuous functions where one part of the graph has an end point marked via a dark solid bullet and the other part of the graph has the end point marked by a hollow bullet. Something like that can be done to distinguish a line segment with and without end points. $\endgroup$
    – Paramanand Singh
    Commented Nov 12, 2017 at 16:46
  • $\begingroup$ @ParamanandSingh Yes, read my comment for my conclusion to this question. $\endgroup$
    – iMath
    Commented Nov 13, 2017 at 1:11

3 Answers 3

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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.

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  • $\begingroup$ Thanks for you answer, but have you ever read the mapping between an interval and a segment in this book mentioned in the post above ? $\endgroup$
    – iMath
    Commented Nov 12, 2017 at 13:41
  • $\begingroup$ I think that the endpoints of the open interval (0,1) are indeed "0" and "1" as bof said , what I haven't know before is that there exists concept named open line segment, different segments can have the same endpoints , so that posted the question . $\endgroup$
    – iMath
    Commented Nov 12, 2017 at 13:41
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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 $x<1$.

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  • $\begingroup$ I have read one opinion on replacing points with locations on a line , so they say : a point on a line is just a location on a line . With your opinion mixed in, do you agree that some line segments do not have end locations? $\endgroup$
    – iMath
    Commented Nov 11, 2017 at 8:45
  • $\begingroup$ @iMath Yes, I agree, this is what I tried to explain, that the $(0, 1)$ segment doesn't have endpoints. $\endgroup$
    – razvanelda
    Commented Nov 11, 2017 at 8:51
  • $\begingroup$ "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 $x<1$.", a recompiling on the opinion using limit is "You can think that the left endpoint is 0 and the right point is 1.", contradict with what you agree $\endgroup$
    – iMath
    Commented Nov 11, 2017 at 8:54
  • $\begingroup$ @iMath That's why I say "You can think", because you can imagine that are infinte numbers that converge to $0$, respectively $1$ using the limit notation, but the endpoints aren't $0$ and $1$. $\endgroup$
    – razvanelda
    Commented Nov 11, 2017 at 9:00
  • $\begingroup$ the endpoints of the open interval (0,1) are indeed "0" and "1" as bof said , what I haven't know before is that there exists concept named open line segment. $\endgroup$
    – iMath
    Commented Nov 12, 2017 at 13:12
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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: $$ f(x) = \left\{\begin{array}{lr} x, & \text{for } x \in \mathbb R \text{ and } x \notin \mathbb N\\ x + (n+1), & \text{for } x \in \mathbb N\\ k, & \text{for } x \in A \text{ and } x = a_k \end{array}\right\} $$

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  • $\begingroup$ Thanks for you answer, but have you ever read the mapping between an interval and a segment in this book mentioned in the post above ? $\endgroup$
    – iMath
    Commented Nov 12, 2017 at 13:42
  • $\begingroup$ I think that the endpoints of the open interval (0,1) are indeed "0" and "1" as bof said , what I haven't know before is that there exists concept named open line segment, different segments can have the same endpoints , so that posted the question . $\endgroup$
    – iMath
    Commented Nov 12, 2017 at 13:42
  • $\begingroup$ So bof answered the question in his comment and the answers here all miss the mark... $\endgroup$ Commented Nov 12, 2017 at 14:30

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