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Since an interval is made up of an infinite number of points, I am considering the relation of the length of an interval and the length of a point, this lead me to ask what is the length of a point on the real number line ?

The nested interval theorem made me feel the length of a point should be $0$ because of $\lim_{{n\to\infty}}(b_{n}\!-\!a_{n})=0$

While if the length of a point on the real number line is $0$, then I get a contradiction : Supposing we remove the point $0$ on the real number line, then there is a gap there, and the width of the gap is $0$ since the length of a point on the real number line is $0$, however I think the width of the gap being $0$ is equivalent to there being no such gap on the real number line, so this leads to a contradiction.

What's wrong here? Does a point on the real number line have a width? If so, what is the length of a point on the real number line? Infinitesimal?

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    $\begingroup$ What do you mean by length? $\endgroup$
    – Paul
    Sep 22, 2016 at 10:21
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    $\begingroup$ If you believe the Cantor ternary set mas measure ("length") zero, you have a potent example of how untrue is "the width of the gap is $0$ is equivalent to that there is no such gap exists on the real number line." :) $\endgroup$
    – Andrew D. Hwang
    Sep 22, 2016 at 10:25
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    $\begingroup$ You implication "width of the gap is 0 => no gap exists" is false. In fact, this logic should have made you to conclude that points don't exist either, since their length is 0. It even looks like you subconsciously replaced the term "length" with the term "width" (for gaps) in order to masquerade the otherwise obvious self-contradiction. $\endgroup$
    – AnT stands with Russia
    Sep 22, 2016 at 16:19
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    $\begingroup$ @iMath: How do you define "gap"? I'd say that a system of axioms which defines "point" as having zero length, defines "gap" as having non-zero width, and links these definitions as you linked them in your question (throgh "removal" procedure), is self-contradictory. Again, why do you accept the existence of points of zero length, yet immediately reject existence of gaps of zero width? What do you see as the qualitative difference between the two? $\endgroup$
    – AnT stands with Russia
    Sep 22, 2016 at 18:07
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    $\begingroup$ " I think the width of the gap being 0 is equivalent to there being no such gap on the real number line" There's utterly no reason for this to be true or assumed. Discard this belief. It's a naive belief and unjustified. $\endgroup$
    – fleablood
    Sep 23, 2016 at 16:39

11 Answers 11

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Much like many other 'paradoxes' you'll find that carefully formalizing the problem makes it go away. Whatever counter intuitive phenomena persist is not paradoxical, but simply counter intuitive.

So, what do you mean by length of a point? Well, whatever it is it seems reasonable to define it to have the following properties: 1) the length of an interval $[a,b]$ is $b-a$; and 2) if the point $p$ is contained in an interval, than the length of the point is $\le $ the length of the interval. We further assume all lengths are measured by non-negative real numbers, so there are no infinitesimals around at all.

Now, from these it follows that the length of any point is $0$. So, any notion of length of a point conforming to the above must assign length $0$ to each point. That is a theorem.

Consequently, the situation you describe, where removing a point having length $0$ results in a broken line is a correct description. It is not a paradox but rather a counter intuitive situation. Well, what shall we do with it? You can examine our assumptions and change them, or hone your intuition. In this case, I suggest the latter.

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    $\begingroup$ +1 For "You can examine our assumptions and change them, or hone your intuition. In this case, I suggest the latter." There is a bunch of examples, where 0-length gaps make sense, and giving it any size would give a contradiction. E.g. Domain of $f(x)=1/x$. There would be much more problems with disallowing 0-length gaps rather than allowing it, if one were to decide. $\endgroup$
    – luk32
    Sep 22, 2016 at 13:19
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What's wrong here ?

This is where you go wrong:

I think the width of the gap is $0$ is equivalent to that there is no such gap exists on the real number line

Much like there are objects with length $0$ (namely, single points), there are real gaps of length zero (namely, the absence of a single point).

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    $\begingroup$ This is a great way to put it. One necessarily implies the other. $\endgroup$
    – Matthew Read
    Sep 22, 2016 at 18:45
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As others have pointed out, the length of a single point must be zero for any reasonable definition of length. Notice however, that "length" and "number of points" are very different ways of measuring size. For a point the first measure gives zero and the second one gives one. For an open interval the first measure gives the distance between the endpoints and the second one gives infinity. A single point is negligible in the sense of length but not in the sense of amount.

If you remove a single point from an interval, the total length does not change. The number of points also stays the same, but only because the interval has infinitely many points to begin with.

These different ways of associating sets with sizes are called measures. The measure corresponding to length is called the Lebesgue measure, and the one corresponding to number of points is called the counting measure. These may or may not mean anything to you at this point, but you will encounter them later on if you continue working with real analysis.

I should probably also point out a possible false reasoning, despite being absent in your question: The interval $[0,1]$ consists of points, so its total length must be the sum of the lengths of its points. (We can make sense of uncountable sums. Especially if all numbers are zero, this is easy: then the sum is indeed zero.) But all points have length zero, so the length of $[0,1]$ is zero, too! This is not a bad argument, but it turns out that lengths — or measures in general — do not have such an additivity property. However, if you only take a finite or countable union of disjoint sets, the length of the union is the sum of the lengths. (I'm ignoring technicalities related to measurability as they are beside the point.) The set $[0,1]$ is uncountable, so this "naive geometric reasoning" fails, and it can be enlightening to figure out why.

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    $\begingroup$ Thanks very much , it is much helpful to a Calculus beginning learner $\endgroup$
    – iMath
    Sep 22, 2016 at 14:24
  • $\begingroup$ @iMath, you are welcome! I'm glad to be able to help. $\endgroup$
    – Joonas Ilmavirta
    Sep 22, 2016 at 14:27
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    $\begingroup$ @iMath, the reason is indeed $\infty-1=\infty$. When you remove a point from an infinite set, you get a strict subset, so in that sense it is smaller. But the size of the set does not change. By size you can mean either counting measure or cardinality. It can be used as a defining property (by Dedekind) of infinite sets that removing a point produces a set which can be bijected to the original one. You might also want read about Hotel Hilbert. I know this is counterintuitive at first. $\endgroup$
    – Joonas Ilmavirta
    Sep 22, 2016 at 14:35
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    $\begingroup$ @iMath, yes and no. Each point has length zero, and summing uncountably many zeroes gives you zero. But this sum doesn't have anything to do with the length of the original interval. (As a rule of thumb, if you need to calculate an uncountable sum, something is probably wrong.) $\endgroup$
    – Joonas Ilmavirta
    Sep 22, 2016 at 14:50
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    $\begingroup$ +1 for this truly remarkable answer. And your last paragraph is really illuminating. $\endgroup$
    – Paramanand Singh ♦
    Sep 23, 2016 at 16:39
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Take for example the rational numbers. They can be considered as having infinitely many gaps, because the don't include the irrationals. However, those gaps have no "width", because you can come arbitrarily close to each irrational from above and below with a rational number. The situation is similar if you remove a single real number from all real numbers.

In other words: There are just to many of them, so you wont notice if a single one is missing ;).

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Strictly speaking the "number line" does not have a "length" automatically associated with it- we have to add that. Typically, we define the length of the interval [a, b] (or (a, b), (a, b], [a, b)) to be b- a. Using that definition of length, the length of a single point is 0.

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Perhaps this helps: When you remove a point from a line, you don't just pick it out with your fingers and throw it away. Instead, you declare that one point not to be a part of the set you are considering. That is, imagine $L$ is a line. Then "$L$ with one point removed" is a new set, $L'$, that contains all of the points of $L$ except for the point being removed. $L'$ is a subset of $L$. So of course the line is now broken into two pieces - to get from one part of $L'$ to another, you have to cross through the removed point.

This is actually the way my high school geometry teacher explained this question to me when I had it a long time ago, and it stuck with me. It doesn't contradict any of the other answers given, I should say, but it may help you think about how to interpret these things.

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  • $\begingroup$ Thanks very much, so obvious and intuitive , it also helped me :) $\endgroup$
    – iMath
    Sep 26, 2016 at 7:56
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While this is not supposed to be a complete, satisfying answer, I would like to draw your attention to an area where your "paradox" has a straightforward interpretation.

"In probability theory, one says that an event happens almost surely (sometimes abbreviated as a.s.) if it happens with probability one." Contrary to what one might expect, probability one does not mean that it will surely happen, for the exact reason that you have pointed out: Individual, but non-negligible gaps of "length" (we ususally set "measure") zero.

Think of drawing a real number from $[-1, 1]$. The probability of drawing $0$ is arbitrarily small; there is an infinity number of other numbers to draw instead. So $P(X = 0)$ is actually $0$, as is the length of the interval $[0, 0]$ - but drawing it is still not impossible. By completeness, $P(X \neq 0) = 1$ - this will almost surely happen (yet not surely).

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  • $\begingroup$ Thanks for answering! However ,as a Calculus beginning learner I cannot understand what you said from the probability perspective $\endgroup$
    – iMath
    Sep 23, 2016 at 16:27
  • $\begingroup$ @iMath just bookmark it and come back to in a couple of year. Much valuable insight comes late, and it's almost never too late :) $\endgroup$
    – bers
    Sep 23, 2016 at 16:29
  • $\begingroup$ thanks! In your opinion, the length of a point is 0, Right ? $\endgroup$
    – iMath
    Sep 23, 2016 at 16:35
  • $\begingroup$ Yes. (Comments must be at least 15 characters.) $\endgroup$
    – bers
    Sep 23, 2016 at 16:37
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The measure of a discrete point is in fact zero ($0$).

For more, see articles: Null set and Lebesgue measure.

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I think the width of the gap is 0 is equivalent to that there is no such gap exists on the real number line, this leads to a contradiction.

You mix up countable and uncountable sets. Uncountable sets do not have this sort of gaps as countable sets have.

You can remove an arbitrary number of points from a line, and it will still be a line. Think of removing all the irritional numbers from $\mathbb{R}$ to have the set of rational numbers $\mathbb{Q}$.

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    $\begingroup$ Does real number line with all the irritional points removing out could still be called a line ? $\endgroup$
    – iMath
    Sep 22, 2016 at 14:11
  • $\begingroup$ A line is just a graphical representation of a set. It is only important that it will remain a function, which in either case is the member function. $\endgroup$
    – rexkogitans
    Sep 24, 2016 at 4:33
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Since an interval is made up of an infinite number of points, I am considering the relation of the length of an interval and the length of a point, this lead me to ask what is the length of a point on the real number line ?

The nested interval theorem made me feel the length of a point should be $0$...

I don't think that the nested interval theorem comes into play here, because it does not require length to be defined. Let's define a set $I$ as a set of all intervals in $\Bbb{R}$, length of an interval is a function $L: I \rightarrow \Bbb{R}$, naturally we can say that if $i$ – an interval with endpoints $a, b$ then $L(i) = d(a, b)$, where function $d: \Bbb{R}x\Bbb{R} \rightarrow \Bbb{R}$ is a distance (not necessarily Euclidean). Now we need to extend the definition of length to a point, because so far L defined on the set of subsets of $\Bbb{R}$ we called $I$. Value $L(x), where \space x \in \Bbb{R}$ is undefined. A natural way to extend the definition of length is to view an element $x$ as a single element set $\{x\}$, which in turn can be expressed as an interval $[x, x]$, then $L([x, x]) = d(x, x)$ which is $0$ according to the definition of distance. Not to confuse with Cardinality of a single element set: $|\{x\}| = 1$.

...however I think the width of the gap being $0$ is equivalent to there being no such gap on the real number line...

I think the closest thing to having no gaps in a metric space is Cauchy Completness. If $\Bbb{R}^x = \Bbb{R} \setminus \{x\},$ then $\Bbb{R}^x$ is not complete, because there is a Cauchy sequence of numbers ${x_n}$ with the following properties: $x_n \in \Bbb{R^x}$ and $\lim_{{n\to\infty}}x_n=x$, but $x \notin \Bbb{R}^x$. Fact that length of a point is $0$ (or more important that distance satisfies the axiom of Identity of indiscernibles) is useful for completion of an uncompleted space.

Note, however, that cardinality is still the same: $|\Bbb{R}^x| = |\Bbb{R}|$

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Yes your are absolutely correct.I argue that length of a point is infinitesimal,so is breadth of a line segment.An interval(a,b) is a disjoint union of points.If point has a length 0, 0+0+.......infinity =b-a-finite is impossible.So it should be infinitesimal.Same argument holds for breadth of a line segment.A rectangle of sides a and b whose area is ab is formed by placing a line segment of length a arbitrarily close to one another.If the line segment has breadth 0 then 0+0+......=ab-finite is impossible. similarly we can prove that thickness of a plane rectangular area is infinitesimal. My final conclusion is that length,breadth, thickness of a point is infinitesimal.point has no length is wrong.it is arbitrarily small . I think this proof clarifies your doubt

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