Maximum separation on an infinite line

  • #1
Andrew Wright
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I am confused about the maximum separation between two points on an infinite line.
If the maximum separation between two points on an infinite line is finite, then what is its value? So the maximum separation is infinite. Does this mean two points on an infinite line can be separated by an infinite distance? Why, why not?
 
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  • #2
fresh_42
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Two points on the real number line have always a finite distance, determined by the definition of the points. This distance, however, can get arbitrarily large, finitely large but arbitrary large. If you want to talk about infinities, then I recommend talking about projective geometry, where an infinitely distant point becomes an entity.
 
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  • #3
jbriggs444
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the maximum separation
We were taught that when one uses the definite article (the word "the"), one must have a guarantee of existence and of uniqueness in hand before doing so.

Unfortunately, the maximum separation does not exist. For any candidate "maximum separation" there is always another candidate that is larger still. There is no maximum

Similarly, there is no largest real number and no largest rational number whose square is less than two.
 
  • #4
Andrew Wright
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That's a couple of great answers and I get it now. My only objection is that the separation on a number line between zero and n = n. This is also equal to the size of the set of positive integers between 1 and n. If you choose the full set of positive integers then the size of the set is aleph null and this now = n, meaning the separation is now aleph null and infinite. Why is this logic flawed please?
 
  • #5
Andrew Wright
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That's a couple of great answers and I get it now. My only objection is that the separation on a number line between zero and n = n. This is also equal to the size of the set of positive integers between 1 and n. If you choose the full set of positive integers then the size of the set is aleph null and this now = n, meaning the separation is now aleph null and infinite. Why is this logic flawed please?
I am worried that you may think I am being argumentative. Actually, I can see that aleph null is not a Real number, being itself infinite cannot appear on your definition of the real number line.

The reason for my stubborness is that 30 years ago when I did my physics degree, I got exam questions asking the limit of say, 2x+5/x as x-> infinity. To my lecturers frustration I imagined x going to the maximum positive integer being infinite. From here I deducted that adding 5 to aleph null left aleph null. Also that 1 line of length aleph null should extend to positive infinity and another to negative infinity. Thus the length of 2 aleph null lines added together was twice aleph null. rendering as 2 |Z+|/|Z+|= 2. The two lines being twice the length of one of them. My lecturers said it was terrible maths but never explained why. So I got through some 1st year maths exams using my "logic" without any satisfactory reason why I shouldn't think this way. So my working would be something like
lim (2x+5/x) x-> inf = 2 |Z+| +5 / |Z+| = 2 |Z+| / |Z+| = 2
 
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  • #6
PeroK
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I am worried that you may think I am being argumentative. Actually, I can see that aleph null is not a Real number, being itself infinite cannot appear on your definition of the real number line.

The reason for my stubborness is that 30 years ago when I did my physics degree, I got exam questions asking the limit of say, 2x+5/x as x-> infinity. To my lecturers frustration I imagined x going to the maximum positive integer being infinite. From here I deducted that adding 2 to aleph null left aleph null. Also that 1 line of length aleph null should extend to positive infinity and another to negative infinity. Thus the length of 2 aleph null lines added together was twice aleph null. rendering as 2 |Z+|/|Z+|= 2. The two lines being twice the length of one of them. My lecturers said it was terrible maths but never explained why. So I got through some 1st year maths exams using my "logic" without any satisfactory reason why I shouldn't think this way. So my working would be something like
lim (2x+5/x) x-> inf = 2 |Z+| +5 / |Z+| = 2 |Z+| / |Z+| = 2
The limit has a precise mathematical definition.

Note also that "logic" is the process of making valid deductions from assumed premises. Logic does not dictate what "infinity" should mean. Instead, once you have the definition of an infinite set, say, then you can apply logical reasoning to it.
 
  • #7
Andrew Wright
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The limit has a precise mathematical definition.

Note also that "logic" is the process of making valid deductions from assumed premises. Logic does not dictate what "infinity" should mean. Instead, once you have the definition of an infinite set, say, then you can apply logical reasoning to it.
Is it bad logic or bad lecturer?
 
  • #9
Andrew Wright
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I learned not to show all my workings in those exams ;)
 
  • #10
PeroK
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I learned not to show all my workings in those exams ;)
Everyone who has ever studied mathematics can imagine plugging infinity into an equation. You can't seriously believe that is an original idea. Anyone can do that!

Learning limits is about learning the more rigorous and subtle approach that allows you to calculate limits where plugging in infinity doesn't work.

Also, it's critical to be able to define limits in terns of finite numbers, which allows a rigorous development of the foundations of Calculus.
 
  • #11
Andrew Wright
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thanks
 
  • #12
Andrew Wright
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thanks
for the record, i did manage to sort out some of the complicated limits back then. i sometimes used the obvious limits to help deduce parts of the solution to more advanced ones i think
 
  • #13
Andrew Wright
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i can honestly say a best guess sometimes popped out of using the largest number my calculator could generate for infinity and then rounding it. Also, workings were sparse :)
 
  • #14
PeroK
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i can honestly say a best guess sometimes popped out of using the largest number my calculator could generate for infinity and then rounding it. Also, workings were sparse :)
That's a good practical approach. Ultimately, the purpose of the limit is not to calculate specific functions, but to know for certain which mathematical results can be rigorously proved.

Modern mathematical physics includes techniques which are based on rigorous proofs and those that are not. It's important to recognise this distinction. It doesn't mean you don't use advanced techniques that haven't yet been proved, but it's better to be aware of where and when you are not on totally secure foundations.

The purpose of teaching mathematical analysis is to give an understanding of this distinction between things that might look good and things that you can rigorously prove.

Short-cutting the easy stuff is all well and good, but it doesn't help you progress to, for example, looking at the convergence properties of Taylor series.
 
  • #15
Andrew Wright
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interesting...
 
  • #16
jbriggs444
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That's a couple of great answers and I get it now. My only objection is that the separation on a number line between zero and n = n. This is also equal to the size of the set of positive integers between 1 and n. If you choose the full set of positive integers then the size of the set is aleph null and this now = n, meaning the separation is now aleph null and infinite. Why is this logic flawed please?
You have correctly observed that the number of integers in the set from 1 to n is always n for any finite n. Now you wish to extend this observation to apply to the infinite set of all integers. You want to say that the largest element of the naturals is equal to the number of naturals.

There are two problems with such an assertion. First, the "largest natural number" does not exist. Second, the way you would have formally proved your statement for finite n (mathematical induction) only works to prove statements of the form: "for all finite n, <this property holds>". It does not help you prove anything about a hypothetical infinite n.

Edit...

You might want to explore the notion of an ordinal number. As the name suggests, ordinals relate to order. An order is something you very much need if you are interested in finding a "first" or "last" element in a set.

The naturals have no last element. But we can speak of a next element, ##\omega## (Omega). Omega is not an element of the natural numbers.
 
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  • #17
WWGD
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A metric on a space S is a function ##d: S \times S \rightarrow \mathbb R ##so d(x,y) must be a Real number.
 
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  • #18
fresh_42
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A metric on a space S is a function #d: S \ times S \right arrow \mathbb R # so d(x,y) must be a Real number.
Not necessarily. The discrete metric only knows ##0## and ##1##.
 
  • #19
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So my working would be something like
lim (2x+5/x) x-> inf = 2 |Z+| +5 / |Z+| = 2 |Z+| / |Z+| = 2
The expression in your limit above is ##2x + \frac 5 x##, which undoubtedly isn't what you meant. If I take what you wrote literally, I get this:
$$\lim_{x \to \infty} 2x + \frac 5 x = \lim_{x \to \infty} 2x + \lim_{x \to \infty} \frac 5 x = \infty + 0 = \infty$$

Although you used parentheses in what you wrote, they are missing where they are needed the most - in the numerator. As linear text, the expression should be (2x + 5)/x.

There is another problem with what you wrote; namely the expression 2 |Z+| /|Z+|, which is ##\frac \infty \infty##, which isn't defined.

The limit you probably intended is this:
$$ \lim_{x \to \infty} \frac {2x + 5}x = \lim_{x \to \infty} \frac {2x}x +\lim_{x \to \infty} \frac 5 x$$
Using the properties of limits at infinity, the above simplifies to 2 + 0 = 2.

The more rigorous approach alluded to in this thread goes something like this:

For any (very large) number M, there exists a positive number ##\epsilon## near 0 such that if x > M, then ##\frac {2x + 5}x - 2 < \epsilon##.
A metric on a space S is a function #d: S \ times S \right arrow \mathbb R # so d(x,y) must be a Real number.
For TeX, use two # signs at each end (inlne) or two $ signs at each end (standalone).
You must be new here. :oldbiggrin:
 
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  • #20
WWGD
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Not necessarily. The discrete metric only knows ##0## and ##1##.
I meant into the Reals, not onto.
 
  • #21
fresh_42
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I meant into the Reals, not onto.
Yes, but any ordered field will do! The order is necessary for the triangle inequality, but ##0\neq 1## is given in any field. Hence we can easily define a metric with values e.g. in ##\mathbb{Q}##.
 
  • #22
WWGD
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Yes, but any ordered field will do! The order is necessary for the triangle inequality, but ##0\neq 1## is given in any field. Hence we can easily define a metric with values e.g. in ##\mathbb{Q}##.
If you define it on the Reals, then the standard ##|x-y|## won't do, as it will be full of holes.
 
  • #23
fresh_42
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If you define it on the Reals, then the standard ##|x-y|## won't do, as it will be full of holes.
The discrete metric is a hole.
 
  • #24
WWGD
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The discrete metric is a hole.
I guess you need an ordered field because of the triangle inequality?
 
  • #25
Andrew Wright
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...Modern mathematical physics includes techniques which are based on rigorous proofs and those that are not. It's important to recognise this distinction. It doesn't mean you don't use advanced techniques that haven't yet been proved, but it's better to be aware of where and when you are not on totally secure foundations.

The purpose of teaching mathematical analysis is to give an understanding of this distinction between things that might look good and things that you can rigorously prove.

Short-cutting the easy stuff is all well and good, but it doesn't help you progress to, for example, looking at the convergence properties of Taylor series.
I went this far so I demand satisfaction. Is it impossible to use projective geometry and allow graphing of points outside reals (with infinite values like our friend Aleph) to render any solutuon to the convergence of Taylor series in a manner that is mathematically useful? Why is this and can you use calculus with projective geometry?

I am currently agnostic to this approach.

Thanks for your time.
 
  • #26
WWGD
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Can you use Calculus with Projective Geometry? Yes. Projective Space is a manifold and you can do Calculus on Manifolds.
 
  • #27
Andrew Wright
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It just felt for a minute that projective geometry might let me graph things that converge at infinity and so "see" a limit converging as it goes to infinity. I have to be honest, I don't get projective geometry.
 
  • #28
Andrew Wright
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I also forgot the mathematics of taylor series years ago. I am not lazy, but I had my chance as an academic and I blew it. I write computer software these days. Any way, your thoughts are much apppreciated.
 
  • #29
pbuk
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I have to be honest, I don't get projective geometry.
Then it is not going to be much use to you in trying to 'get' limits.

But you don't need it anyway: a simple plot in Cartesian coordinates should be enough. When we say ## \displaystyle \lim_{x \to \infty} f(x) = g(x) ## we mean that for large enough ## x ## the graph of ## f(x) ## is as close as you like to the graph of ## g(x) ##.
 
  • #30
benorin
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While there's no maximum distance between points on two infinite lines, there's a minimum. I encourage you to think what the situation is geometrically, in the case of looking for a maximum either two lines with arbitrary slopes and intercepts, no maximum distance exists because for any two arbitrarily far apart pair of points we may find another pair farther apart by moving away from the origin along either line; and this new pair of points can't be the maximal distance either for the same reason as the first pair of points (ie moving one point further away from the origin along one of the lines produces a greater maximum) and so the set of distances between points on the lines has no least upper bound hence the maximum d.n.e. (does not exist).

These other people responding to this thread are lightyears beyond my capabilities in math, they might even need to correct my work here lol @fresh_42 often has good points to correct my work. Thx for training me up you guys!
 

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