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This may be considered a philosophy but is the number "10" closer to infinity than the number "1"?

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It is infinitesimally closer. – sureshvv Nov 4 '13 at 6:47
Slight aside: in my opinion, 99% of all the "philosophical" problems people have with infinity/infinite/infinitely arise simply from a reluctance to try and actually give precise meanings to things. – Hurkyl Nov 4 '13 at 7:00
@MarcvanLeeuwen: That's why I prefer the phrase "increases without bound". – Blue Nov 4 '13 at 8:29
Does this question mean "Is 10 closer to infinity than 1 is to infinity" or "Is 10 closer to infinity than 10 is to 1"? – LarsH Nov 4 '13 at 11:08

16 Answers 16

There is more than one notion of "closer than."

Metric: We are not able to extend the metric on $\Bbb R$ to the extended line $[-\infty,\infty]$ in a meaningful way (i.e. where bigger elements of $\Bbb R$ are closer to $+\infty$ than smaller elements). Even if we extend the concept of "metric" to include infinite distances, this doesn't work.

Order-theoretic: A partial order $\le$ on a set is a relation satisfying certain properties. It is a way of ordering the elements of the set, put simply. A linear ordering is one in which every two elements are comparable: for any $x$ and $y$, one of $x\le y$ or $y\le x$ is true at least. The extended real line gets its linear ordering from $\Bbb R$, with $+\infty$ made maximal and $-\infty$ made minimal. If $x,y\le z$ then we can say that $y$ is closer than $x$ to $z$ if $x\le y\le z$ and $x\ne y$. On this view, $10$ is indeed closer to $+\infty$ than $1$ (and we can reverse directions to say $1$ is closer to $-\infty$ than $10$).

One observation to be made here is that we essentially decreed larger numbers closer to $+\infty$ purely by fiat. Rather than making this whole discussion trivial, though, it actually illustrates the power of modern algebra in converting imagination into (mathematical) reality. It makes intuitive sense for bigger numbers to be closer to $+\infty$, and there is a consistent way to encode this mathematically, so we simply make it so and therefore have what we want.

Topological: A metric on a space induces a topology on that space. In particular, a metrizable topology. There are more general topologies though, so we can do and allow more things. However not everything can be compared in terms of "closer than." One possible interpretation is that $x$ is closer than $y$ to $z$ if neighborhoods of $z$ containing $y$ also contain $x$ but not vice-versa. It is generally too optimistic to expect elements to be comparable in this way, but the topology on $\Bbb R$ and the extended line is nice enough to make this an almost useful concept.

The problem with this approach is that neigborhoods can be "disconnected," so we can simply put a neighborhood around $+\infty$ containing both $1$ and $10$ and then delete a patch around $10$, thus telling us that neither $1$ nor $10$ is closer than the other. But if we strengthen our notion to only connected neighborhoods, then we get what we want: every connected neighborhood of $+\infty$ containing $1$ must also contain $10$, but not vice-versa. [This originally slipped my mind, but Mike helpfully pointed it out.]

It is worth noting that the topology on the extended line is induced by the ordering on it (topologies are induced by orderings by considering "intervals" to be a topological base). So this actually isn't fundamentally any newer information than the previous context. Furthermore, different metrics can induce the same topology, and it is possible to make $[-\infty,+\infty]$ a metric space in which $10$ is closer than $1$ to $+\infty$. For example, $d(a,b):=|\tan^{-1}(a)-\tan^{-1}(b)|$ does this, as Rahul points out

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I don't get it. Surely, $U = (0,5) \cup (100,\infty]$ and $V = (5, \infty]$ are open neighbourhoods of $\infty$ such that $1 \in U, 10 \notin U$, $1 \notin V, 10 \in V$? – Mike F Nov 4 '13 at 7:30
@Mike Oops, yes. Amended. – anon Nov 4 '13 at 7:41
Perhaps you can add a section on Surreal Numbers where we have both the order-theoretic approach and addition-subtraction between infinite numbers. And the (simplest) infinite number ω is well defined and so are ω-10 and ω-1 and we can prove that ω-10 < ω-1. – ypercube Nov 4 '13 at 10:12
For the topological version, I think it's important to distinguish between topological approaches. If, instead of two point compactification with your (excellent) choice of metric, we chose a one-point compactification in which both infinities are identified and decided we liked arclength as our measure of distance, then we could choose different metrics on the circle which could put 1 closer to the infinite point than 10 and vice-versa. I don't think that using $[-\infty, \infty]$ is canonical in any sense other than it being the most "intuitive" idea one can use with "infinite measurements." – Adam Hughes Nov 4 '13 at 22:38

If you give a suitable precise meaning to "closer", then the answer can be yes.

Here are two things that true for the extended real number line, both of which could reasonably be informally described the way you suggest:

  • $10$ lies between $1$ and $+\infty$.
  • There are intervals containing $+\infty$ and $10$ that do not contain $1$. However, every interval containing $+\infty$ and $1$ also contains $10$
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Exactly. However, the OP didn't say +∞. My maths teacher always said that +∞ is really the same as -∞, just look at the graph for y=1/x; what happens there in the neighbourhood of zero. So... if you go the other way, then, informally of course, 1 is closer to infinity than 10. – Mr Lister Nov 4 '13 at 14:46
@MrLister Then your math teacher was doing you a disservice. That's not to say that there isn't a sensible notion to be made of positive and negative infinity being 'the same', but that's a far cry from saying that they have to be the same, and your teacher's argument is missing several layers of justification. – Steven Stadnicki Nov 4 '13 at 15:29
@StevenStadnicki I'm sure his explanation was more elaborate than my comment here. – Mr Lister Nov 4 '13 at 15:31
In my (very limited) experience, an unsigned number is understood to be positive, so ∞ implies +∞ where -∞ would require explicit declaration. – TecBrat Nov 4 '13 at 15:35
@MrLister: The projective real line (the one where $\infty$ is on "both ends") is extremely useful if you work with polynomials and rational functions, or with analytic functions. The extended real line, however, tends to be more useful for the purposes of geometry and calculus. If we are talking projective line, the we don't have a notion of "between" -- we can't even say $0$ is between $-1$ and $1$ (because you could go the other way through $\infty$ to get from $1$ to $-1$). There is a notion of "separates", though: $0$ and $\infty$ separate $1$ and $-1$. (and vice versa) – Hurkyl Nov 4 '13 at 16:41

"Just follow that line forever," said the Mathemagician, "and when you reach the end, turn left. There you'll find the land of Infinity, where the tallest, the shortest, the biggest, the smallest, and the most and the least of everything are kept."

"I really don't have that much time," said Milo anxiously. "Isn't there a quicker way?"

"Well, you might try this flight of stairs," he suggested, opening another door and pointing up. "It goes there, too."

Milo bounded across the room and started up the stairs two at a time. "Wait for me, please," he shouted to Tock and the Humbug. "I'll be gone just a few minutes."

―Norton Juster, The Phantom Tollbooth

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How nice, and very allusive to the "philoshophy" word in the post. :) – J. W. Perry Nov 4 '13 at 6:38
Would anyone care to explain the punch line for me? (!) – The Chaz 2.0 Nov 4 '13 at 17:54
@TheChaz2.0 In the book, the stairs go up forever as well. – Chris Cudmore Nov 4 '13 at 19:15
… but because they are readily quantifiable, it's easy to take them two at a time :) – Potatoswatter Nov 5 '13 at 3:12
+1: this is probably my favorite childhood book. It does not have the universal acclaim that it deserves. – Pete L. Clark Nov 8 '13 at 22:38

As other has already mentioned:

  • $\infty$ is not a point
  • We don't have well-defined distance-function for $\infty$

But what if $\infty$ was a point and we did have a well-defined distance function that preserved the usual ordering? Then we could answer the question in that frame of reference.

Take the positive part (to avoid extra detail for $-\infty$) of the real line together with infinity $\mathbb{R^+}\cup{\{\infty\}} $ and wrap it up into a half-circle:

to circle

Let $\hat{x} \in \mathbb{R^2}$ be the point on the circle corresponding to the point $x$ on our extended positive line. Then we can define our distance function as: $$p(x,y) = \text{arc-length from $\hat{x}$ to $\hat{y}$}$$

by simple visual inspection we see that 10 is closer than 1 to infinity, and we can calculate it too:

$$p(1, \infty) = \frac{\pi}{2} + 2\arctan(\frac{1}{3}) \approx 2.2 $$ $$p(10, \infty) = \frac{\pi}{2} - 2\arctan(\frac{2}{3}) \approx 0.4 $$

More or less boring details:

The circle used is $B( (0, 1), 1)$ and the mapping from $x \in \mathbb{R^+}$ to the circle is given by the point on the line from the top of the circle $(0,2)$ to $(x,0)$ that intersects the circle. It's easiest to represent everything in terms of the angle $\theta_x$ between the center of the circle and the intersection point $\hat{x}$.

The line can be parametrized as $$ t \left\{\cos \left(\theta _x\right), \sin \left(\theta _x\right)+1\right\}+(1-t) \{0,2\}$$ For given $\theta_x$ we can deduce that the line intersects with the real axis when $ t = \frac{2}{1-\sin(\theta)} $ and further that $\theta_x = 2 \arctan(\frac{-2+x}{2+x})$

Since the arc-length between two points with angles $\theta_x$ and $\theta_y$ on a unit-radius circle is simply $|\theta_x - \theta_y|$ the distance function is defined as:

$$p(x, y) = |\theta_x - \theta_y| $$

And since it was defined via arc-length and the mapping from $x$ to $\theta_x$ is is increasing $p$ clearly satisfies the conditions for being a distance-function.

(We don't actually need $\infty$ to be a point for this, as the distance function will converge to $\pi/2$)

Mathematica code

fromCircle[θ_] := 2/(1 - Sin[θ]) Cos[θ]
fromCircle[Pi/2] = Infinity;
toCircle[x_] := 2 ArcTan[(-2 + x)/(2 + x)]
toCircle[Infinity] = Pi/2;
dist[x_, y_] := Abs[toCircle[x] - toCircle[y]]

showMapping[θ_] := 
    Circle[{0, 1}],
    {Arrow[{{fromCircle[θ], 0}, {Cos[θ], 1 + Sin[θ]}}]},
    {Red, Point[{Cos[θ], 1 + Sin[θ]}]},
    {ColorData[1][1], Line[{{0, 0}, {10, 0}}]},
    Dashed, Line[{{0, 2}, {Cos[θ], 1 + Sin[θ]}}]
   Axes -> {True, False},
   PlotRange -> {{0, 10}, {0, 2}},
   AspectRatio -> Automatic];
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Great stuff, and great starting point to fiddle around with Mathematica – TheBlastOne Nov 5 '13 at 13:02
Which programming language did you use ? – Fardad Pouran May 26 at 7:00

Not only is 10 not closer to infinity than 1, but they are the same distance from infinity as each other.

Technically speaking, the question is not well defined, as infinity isn't a number, but rather a sort of numerical shorthand for "beyond any number". One literal definition is "larger than K for any value of K".

K could just as easily be substituted for L + 9 without changing the definition at all, because K is just any big number (let's assume it's bigger than 9, to simplify things). We rewrite the definition:

Infinity is larger than L + 9 for any value of L + 9.

We can also rewrite "10" as "1 + 9". Now obviously, the "distance" from any number to infinity would be the same if we subtracted 9 from everything (including infinity), so we subtract 9 from our definition of infinity using L, and we get

Infinity - 9 is larger than L for any value of L.

However, that makes the definition of (infinity - 9) the same as the definition of infinity:

Infinity is larger than K for any value of K.

So the distance from 10 to infinity, which is the same as the definition of 1 to infinity - 9, is the same as the distance from 1 to infinity.

It's not the same thing, but it might help you to think about it: if you walk nine feet towards the horizon, the horizon is the same distance away.

In conclusion: infinity is not a number, and it is insane.

(Incidentally, the argument works exactly the same way if you substitute "multiply by ten" for "add nine", or any one of a bunch of other ways of getting from 1 to 10.)

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Did I ever tell you the definition of insanity? It is doing the same thing over and over again expecting things to change. So taking steps toward the horizon expecting it to be closer would make you a crazy person... – Sebastien Nov 4 '13 at 18:53
That's not a good definition of insanity. If you walk in any direction on Earth, you will eventually reach water, but you might walk over uninterrupted land for a long time before you did (would it be insane to expect to reach water after a week of land? a month?) but if you walk in any direction on the moon, you will not reach water. In any case, I'm simply using the horizon as an illustration. – Andrew Wyld Nov 8 '13 at 11:31

The answer depends on what you consider to be "closer". I assume you want to consider this in the one-point compactified real line with the standard Euclidean topology (i.e. it is the limit point of the sequence $n$). In this case I would say no it is not "closer", $\infty$ is a concatenated element to the set, the distance function is not defined on it.

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There are multiple ways to translate "closer than" into math. For finite quantities these end up all meaning the same thing (usually), but once you involve infinity they aren't the same anymore.

Three Definitions of 'Closer'

I see a three "obvious" ways to define "being closer", that will work on the real number line:

  • Distance: P is closer to C than Q is, when the distance from P to C is less than the distance from Q to C.
  • Ordering: P is closer to C than Q is, when Q < P <= C or Q > P >= C.
  • Occlusion: P is closer to C than Q is, when Q must go through P to reach C.

All three definitions agree when Q, P, and C are finite real numbers. For example, try out Q=1, P=10, and C=15.

Things aren't so agreeable when Q, P, or C can be infinity.

The Infinite Case

Let's consider Q=1, P=10, and C=infinity for each of our definitions:

  • Distance: The distance from 1 to infinity is infinity. Same as the distance from from 10 to infinity. So 10 is not closer under this definition.
  • Ordering: It is the case that 1 < 10 < infinity. So 10 is in fact closer under this definition.
  • Occlusion (extended reals): The only way to get to +infinity from 1 is to head positive-ward and smack into 10. So 10 is closer under this definition.
  • Occlusion (projective reals): On the projective real line, +infinity=-infinity. We can reach infinity from 1 by heading negative-ward, without passing through 10. So 10 is not closer under this definition.

So the answer is... yes, no, and yes/no. How disappointing interesting.

Again, allowing infinities changes how things work. Details that didn't matter before start to matter, so different definitions give different answers when you extend them to apply to infinity.

Coming to grips with the fact that there are multiple types of 'close' and that's okay is the real trick to intuitively understanding what's going on.

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This is definitely a philosophy question as long as the notions of "infinity" and "distance" are left undefined. There are many ways of giving these terms precise mathematical meanings—none entirely faithful to their original meanings in natural language, I think, including mine below—leading to a great variety of mathematical answers.

Here is one interpretation that I didn't see in the other answers yet.

One way to think of the numbers 1 and 10 is as cardinals; that is, as objects that measure the sizes of sets. This is one of the most useful interpretations of numbers, and for the question at hand it is useful because it allows us to consider "infinite numbers" on a logically sound footing, as cardinal numbers. (There are many different infinite cardinal numbers in mainstream set theory, rather than a single "infinity", but this won't affect our answer much.)

We could interpret the "distance" between two cardinal numbers $m$ and $n$ as being the least number of elements one needs to add to a set of cardinality $m$ to get a set of cardinality $n$ or vice versa. This notion of distance agrees with the usual one for finite cardinal numbers; that is, natural numbers. Moreover it can be extended in to infinite cardinal numbers in a reasonable way (satisfying the triangle inequality for instance.)

Under this particular interpretation of the question, the answer is no.

If $\Omega$ is any infinite cardinal number then the distance between 1 and $\Omega$ is equal to $\Omega$ itself. That is, having one element to start with does not bring you any closer to having $\Omega$ elements! In fact, for any cardinal number $\Omega$, having this property is equivalent to being infinite; in our interpretation we could say that a cardinal number is infinite if and only if it is as far away from 1 as it is from 0.

Now for any infinite cardinal number $\Omega$ one can show by induction that for any finite cardinal number $n$, the "distance" between $n$ and $\Omega$ is equal to $\Omega$ itself. So in particular, 10 is no closer to $\Omega$ than 1 is.

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It all depends on how you define the distance between a number and $\infty$, which you can do in a consistent way. However, you can surely do this in different ways so that in one case $10$ was close and in the other $1$ was closer. Just imagine taking the "ends" of the real line and connecting them together and calling that point $\infty$. This new space should look like a circle. The most obvious way to put any notion of distance on this space makes it so that $0$ is the furthest from infinity, $a$ and $-a$ are equally distant from $\infty$, etc. In this case, $10$ is definitely closer. Now imagine taking this circle as if it were made of a rubber material pinching it at $10$ and $1$, and holding the $10$ still while you moved the $1$ closer to $\infty$, now $1$ is closer to infinity.

What I've basically done is taken a metric space where $10$ was closer to $\infty$ and deformed the metric so that $1$ is now closer. I can clarify if necessary.

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I am really confused, about what you mean by the "ends" of the real line, I mean as far as I know real line does not have have a biggest or smallest element, can you straighten that out for me? – Adam Nov 4 '13 at 11:51

To me your question is one of distance. More specifically, for some metric $\operatorname{d}$ is the distance $\operatorname{d}(10,\infty)$ less than $\operatorname{d}(1,\infty)$? Since $\infty$ is not actually a number like $1$ or $10$ (not consistent elements) we need a metric to define this. Without this metric the question cannot be answered.

If one could devise an actual metric to incorporate what would otherwise be inconsistent units like real numbers (or any type of number for that matter) and $\infty$, then we could broach the issue assuming it met the definition of a metric.

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We could use the hyper-real number system. Now $\infty$ is ambigous, since they are many infinite numbers. I shall choose an arbitrary one and call it Harry, or $H$ for short. The distance from 1 to $H$ is $H-1$. The distance from 10 is $H-10$. $$10>1$$ $$-10<-1$$ $$H-10<H-1$$ $\therefore$ We conclude $10$ is less distant, and closer.

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The word "closer" is the philosophy part . Now , the most basic explanation can be obtained by looking at the number line Number line . Now look on the left side you have numbers that are getting bigger and at last if you look you will have $\infty$(Quite philosophically) . Now in theory - $10$ appears to be more on the left side than $1$ does. Thus $10$ is closer to $\infty$ than $1$

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My thought is that it is closer to both, dependent on your origin of view. To better define this question we would need a third variable, say PoV for Point of View.

Take an infinite length (aka: a Line) in a straight line. We must therefore assume there is no starting point as this fundamentally destroys our idea of 'infinite'. So where on this infinite length do we place the 1 and the 10? They must be relative to each other given our laws of mathematics i.e. 10 comes after 1 and both are equal distance either side of 5 and so on.

But either side of 10 and 1 is infinite and so infinite has two ends (in the case of a line) and therefore is the same distance to both -infinity (meters or whatever).

Now this raises another question. if we are measuring in a finite length on an infinite line then how do we decide where 1 and 10 must be placed. It could be placed at any length and at any distance apart and the line either side must always be infinite. So then, could the distance between the two points be infinite? Yes.

Personally I think the word 'infinite' is a cause of unnecessary philosophical questions :D

I also think The word infinite should be labelled as a vector or mathematical expression, rather than a number which most people treat it so. 1 and 10 are plotted points on a virtual graph whereas infinity is a mathematically defined expression (whether it be a curve or a straight line is irrelevant).

My 2 cents. Take it with a grain of salt as I am not a mathematician; I'm just a software developer!

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I was just struck with an interesting thought. A circle cannot be infinite in size at it would suggest an infinite diameter as well as an infinite circumference and therefore it would become a straight line. I guess this brings us back to the word 'undefined'. – Le-roy Nov 4 '13 at 7:59
Why can't a circle be infinite in size? For all ε you care to mention, I'll just define my infinitely large circle to have diameter d equal to max(ε)+1 and circumference to be πd... – James World Nov 4 '13 at 21:34
@James: You are saying circles can be arbitrarily large, while Le-roy is saying no particular circle can have infinite radius. Le-Roy: Here's some interesting background. The usual definition of circle disallows infinite radius. But "generalized circles" in geometry (in particular inversive geometry) do include straight lines as circles of infinite radius. This idea is useful when considering Mobius transformations and the Riemann sphere in complex analysis and beyond. – anon Nov 4 '13 at 21:51
@anon: Interesting! You lost me at Mobius transformations but I shall Google it now :) – Le-roy Nov 6 '13 at 8:51

Depends on whether you define distance in absolute terms (as a difference), or in relative terms (as a ratio or percentage). In the first case, we have $$\lim_{n\to\infty}[(n-1)-(n-10)]=9>0\iff10\text{ is apparently closer to }\infty\text{ than }1.$$ In the latter case, we have $$\lim_{n\to\infty}{n-1\over n-10}=1\iff\text{1 and 10 are apparently equally distanced from }\infty.$$

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Geometrically speaking...(sketch)

I would say no.. since if we need to speak about a distance to $\infty$ then it must be a point in our space..from the context I assume we want to work with $\mathbb{R}$ and infinity. Naturally, lets take $\hat{\mathbb{R}}$ of the one point compactification of the reals. This is the same as the circle $S^1$.
Fixing one point on the circle (lets call it $\infty$ then its opposing point.. lets call it $0$ is the maximal distance (wrt arc length from $\infty$). Any two equidistant point must find themselves on either side of this partition of the circle from $\infty$ to $0$ making the approach along the left, or right arcs implies that if 10 and 1 were equidistant... then one of them would be negative. a contradiction...

(of course this is my "natural" interpretation of question) :P

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Another sense in which $10$ is closer to $\infty$ than $1$ is to $\infty$ is that $\frac 1{10} \lt \frac 11$. This comes into play if you do Taylor series "around infinity" by making use of the variable $\frac 1x$. The powers of $\frac 1x$ get small more quickly at $x=10$ than at $x=1$.

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