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Seeking explanation for this apparent paradox shaking the grand pillars of mathematics.

If I stand on the unit circle (of circumference $2\pi$) and I take steps of arclength $s$, then what is required of $s$ such that I eventually come within length $\delta$ of any point?

And I have a device that makes a gong at time t_0=0 seconds, then at t_1=1/2 sec, next at t_2=3/4 sec, next at t_3=7/8 sec, the n'th gong happening at time $t_n=1-1/2^n$ sec. And at the $n$'th gong i take f(n) steps of length $s$. After 1 second has passed, with f(n)=1, is there an $s$ such that I visited every point on the circle? If not, which points did I not visit?

Same questions at t=1, except now I take f(n)=3^n steps at the n'th gong.

Same questions at t=1, except now at the n'th gong, I take f(n)=n^n*(number of subsets of the natural numbers (1 to n)) steps.

Apparently, already after 1 second, I have made a step for every (finite and infinite) subset of the natural numbers, which has cardinality continuum!?

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For your very first question: If $s$ is not a rational multiple of $\pi$ then the points you visit will be a dense subset of the circle. –  Ragib Zaman Oct 15 '11 at 14:08
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Given that you believe it shakes the pillars of mathematics, it would make sense to point out where in what you wrote you see a paradox. –  joriki Oct 15 '11 at 16:26
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And most of all, you need to be able to be really open for the possibility that you completely misunderstood something. Don't dismiss automatically anything that is said to imply you were wrong. Mathematics is nontrivial and there is a reason most people don't sit all day and contemplate about theorems and infinite objects, and whatnot. –  Asaf Karagila Oct 15 '11 at 16:40
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No you don't. Each particular step along the way is still finite. There is none of your steps that counts, for example, the set of all primes. –  Henning Makholm Oct 15 '11 at 18:15
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@Henning How do you go to job every day? Well you first walk half the way, then 3/4 of the way, then 7/8 of the way, infinite steps. Yet you DO get there in finite time. And none of the previous steps before t=1 got you there. –  coffeemachine Oct 15 '11 at 19:24
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4 Answers

up vote 1 down vote accepted

Yes you are right, you made only countably many steps and yet you visited uncountably many intervals.

But in every step you stopped at a point which belongs to uncountably many intervals. Since with every step you visited uncountably many intervals, in the end you visited uncountably many intervals.

From a mathematical point of view look to it this way:

Let $A$ be the set of subintervals of $[0,1)$. Let $N$ be the set of positive integers.

Let $f: A \rightarrow B$ be the function which figures the first time you step in that particular interval.

Then $f$ is a function from an uncountable set to a countable set which is not 1-1. There is no contradiction/paradox there.

What you miss when you think this is a aparadox, is the fact that at every step you visit uncountably many different intervals. If your first step is $a$ then you visit ALL intervals containing $a$ in that first step.

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I'll assume that by "every subset of the natural numbers (1,n)", you mean every subset of the set $\{1,2,\dotsc,n\}$.

The answer to all versions of the question is the same. In each case, you take a countably infinite number of steps, since you take a finite number of steps for each natural number in turn. Thus you cannot visit every one of the uncountably many points on the circle in any of these ways. The points that you didn't visit are the uncountably many points whose angle is not of the form $ks+2\pi l$ with non-negative integer $k$ and integer $l$.

The arc length $s$ must be incommensurate with $\pi$ for you to evenually come arbitrarily close to every point; that is, it must not be a rational multiple of $\pi$.

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No, at t=1, n=inf so i take an infinite steps –  coffeemachine Oct 15 '11 at 14:32
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@coffeemachine: Please familiarize yourself with the standard mathematical treatment of infinities before you go downvoting answers. You may not have intended your question to refer to this standard treatment; in that case, you would need to define phrases such as "every point on the circle". Under the standard meaning of such expressions accepted throughout standard mathematics, you have not visited all of the uncountably many points despite $n$ going to (countable) infinity. –  joriki Oct 15 '11 at 15:10
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The power set of the integers has cardinality continuum. I win, no sorry, Math wins –  coffeemachine Oct 15 '11 at 15:38
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You certainly aren't trying to make any friends, are you? –  Ragib Zaman Oct 15 '11 at 15:41
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@coffeemachine: There are only countably many finite subsets of natural numbers. –  Asaf Karagila Oct 15 '11 at 17:24
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The number of points on the circle is an uncountable infinity, therefore for your second question, you will never visit all the points.

As a variant of the diagonalisation argument - let the circumference of the circle be unit length 1. For your first step you visit point .1, the second .11, the third .111 and so on - so we have:

Step Point 0 0.000000000000000...

1 0.100000000000000...

2 0.110000000000000...

3 0.111000000000000... and so on...

But such a sequence of steps, no matter how constructed could never cover all the points on the circle - as if we can always find an unvisited point by taking the diagonal and reversing the numbers eg

here the diagonal is (0, 0, 0, 0....)

So the reverse would be (1, 1, 1, 1...) ie 0.11111111111

But that cannot be the first item, the second, the third and so on...

So let us say we add this point, then the new diagonal is

1, 0, 0, 0, 0 etc, and the new, unvisited point is 0.011111111....

Again, this cannot be the first, the second, the third and so on point...

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If you write down rational numbers in interval $[0,1]$:

$0, 1, \frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{3}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5},\frac{1}{6}, \frac{5}{6}, \frac{1}{7}, \frac{2}{7}, \dots$

and on step $i$ visit the $i$-th number in this sequence, then every point in the segment will be within $\delta$ of some visited point. Rational numbers are countable but arbitrary close to real numbers which are uncountable. There is no contradicion here. For any real number and desired approximation you can find a rational number which is close to the real number.

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