# Thread: Density in the unit circle

1. Originally Posted by Hartlw
I sense FernandoRevilla is correct though I can't follow it (I am not qualified to judge). Defunkt states, in effect, a countably infinite set of points in a real interval doesn't include all the points. That is not the question at hand, though I admit I was a little sloppy in some previous phrasing about "all points."

From an Engineering perspective:

e^ik is a point on the unit circle at an angle of k radians.

If I proceed around the unit circle in steps of k radians do I ever hit the same point again, ie, after m steps do I move a multiple of 2pi? Or, does k exist such that m=2pik? Is pi rational? No.
So far, correct.

So I have a countably infinite number of points on the circle. Are they dense?

Somehow this has to get to the equivalent question, are the rational points on [0,1] dense? Yes, because every real point of the interval is a limit point of the rational numbers on the interval.
I don't see how this question reduces to the density of the rationals in [0,1] - surely not every dense set in [0,1] is $\displaystyle \mathbb{Q} \cap [0,1]$, right? I mean, for example, $\displaystyle cos(1)$ is not rational..

An infinite number of points on the circle must indeed have a limit point, but is every point hit as per above a limit point, in terms I could understand? In other words, given any point on the circle, are there points constructed as per above on the circle within an arc length epsilon?

EDIT: So I have the question, given any point on the unit circle, how many times do I have to go around to get to within a distance epsilon? Getting close.
Fernando's proof essentially seals the deal. What do you not understand about it?

2. Sorry you missed my edit. Let me summarize:

e^ik is a point on the unit circle at an angle of k radians.

If I proceed around the unit circle in steps of k radians do I ever hit the same point again, ie, after m steps do I move a multiple of 2pi? Or, does k exist such that m=2pik? Is pi rational? No.

So I have a countably infinite number of points on the circle. Are they dense?

An infinite number of points on the circle must indeed have a limit point, but is every point hit as per above a limit point, in terms I could understand? In other words, given any point on the circle, are there points constructed as per above on the circle within an arc length epsilon?

EDIT: So I have the question, given any point on the unit circle, how many times do I have to go around to get to within a distance epsilon? Getting close.

OK: For any angle a and any desired degree of accuracy (couched in terms of epsilon),
m/n<=a<=(m+1)/n
m<=na<=(m+1)
k=m

I don't see where Fernando proves you can get to within a distance epsilon of any point other than "..there must be integers m and n ..."
But pick whichever proof you like.

3. I don't really understand your proof. From what I gather, you say that there exist integers $\displaystyle m,n \in \mathbb{N}$ such that $\displaystyle \frac{m}{n} \le a \le \frac{m+1}{n}$, but what is k? and what does choosing k=m prove?

Fernando showed that for any $\displaystyle \epsilon > 0$, you can 'cover' the circle with arcs of length smaller than $\displaystyle \epsilon$, each of them having a point of the form $\displaystyle e^{ik}$ inside. Do you see why this proves the result?

4. From Rosenlicht, Intro to Analysis, pg26, LUB4

LUB 4. For any x in R and positive integer N, there is an integer n st n/N<=x<(n+1)/N

OK: For any angle a and any desired degree of accuracy (couched in terms of epsilon),
m and n exist st (see Rosenlicht above):
m/n<=a<(m+1)/n
m<=na<(m+1)
k=m

If e^ik = e^im, then k/n<=a<(k+1)/n
ie, e^ik differs from e^ia by at most 1/n, n arbitrary..

e^ia is the point on the unit circle at angle a. e^ik is the point on the unit circle at angle k (in radians). The arc distance between them is a-k/n < 1/n

Edit: Need to clean up algebra. No. It's OK. A little tough to see.

5. The purpose of this post is to alert people following this thread of an Edit in previous post and to answer a question of Defunkt's.

Originally Posted by FernandoRevilla
Theorem

Let $\displaystyle \lambda\in\mathbb{R}$ and consider:

$\displaystyle T_{\lambda}:S^1\rightarrow S^1,\quad T_{\lambda}(\theta)=\theta +2\pi\lambda$

Then, each orbit of $\displaystyle T_{\lambda}$ is dense in $\displaystyle S^1$ if $\displaystyle \lambda$ is irrational

Proof

Let $\displaystyle \theta\in S^1$ . The points on the orbit of $\displaystyle \theta$ are distinct for if $\displaystyle T_{\lambda}(\theta)^n=T_{\lambda}^m(\theta)$ we would have $\displaystyle (m-n)\lambda\in \mathbb{Z}$, so that $\displaystyle n=m$ . Any infinite set of points on the circle must have a limit point. Thus, given any $\displaystyle \epsilon >0$ there must be integers $\displaystyle n$ and $\displaystyle m$ for which $\displaystyle |T_{\lambda}^n(\theta)-T_{\lambda}^m(\theta)|<\epsilon$ . Let $\displaystyle k=n-m$ . Then $\displaystyle |T_{\lambda}^k(\theta)-\theta|<\epsilon$ .

Now $\displaystyle T_{\lambda}$ preserves lengths in $\displaystyle S^1$ . Consequently, $\displaystyle T_{\lambda}^k$ maps the arc connecting $\displaystyle \theta$ to $\displaystyle T_{\lambda}^k(\theta)$ to the arc connecting $\displaystyle T_{\lambda}^k(\theta)$ and $\displaystyle T_{\lambda}^{2k}(\theta)$ which has length less than $\displaystyle \epsilon$ . In particular it follows that the points $\displaystyle \theta,T_{\lambda}^k(\theta),T_{\lambda}^{2k}(\the ta),\ldots$ partition $\displaystyle S^1$ into arcs of length less than $\displaystyle \epsilon$ . Since $\displaystyle \epsilon$ was arbitrary, this completes the proof.

Fernando Revilla
The second underlined sentence does not follow from the first. In other words, the fact that a limit point exists does not prove that every point is a limit point.

Previous post by Hartlw has correct proof:

OK: For any angle a and any desired degree of accuracy (couched in terms of epsilon),
m and n exist st (see Rosenlicht above):
m/n<=a<(m+1)/n
m<=na<(m+1)
k=m

If e^ik = e^im, then k/n<=a<(k+1)/n
ie, e^ik differs from e^ia by at most 1/n, n arbitrary..

e^ia is the point on the unit circle at angle a. e^ik is the point on the unit circle at angle k (in radians). The arc distance between them is a-k/n < 1/n

6. Actually, it does follow precisely from the first sentence, by the definition of a limit point.

If you're honestly not going to try to understand his proof and claim that it's wrong when you see a step which you don't understand, then I don't see any point continuing this discussion. Also, as far as I see, the original question has been answered in Fernando's post (well, for anyone but you, but you're not really trying to understand it, only refute it).

7. I agree that the symbology of Fernando's proof is unintelligible to me, which is why I proved it myself in intelligible (my opinion) terms.

But I did follow it enough tio know that the existence of a limit point on an interval does not prove every point is a limit point, as claimed.

I also note that agreement or approval does not imply understanding.

8. Nowhere did he assert that every point is a limit point.

And a comment regarding your attitude - when someone tries to help you, and you tell them "no, you're wrong and this is how you do it" then don't expect them to help you again. If you don't understand a part of the proof, some of the ideas or some of the notation used - then say so.
Telling someone that their proof is wrong when, in fact, it is not, will make them extremely reluctant to help you in the future. Asking them to explain more elaborately is a common thing to do - nobody's going to bite you if you do that. On the other hand, if you simply claim that the proof is false, then you probably will be bitten

Common courtesy can't ever hurt you.

9. Originally Posted by Hartlw
I agree that the symbology of Fernando's proof is unintelligible to me,

Fair enough, but then stop claiming that his proof is wrong only because you don't understand some symbols in it!

Instead, why won't you nicely ask Fernando to explain "his" symbols or anything else in his proof?

I think also that the proof in the link I gave in my message, with the interval [0,1) modulo 1 instead of $\displaystyle S^1$

is pretty straightforward. Did you give it a look and a thought?

Tonio

which is why I proved it myself in intelligible (my opinion) terms.

But I did follow it enough tio know that the existence of a limit point on an interval does not prove every point is a limit point, as claimed.

I also note that agreement or approval does not imply understanding.
.

10. Originally Posted by Hartlw
The purpose of this post is to alert people following this thread of an Edit in previous post and to answer a question of Defunkt's.

The second underlined sentence does not follow from the first.

Of course it does: it's been shown that $\displaystyle \{T^k_\lambda(\theta)\}$ is infinite, so it has (by Bolzano-Weierstrass,

since $\displaystyle S^1$ is compact) a limit point (i.e., a partial limit), which means that some infinite subsquence of $\displaystyle \{T^k_\lambda(\theta)\}$

converges to some point in $\displaystyle S^1$ itself, which means that thus this subsequence is Cauchy and

that's what's written there...

In other words, the fact that a limit point exists does not prove that every point is a limit point.

Why do you think the first part is the same as the above? Where was implied, let alone said, that "every point is a limit point"?

Tonio

Previous post by Hartlw has correct proof:

OK: For any angle a and any desired degree of accuracy (couched in terms of epsilon),
m and n exist st (see Rosenlicht above):
m/n<=a<(m+1)/n
m<=na<(m+1)
k=m

If e^ik = e^im, then k/n<=a<(k+1)/n
ie, e^ik differs from e^ia by at most 1/n, n arbitrary..

e^ia is the point on the unit circle at angle a. e^ik is the point on the unit circle at angle k (in radians). The arc distance between them is a-k/n < 1/n
.

11. Defunkt is right. I do not understand Fernando's proof. As far as I can tell, it assumes what we are trying to prove. It does not prove that every point of e^ik is a limit point. Furthermore, the existence of a limit point, which Bolzano Weirstrass shows, does not prove that every point of e^ik is a limit point.

The problem is straight-forward. I do not consider an obscure theorem in obscure symbolism "help." I consider that a game. By the way, I couldn't solve it.

If someone wants to help, this is what I would cosider help:

Given any point at angle alpha <= 2pi on the unit circle. What are k and n st:

k-n2pi-alpha < epsilon

An intelligible algebraic solution is sought. I'll start it as a thread since this one is hopelessly cluttered, unless some one gives an intelligible answer.

EDIT: By the way, I assume this is the Jacobi theorem Fernando is referring to, in case anyone is interested.
Elements of the theory of elliptic ... - Google Books

12. How can you say in one breath that:

1) You don't understand Fernando's proof
2) Even though you don't understand it, it has to be wrong

????

The topic creator asked for a proof of his proposition. A proof was given by both tonio and Fernando, and you're the only one who complained about the clarity of the proofs, but instead of asking for help understanding them - you're simply claiming that they're wrong, without giving any argument as to why they are! You're not even trying to read our elaborations which are trying to explain to you why the proofs are correct.

The problem is straight-forward. I do not consider an obscure theorem in obscure symbolism "help." I consider that a game. By the way, I couldn't solve it.
You are thinking that because this problem can be easily stated, it can be easily proven. This is a common misconception - for example, see Fermat's Last Theorem - which was conjectured by Fermat on 1637, and it took many generations (more than 350 years) and hundreds, if not thousands of mathematicians to solve - and it can be easily stated: if $\displaystyle n>2$ is an integer, then there are no integers $\displaystyle a,b,c$ such that $\displaystyle a^n + b^n = c^n$. It takes zero mathematical knowledge to understand what this conjecture tries to say, but it took the development of an extremely rich and sophisticated field of mathematics to solve it.

Also, if you'd cared to actually ask us what the notation in Fernando's proof means and how the proof works, we would have gladly elaborated. But I guess you're not even reading anybody's posts but your own, so have fun with that.

I will ask for a moderator to lock this topic (if the topic creator agrees, of course).

13. Originally Posted by Hartlw
Defunkt is right. I do not understand Fernando's proof. As far as I can tell, it assumes what we are trying to prove.

No, it doesn't

It does not prove that every point of e^ik is a limit point.

Why do you think this is what was to be proven? And "a limit point"...of what??

Furthermore, the existence of a limit point, which Bolzano Weirstrass shows, does not prove that every point of e^ik is a limit point.

It also does not prove that Saturn has more than 15 satellites...I think you didn't even understand what was to be proved.

The problem is straight-forward. I do not consider an obscure theorem in obscure symbolism "help."

The problem was obviously not that straightforward for you as you apparently
didn't even understand what had to be proved.
Certainly that Jacobi's theorem may be not that well-known, but its symbolism is completely
standard and its difficulty rather medium/low for anyone with 2-3 years of undergraduate studies in maths.

I consider that a game.

Much of mathematics is a game: a rather wonderful, intelectually challenging and

mind-absorbing game. What with this?

By the way, I couldn't solve it.

After you belittled someone who tried to help and in the same instance you

claimed you did prove it? Well, go figure...

If someone wants to help, this is what I would cosider help:

Given any point at angle alpha <= 2pi on the unit circle. What are k and n st:

k-n2pi-alpha < epsilon

You surely meant: "Given any $\displaystyle \epsilon>0$....etc", and this is exactly what the proof

presented by Fernando did

An intelligible algebraic solution is sought.

It has been given already, both by Fernando and by myself, at least. That you didn't understand

them is , perhaps, deplorable and sad, but you could have asked for further explanations.
Instead, you chose to behave in an even more deplorable way.

Tonio

I'll start it as a thread since this one is hopelessly cluttered.
.

14. Originally Posted by Defunkt
Also, if you'd cared to actually ask us what the notation in Fernando's proof means and how the proof works, we would have gladly elaborated. But I guess you're not even reading anybody's posts but your own, so have fun with that.

I will ask for a moderator to lock this topic (if the topic creator agrees, of course).
I note that your offer to explain Fernando's proof is followed immediateley by the suggestion the thread be locked. You mean no one has picked up that I would like an (intelligible) explanation of Fernando's proof? OK, I will make it official. Please explain Fernando's proof in intelligible terms. In particular, how he shows a limit point exists and why every point of e^ik is a limit point.

PS You need to show every point of e^ik is a limit point on the unit circle in order to show e^ik is dense. But that was all gone over at the beginning. To do that you have to solve: k-n2pi-alpha < epsilon for n and k. Unbelievably, someone asked me to explain epsilon.

15. The suggestion that the thread be locked was because you were not reading the posts explaining the proofs.

Here is a detailed explanation of Fernando's proof. If there is something you do not understand, say exactly what is not clear.

Let $\displaystyle \lambda \in \mathbb{R}$ be an arbitrary irrational number, and let $\displaystyle T_{\lambda} : S^1 \to S^1$ be a function which maps an element of the unit circle to an element of the unit circle.
Define the mapping by $\displaystyle T_{\lambda} (e^{i \theta}) = e^{i ( \theta + 2 \pi \lambda ) }$ and note that, since each element of $\displaystyle S^1$ is uniquely determined by its angle $\displaystyle \theta$ with the positive x axis, this mapping is the same as the mapping that Fernando has defined.

Note that we will use the notation $\displaystyle T_{\lambda}^k(e^{i \theta})$ to abbreviate $\displaystyle \left( T_{\lambda}(e^{i \theta}) \right)^k =$ $\displaystyle = T_{\lambda}(e^{i \theta}) \circ T_{\lambda}(e^{i \theta}) \circ \overbrace{\ldots}^{\text{k times}} \circ T_{\lambda}(e^{i \theta})$

First, we will show that if $\displaystyle m,n$ are distinct integers ($\displaystyle m \ne n$), then $\displaystyle T_{\lambda}^m(\theta) - T_{\lambda}^n(\theta) \ne 0 \text{ or } 2 \pi k, \ k \in \mathbb{N}$:
$\displaystyle T_{\lambda}^m(\theta) - T_{\lambda}^n(\theta) = (\theta + 2 \pi m \lambda) - (\theta + 2 \pi n \lambda) = \theta - \theta + 2 \pi (m - n) \lambda = 2 \pi (m-n) \lambda$
Now, that equals 0 or $\displaystyle 2 \pi k$ iff $\displaystyle m=n$ or $\displaystyle m - n$ is irrational, but we chose them so that $\displaystyle m \ne n$ and $\displaystyle m - n \in \mathbb{N}$, and so $\displaystyle T_{\lambda}^m(\theta) - T_{\lambda}^n(\theta) \ne 0 \text{ or } 2 \pi k, \ k \in \mathbb{N}, \ \forall m \ne n \in \mathbb{N}$.

Now, this means that all the elements of the sequence $\displaystyle \{T_{\lambda}^k(\theta)\}_{k=1}^{\infty}$ are distinct, and therefore this sequence has a limit point (by sequential compactness), ie. a convergent subsequence.

Since that subsequence is cauchy, for any $\displaystyle \epsilon>0$ there exist integers $\displaystyle m<n$ such that $\displaystyle |T_{\lambda}^n(\theta) - T_{\lambda}^m(\theta)| < \epsilon$, but then note that $\displaystyle T_{\lambda}^n(\theta) - T_{\lambda}^m(\theta) = \theta + 2 \pi n \lambda - \theta - 2 \pi m \lambda = 2 \pi (n-m) \lambda = T_{\lambda}^{n-m}(\theta) - \theta$, so let $\displaystyle k = n-m$ to have $\displaystyle |T_{\lambda}^k(\theta) - \theta| < \epsilon$.

Now, note that $\displaystyle T_{\lambda}$ is a length-preserving map (ie. it maps an interval of length $\displaystyle t$ to an interval of length $\displaystyle t$).

Also, note that $\displaystyle T_{\lambda}^k$ maps the arc connecting the point $\displaystyle \theta$ with $\displaystyle T_{\lambda}^k(\theta)$ to the arc connecting $\displaystyle T_{\lambda}^{k}(\theta)$ with $\displaystyle T_{\lambda}^{2k}(\theta)$.
Consequently, since $\displaystyle |T_{\lambda}^k(\theta) - \theta| < \epsilon$ and $\displaystyle T_{\lambda}$ preserves lengths, we have that $\displaystyle |T_{\lambda}^{2k}(\theta) - T_{\lambda}^k(\theta)| < \epsilon$, and by the same fashion $\displaystyle |T_{\lambda}^{mk}(\theta) - T_{\lambda}^{(m-1)k}(\theta)| < \epsilon$ for any $\displaystyle m \in \mathbb{N}$.

Now, since all elements of $\displaystyle \{T_{\lambda}^k(\theta)\}_{k=1}^{\infty}$ are distinct, we have that for every $\displaystyle \theta \in S^1$ there exists an integer $\displaystyle n \in \mathbb{N}$, such that $\displaystyle |\theta - T_{\lambda}^{nk}(\theta)| < \epsilon$, and so we have that $\displaystyle \{T_{\lambda}^{nk}(\theta)\}_{n=1}^{\infty}$ is dense in $\displaystyle S^1$, and in particular any set containing it is dense in $\displaystyle S^1$ - and such is $\displaystyle \{T_{\lambda}^m(\theta)\}_{m=1}^{\infty}$.

Now, if we take $\displaystyle \lambda = \frac{1}{2 \pi}$ (which is clearly irrational) and take $\displaystyle \theta = 0$, we get that $\displaystyle \{T_{\lambda}^m(\theta)\}_{m=1}^{\infty} = \{e^{im}\}_{m=1}^{\infty}$ is dense in $\displaystyle S^1$, as conjectured.

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