# Thread: Density in the unit circle

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

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

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

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

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

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

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

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

Now, if we take $\lambda = \frac{1}{2 \pi}$ (which is clearly irrational) and take $\theta = 0$, we get that $\{T_{\lambda}^m(\theta)\}_{m=1}^{\infty} = \{e^{im}\}_{m=1}^{\infty}$ is dense in $S^1$, as conjectured.
Everything up to the first underline was covered by me in post #15:
"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." With the obvious application of Cauchy to exprdess the limit

After the second underline you make the statement: we have that for every $\theta \in S^1$ there exists an integer $n \in \mathbb{N}$, such that...
You do not know that such an n exists for an arbitrary theta. You are doing the same thing that I have been complaining about all along, you are assuming that in a bounded infinite collection of distinct points, every point is a limit point. In such a situation all that Bolzano Weirstrass guarantees is the existence of a limit point, not that every point is a limit point. If you can't grasp that, then this discussion is indeed futile.

(This post was particularly difficult because of the small edit box. Is there a way to enlarge it?)

EDIT: Whoops! Almost forgot. Thanks for taking the trouble to write out the above proof in more detail. Some of it was quite helpful

2. You are doing the same thing that I have been complaining about all along, you are assuming that in a bounded infinite collection of distinct points, every point is a limit point.
I did not assume that.

To understand why that argument is correct, note that:

(1) The arcs connecting $T_{\lambda}^{nk}(\theta)$ with $T_{\lambda}^{(n-1)k}(\theta)$ and $T_{\lambda}^{mk}(\theta)$ with $T_{\lambda}^{(m-1)k}(\theta)$ are pairwise disjoint, for any $m \ne n \in \mathbb{N}$ (this is because $T_{\lambda}^k$ maps the arc connecting the point $\theta$ with $T_{\lambda}^k(\theta)$ to the arc connecting $T_{\lambda}^{k}(\theta)$ with $T_{\lambda}^{2k}(\theta)$, and so on)

(2) The arc connecting $T_{\lambda}^{nk}(\theta)$ with $T_{\lambda}^{(n-1)k}(\theta)$ has the same length as the arc connecting $T_{\lambda}^{mk}(\theta)$ with $T_{\lambda}^{(m-1)k}(\theta)$, for any $m \ne n \in \mathbb{N}$, which has the same length as the arc connecting $T_{\lambda}^{k}(\theta)$ with $\theta$, which is $< \epsilon$.

From (1) we can gather that the set consisting of all arcs connecting $T_{\lambda}^{nk}(\theta)$ with $T_{\lambda}^{(n-1)k}(\theta)$ for any $n \in \mathbb{N}$, will cover the whole unit circle (since the original sequence is infinite and acyclic)

That means that for any $\alpha \in S^1$ there exists an integer $p \in \mathbb{N}$, such that $\alpha$ is contained in the arc connecting $T_{\lambda}^{pk}(\theta)$ with $T_{\lambda}^{(p-1)k}(\theta)$

But from (2), we have that the length of that arc is less than $\epsilon$, and therefore (triangle inequality) we have that also the arc connecting $\alpha$ with $T_{\lambda}^{pk}(\theta)$ has length less than $\epsilon$, and this finishes the proof of the argument.

Note that there was a mistake in the original post -

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

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

3. You write:

"....there exists an integer , such that ..."

You are assuming such an integer exists. You have to prove it.

4. Originally Posted by Hartlw
Everything up to the first underline was covered by me in post #15:
"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." With the obvious application of Cauchy to exprdess the limit

After the second underline you make the statement: we have that for every $\theta \in S^1$ there exists an integer $n \in \mathbb{N}$, such that...
You do not know that such an n exists for an arbitrary theta. You are doing the same thing that I have been complaining about all along, you are assuming that in a bounded infinite collection of distinct points, every point is a limit point. In such a situation all that Bolzano Weirstrass guarantees is the existence of a limit point, not that every point is a limit point. If you can't grasp that, then this discussion is indeed futile.

You see? You continue with your disgusting attitude: it is you that can't grasp stuff here! You, in an almost unbelievable fashion,

continue to claim that someone (this time Defunkt) "assumes" that in an infinite collection of points every point is a limit point.

I'm not sure anymore how to break the news for you, but NOBODY has ever claimed such a thing except. perhaps, you, and thath thing is neither needed nor

even mentioned anywhere in the original post. Please do stop claiming the same nonsense over and over again.

The same can be said about what the B-W theorem states: nobody thinks it promises that EVERY point of whatever is a limit point. For your own educational

sake stop insisting in this.

Tonio

(This post was particularly difficult because of the small edit box. Is there a way to enlarge it?)

EDIT: Whoops! Almost forgot. Thanks for taking the trouble to write out the above proof in more detail. Some of it was quite helpful
.

5. Originally Posted by Hartlw
You write:

"....there exists an integer , such that ..."

You are assuming such an integer exists. You have to prove it.

Well, it seems obvious you can't read mathematical proofs and all this is poinless. Kudos to Defunkt, and anyone else, wasting their time trying

to make you understand something instead of advicing you to go and study some basic maths first.

Tonio