Let $\alpha$ a irrational number, and $R:S^{1}\rightarrow S^{1}$ the irrational rotation, i.e., $[x]\rightarrow[x+\alpha]$. I need to prove that, for all $[x]\in S^{1}$, the set $\{R^{n}([x])\}$ is dense in $S^{1}$.  First, I can write $R$ by  $$R(e^{2\pi ix})=e^{2\pi i (x+\alpha)}.$$  So, I can write $R^{n}(x)$, $n\in\mathbb{Z}$ by  $$R^{n}(e^{2\pi i x})=e^{2\pi i(x+n\alpha)} $$  I need to prove that, for all $e^{2\pi i x}\in S^{1}$ and for all $[y]=e^{2\pi i y}\in S^{1}$, every neighborhood $V$ of $[y]$ contains a point $[z_{y}]=e^{2\pi i z}$ such that $[z_y]=R^{n}([x])$ for some $n\in\mathbb{Z}$. That is,  $e^{2\pi i z}=e^{2\pi i (x+\alpha n)}\Rightarrow 2\pi iz=2\pi i(x+\alpha n)+2ik\pi\;\textrm{for some}\;k\in\mathbb{Z}\Rightarrow z=x+\alpha n+k.$  Is my way correct? If it does, how can I proceed now? If it doesn't, what I need to do?  I don't think this question is duplicate. I'm showing my attempt to proof, that is different from other proofs.
122250
This is exercise 1.9 in Lie Groups, Lie Algebras and Representations - Hall.     Suppose $a$ is an irrational real number.  Show that the set $E_a$ of the numbers of the form $e^{2\pi i n a}$, $n \in \mathbb{Z}$, is dense in the unit circle $S_1$.  Hint: Show that if we divided $S^1$ into $N$ equally sized "bins" of length $2\pi/N$, there is at least one bin that contains infinitely many elements of $E_a$.  Then use the fact that $E_a$ is a subgroup of $S^1$.   My proof of this proposition is as follows.  Since $a$ is irrational, you can determine that the set of rotations $E_a$ is infinite.  Since $S^1$ is compact we can find two $r_1, r_2$ that are within $\epsilon$ of each other.  Then $r_1^{-1}r_2$ is a small rotation of size $\epsilon$.  Now, $r_1^{-1}r_2$ generates rotations that are within $\epsilon$ distance of any point of $S^1$.    That said, I don't believe that the hint suggested in the problem uses that technique.  My knowledge of algebra is not all that strong so I was hoping someone could shed some light on what is being suggested there.
20485
Let $x\in \mathbb{R}$ an irrational number. Define $X=\{nx-\lfloor nx\rfloor: n\in \mathbb{N}\}$. Prove that $X$ is dense on $[0,1)$.    Can anyone give some hint to solve this problem? I tried contradiction but could not reach a proof.  I spend part of the day studying this question   and its answers. Only one answer is clear and give clues to solve the problem. This answer is the first one. However, this answer does not answer the question nor directly, nor the proof follows from this answer.   This answer has some mistakes, he use that $[(k_1-k_2)\alpha]=[k_1\alpha]-[k_2\alpha]$ which is not true. Consider $k_1=3, k_2=1, \alpha=\sqrt{2}$ we have $[(k_1-k_2)\alpha]=2\not= 3=[k_1\alpha]-[k_2\alpha] $. We only can assure that $[k_2\alpha]-[k_1\alpha]-1\leq [(k_2-k_1)\alpha]\leq[k_2\alpha]-[k_1\alpha]$.   Who answered said something interesting about additive subgroups of $\mathbb{R}$, but unfortunately the set $X=\{nx-[nx] : n\in \mathbb{N} \}$ is not a subgroup. Considering the additive subgroup $G=\langle X \rangle$, if we prove the part (a) of the link, we get that indeed $G$ is dense on $\mathbb{R}$ but we can not conclude that $X$ is dense on $[0,1)$.  I think this problem has not been solved.  Thanks!
54666
Possible Duplicate:          I have to show that show that $A=\{ \frac{m}{2^n}:m\in \mathbb {Z},n\in \mathbb {N}\} $ is dense in $\mathbb {R}$.  A set A is dense in $\mathbb {R}$ if $\overline A=\mathbb {R}$.  But also $Y$ is a subset of $X$, we say that $Y$ is dense in $X$, if for every $x\in X$ , there is $y \in Y$  that is arbitary close to $x$.  So ,I have to prove that for every $x \in \mathbb {R}$ ,there is a number $\frac{m}{2^n}$ arbitrarily close to $x$.So $\forall \epsilon,x ,\exists y$ such that $|y-x|&lt;\epsilon$.   I got a little stuck at this point...Could anyone give me a hint?Thanks a lot!
17281
Does there exist a positive irrational number $\alpha $, such that for any positive integer $n$ the number $\lfloor n\alpha \rfloor$ is not a prime? My try if $\alpha=\sqrt{17}$ then $\lfloor n\alpha \rfloor=4n$
33631
Let $T=\{z:|z|=1 \mathbin{\text{and}} z \in \mathbb C\}$ be the unit circle in the complex plane, considered as a topological group under complex multiplication and the usual topology. Show that every subgroup of $T$ is either dense in $T$ or finite.  How to prove the denseness?
101174
Prove $\alpha \in R[[x]]$ is a unit iff $a_0 \in R$ is a unit.   $"\Rightarrow"$ suppose $\alpha \in R[[x]]$ is a unit, where $\alpha = \sum_{n=0}^{\infty} a_nx^n$  then there exists an inverse $\beta = \sum_{n=0}^{\infty} b_nx^n$   stuck at this spot,  $(\sum_{n=0}^{\infty} a_nx^n)(\sum_{n=0}^{\infty} b_nx^n) = \sum_{n=0}^{\infty}(\sum_{k=0}^{n}(a_kb_{n-k}))x^n$  A little mixed up on how this becomes $1 = \sum_{n=0}^{\infty}e_nx^n$ where $e_0 =1, e_{n&gt;0} = 0$ and why that would make $a_0b_0 = 1$ which would make $a_0 \in R$ a unit. I haven't tried the other direction yet.  EDIT  Maybe I get it, because we assumed this sum is a unit then its inverse exists and so the $x^0=1$ and all $a_ib_i=1$ thus $a_0b_0=1$ which gives $a_0$ is a unit in $R$
17440
Let $S^{1} := \lbrace(\cos\alpha, \sin\alpha) \subset \mathbb{R}^{2} | \alpha \in \mathbb{R}\rbrace$ be the circumference of radius $1$ and $f: S^{1} \to \mathbb{R}$ a continuous function. Prove that there exist two points diametrically opposed at which $f$ assumes the same value.  My idea for the solution: Define $\varphi: S^{1} \to \mathbb{R}$ as$$\varphi(\cos\alpha, \sin\alpha) = f(\cos\alpha, \sin\alpha) - f(-\cos\alpha, -\sin\alpha).$$ If $f(\cos\alpha, \sin\alpha) = f(-\cos\alpha, -\sin\alpha)$ for all $\alpha \in \mathbb{R}$, the result follows. Otherwise, there exist $\alpha_{1},\alpha_{2}$ such that $$f(\cos\alpha_{1}, \sin\alpha_{1}) - f(-\cos\alpha_{1}, -\sin\alpha_{1}) &gt; 0$$ and $$f(\cos\alpha_{2}, \sin\alpha_{2}) - f(-\cos\alpha_{2}, -\sin\alpha_{2}) &lt; 0.$$ Applying the intermediate value theorem, we prove the result. Is this a correct idea? I appreciate your commentd. Thanks!
66674
I am not sure if what I am looking for even makes sense (or) exists. Anyway I would be happy if someone can clear my confusion.  The set of real numbers $\mathbb{R}$ is obtained as completion of $\mathbb{Q}$. However, $\mathbb{Q}$ is not the only set which is dense in $\mathbb{R}$. $\mathbb{Q} \backslash \mathbb{Z}$ is also a dense subset of $\mathbb{R}$. I am wondering if it makes sense to talk of the "smallest" dense subset of $\mathbb{R}$. To phrase what I am looking for precisely and what I mean by smallest, I am looking for a dense set $A$ of $\mathbb{R}$ such that if $B$ is a proper subset of $A$ then $B$ is not dense in $\mathbb{R}$. I am able to "see" that the set $A$, I am looking for doesn't exist since any open interval consists of infinite rationals. But I am unable to precisely argue out to myself and convince why $A$ doesn't exist. Could some one throw more light on this?
55183
I am trying to prove the following:  In $C[0,1]$ the functions that preserve the rationals (i.e. $f(\mathbb{Q})\subseteq \mathbb{Q}$) are dense.  So far, I have not made much progress- I am aware that $A$ is dense in $B$ iff $A$ has non-empty intersection with each open set $U$ in $B$ but am not sure how to apply this definition. I have also thought about showing that the closure of $X$ (where $X$ is the set of functions that preserve the rationals) is equal to $C[0,1]$ by a double-inclusion proof but I have also made no progress with this approach. Am I on the right track or should I change my approach?
75832
Show that every irrational number in $\mathbb{R}$ is the limit of a   sequence of rational numbers. Every rational number in $\mathbb{R}$ is   the limit of a sequence of irrational numbers.   How can I prove this?
73622
The Weyl equidistribution theorem states that the sequence of fractional parts ${n \xi}$, $n = 0, 1, 2, \dots$ is uniformly distributed for $\xi$ irrational.   This can be proved using a bit of ergodic theory, specifically the fact that an irrational rotation is uniquely ergodic with respect to Lebesgue measure. It can also be proved by simply playing with trigonometric polynomials (i.e., polynomials in $e^{2\pi i k x}$ for $k$ an integer) and using the fact they are dense in the space of all continuous functions with period 1.  In particular, one shows that if $f(x)$ is a continuous function with period 1, then for any $t$,  $\int_0^1 f(x) dx = \lim \frac{1}{N} \sum_{i=0}^{N-1} f(t+i \xi)$. One shows this by checking this (directly) for trigonometric polynomials via the geometric series.  This is a very elementary and nice proof.  The general form of Weyl's theorem states that if $p$ is a monic integer-valued polynomial, then the sequence ${p(n \xi)}$ for $\xi$ irrational is uniformly distributed modulo 1.  I believe this can be proved using extensions of these ergodic theory techniques -- it's an exercise in Katok and Hasselblatt.  I'd like to see an elementary proof.  Can the general form of Weyl's theorem be proved using the same elementary techniques as in the basic version?
106467
Good evening, I need to compute $\pi_1(\mathbb{R}^2 \setminus (\mathbb{Z} \times \{0\}))$.  I see that the set $\mathbb{R}^2 \setminus \mathbb{Z} \times \{0\}$ has a countable set of points less than whole $\mathbb{R}^2$, so at each point we can retract the space into circles, and finally the space can retract to a wedge of infinite countable circles. Aplying Seifert-van Kampen theorem we can conclude that the fundamental group is free product of $\mathbb{Z}$ with itself countable times. Is this idea correct? Thanks a lot.
105679
Let $\mathbb{RP}^n=\frac{\mathbb{R}^{n+1}\setminus\{0\}}{\sim}$ be the real projective space of dimension $n$, where $\sim$ is the proportionality relation. Then  $\mathbb{RP}^n$ has the quotient topology induced by the canonical surjection $$\pi\colon\mathbb{R}^{n+1}\setminus\{0\}\to \mathbb{RP}^n, \quad x\mapsto [x]_{\sim}$$ Let $p$ and $q$ be distinct element of $\mathbb{RP}^n$, and let be $\pi_{|S^n}^{-1}(\{p\})=\{x,-x\}$ and $\pi_{|S^n}^{-1}(\{q\})=\{y,-y\}$. Let be $\varepsilon\leq \frac{1}{2}\text{min}\{\lVert x-y\rVert,\lVert x+y\rVert\}$. Let be $U=B_\varepsilon^{n+1}(x)\cup B_\varepsilon^{n+1}(-x) $ and $V=B_\varepsilon^{n+1}(y)\cup B_\varepsilon^{n+1}(-y) $. Then $U$ and $V$ are open and disjoint. Let be $U'=\pi(U)$ and $V'=\pi(V)$. Since $\pi$ is an open map, then $U'$ and $V'$ are open neighborhood of $p$ and $q$ respectively.  If I show that $U'$ and $V'$ are disjoint, then I have completed the proof that  $\mathbb{RP}^n$ is Hausdorff.  Suppose there are $u\in U$ and $v\in V$ such that $\pi(u)=\pi(v)$.  My book says that I can consider $u$ and $v$ of norm${}=1$.  Then, by this assumption, from $\pi(u)=\pi(v)$ follows that $u=\pm v \in U\cap V$ which is absurd.  My question is: why can I consider $u$ and $v$ of norm${}=1$? For example, suppose that $u\in B_\varepsilon^{n+1}(x)$. According to my book, I can replace $u$ with $\frac{u}{\lVert u\rVert}$ since $\pi(u)=\pi(\frac{u}{\lVert u\rVert})$. But form $u\in B_\varepsilon^{n+1}(x)$ doesn't follow that $\frac{u}{\lVert u\rVert}\in B_\varepsilon^{n+1}(x)$. So $\frac{u}{\lVert u\rVert}$ could not be in $U$.
100789
$\mathbb{Q}$ is dense in $\mathbb{R}$ (with the standard topology). I'm pretty sure that $\mathbb{Q}^n$ is dense in $\mathbb{R}^n$ too. Is there an easy argument to prove that without reproducing the same argument to show "$\mathbb{Q}$ dense in $\mathbb{R}$"?
63191
Prove the following in $S_n$:  Let $\alpha = (a_1,a_2,\ldots,a_s)$ be a cycle and let $\pi$ be a permutation in $S_n$. Then $\pi\alpha\pi^{-1}$ is the cycle $(\pi(a_1),\ldots,\pi(a_s)).$     I'm not sure where to even start.
4062
Consider the function $f: [0,1] \to \mathbb{R}$ where f(x)= \begin{cases} \frac 1q &amp; \text{if } x\in \mathbb{Q} \text{ and } x=\frac pq \text{ in lowest terms}\\ 0 &amp; \text{otherwise} \end{cases}  Determine whether or not $g$ is in  $\mathscr{R}$ on $[0,1]$ and prove your assertion. For this problem you may consider $0= 0/1$ to be in lowest terms.  Here's an attempt. I may have abused a bit of notation here, but the ideas are there.   Proof: Let $M_i = \sup \limits_{x \in [x_{i-1},x_i]} f(x)$.  Notice first that the lower Riemann sums are always $0$, since every interval contains an irrational number. Thus, to prove $f \in \mathscr{R}$, it suffices to prove that, given any $\epsilon &gt;0$, $\sum \limits_{i \in P} M_i \Delta x_i &lt; \epsilon$ for some partition.     Let $\epsilon &gt; 0 $ and $M  &gt; \frac{2}{\epsilon}$. We first show that there exists $\eta(x,\frac{1}{M})$ so that $|f(x) - f(y)| &lt; \frac{1}{M}$ if $|x-y| &lt; \eta$. Fix $x \in (\mathbb{R} \setminus \mathbb{Q}) \cap [0,1]$. Now, consider the set $$R_{M} := \{ r \in \mathbb{Q} : r = \frac{p}{n}, n \leq M, p \leq n, p \in \mathbb{N} \}.$$ Clearly this set is finite, enumerate it as $\{q_1,\ldots, q_m\}$. So, let $$\eta(x,\frac{1}{M}) = \min_{i=1,\ldots, m} |x- q_i|.$$ We see then, $|f(x) - f(y)| &lt; \frac{1}{M}$ on this $\eta$-neighborhood.    After we choose that $\eta$ so that $x \in (\mathbb{R} \setminus \mathbb{Q}) \cap [0,1]$, is continuous in a $\eta$-neighborhood, we see  $$ A:= [0,1] \setminus R_M \subset \left( \bigcup_{ x \in ( \mathbb{R} \setminus \mathbb{Q}) \cap [0,1]} B_{\eta(x)} (x) \right) \cap [0,1].$$  Since $A$ is compact, we may take finite sub-covering, and let $\delta = \min \limits_{i=1,\ldots,n} \{\eta(x_i)\}$. Take a partition $P_1$ of $A$ so that $\Delta x_i &lt; \delta$. Since $R_M$ is non-empty, we can take a partition $P_2$ of $R_M$ so that $\Delta x_i &lt; \frac{\epsilon}{2m}.$ Moreover, we see that,  on $[0,1]$, $f$ is at most $1$. Let $P = P_1 \cup P_2$. Thus,   \begin{eqnarray*} \sum_{i \in P} M_i \Delta x_i &amp;=&amp; \sum_{i \in P_1} M_i \Delta x_i + \sum_{i \in P_2} M_i \Delta x_i \\ &amp;\leq&amp; \frac{1}{M} \sum_{i \in P_1} \Delta x_i + \sum_{i \in P_2} \Delta x_i \\ &amp;&lt;&amp; \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon  \end{eqnarray*}    Comments?    EDITED I think I resolved the issue.
123213
Let $S^1\subseteq\mathbb{C}$ be the unit circle and let $U\subseteq S^1$ be open. How to show that there exist a continuous function $$\theta:U\to\mathbb{R}$$ such that $$e^{i\theta(z)}=z$$ for all $z\in U$ if and only if $U\neq S^1$?  I have been able to show the $\Rightarrow$ direction but not the other one.  $(\Rightarrow)$ If such a continuous $\theta:S^1\to\mathbb{R}$ exist, then it is injective and so $\theta(S^1)$ is a subset of $\mathbb{R}$ that is homeomorphic to $S^1$. Now, $S^1$ is compact and connected so $\theta(S^1)$ is a closed interval $[a,b]$. But removing a point from $[a,b]$ gives two connected components, while removing a point from $S^1$ still gives a connected space. Hence, $[a,b]\not\cong S^1$, so $\theta$ doesn't exist.  How would you show the $\Leftarrow$ direction?  Edit:  This is an exercise in topology with no assumed prior knowledge of complex analysis. Note that we only need to show continuity of $\theta$ (not analyticity). The question is phrased in terms of complex numbers for simplicity of notation, but it is to be interpreted in terms of subspace of $\mathbb{R}^2$.