The interval [0,1] with additional modulo 1

Question

The interval $[0, 1]$ is an abelian group with addition modulo $1$. Let $H$ be a proper subgroup of $[0, 1]$, which is closed as subset of $[0,1]$. Show $H$ is finite.

I assumed $H$ is infinite: since $H$ is closed, all limit points of $H$ are in $H$.

Then let $x$ be in $[0, 1]\setminus H=S$ ($S$ is open since $H$ is closed) so there exists an open set $U_x$ which contains $x$ in it and $U_x \subset S$.

Let $\{x_n \}_n$ be a sequence converging to $x$. Thus for all $\epsilon>0$ there exists a $N$ natural number so that for all $n>N$ $|x_n-x|<\epsilon$.

For some $\epsilon_1 >0$ $B(x,\epsilon_1)$ is contained in $U_x$, then if I prove that $x$ is an limit point of $H$ then I can say $H$ is dense (because I express a generic element of $[0, 1]$ as belonging to $H$ or as limit point of $H$) then closure of $H$ will equal to $[0, 1]$ and this would contradict $H$ being proper subgroup.

But I couldn't find a way to show that.

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score 2 · Answer 1 · answered Nov 16 '20 at 11:48

2

A canonical answer.

It suffices to show that each closed infinite subgroup $H$ of $G=[0,1]$ coincides with $G$. Suppose to the contrary that $G\setminus H$ is non-empty. Since $H$ is closed, $G\setminus H$ contains an interval $I$ of length $\varepsilon>0$. By the pigeonhole principle there exist elements $x<y$ of $H$ such that $y-x<\varepsilon$. Then $n(y-x)\in I$ for some natural $n$. But $n(y-x)\in H$, a contradiction.

answered Nov 16 '20 at 11:48

Alex Ravsky

90,434

How exactly this happens ? " By the pigeonhole principle there exist elements $x<y$ of $H$ such that $y−x<\epsilon$." – Jale'de jaled Nov 22 '20 at 07:30
1

Is it because I have infinite element from $H$ and if I divide $[0,1]$ into nonzero $\epsilon$ intervals, some $x<y$'s must be in one of these intervals? but still cannot be sure – Jale'de jaled Nov 22 '20 at 07:32
@Jale'dejaleuffnejale Right. ${}{}{}$ – Alex Ravsky Nov 22 '20 at 07:59

N.B. · Answer 2 · 2018-11-06T11:04:38.630

If you did not know you are asking about the subgroups of the circle $S^1$ in the complex plane $\mathbb{C}$. The exponential $e^{ 2\pi i}\colon [0, 1]_{/\mathbb{Z}}\rightarrow S^1$ defines a group isomorphism that is actually an homeomorphism of spaces.

The problem is widely explained in other questions, for example this one should contain all you need: Subgroup of the unit circle under complex multiplication

Anyway I see a problem in your proof: basically you are not using the fact that $H$ is a subgroup of $S^1$. To get that the $x$ is limiting point of $H$ you have to use this, otherwise you would show that every infinite subset of $S^1$ is dense (clearly false). You could use an argument similar to the one suggested by Seirios in the question I linked: by taking the preimage along the projection get a subgroup of $\mathbb{R}$ and consider the infimum of its positive elements.

score 1 · Answer 3 · answered Nov 16 '20 at 11:57

Hint : Consider the function $\pi : \mathbb{R} \rightarrow [0,1)$ defined for all $x \in \mathbb{R}$ by $\pi(x)=x-\lfloor x \rfloor$. Then :

Show that $\pi^{-1}(H)$ is a subgroup of $\mathbb{R}$.
Deduce that either $\pi^{-1}(H)=a\mathbb{Z}$ for a real number $a$, either $\pi^{-1}(H)$ is dense in $\mathbb{R}$.
Show that the fact that $H$ is closed in $[0,1)$ implies that $\pi^{-1}(H)$ cannot be dense, so it has to be of the form $a \mathbb{Z}$.
Deduce that $H$ is finite.

The interval [0,1] with additional modulo 1

3 Answers3