How to find limit points of the following sets of real numbers, for irrational $\alpha$ ?

$(1)$ $\{ m+n\alpha:m,n\in\mathbb{Z}\}.$

$(2)$ $\{ m+n\alpha:m\in\mathbb{N},n\in\mathbb{Z}\}.$

$(3)$ $\{ m+n\alpha:m\in\mathbb{Z},n\in\mathbb{N}\}.$

$(4)$ $\{ m+n\alpha:m,n\in\mathbb{N}\}.$

There is a result, which says that every additive subgroup of the group $(\mathbb{R},+)$ is either discrete or dense in $\mathbb{R}.$ Therefore using this fact we can say that set in $(1)$ is dense in $\mathbb{R}.$ I think set in $(4)$ has no limit point because of $m,n\in\mathbb{N}.$ But I don't know how to find limit points of there sets. Please explain. Thanks in advance.


1 Answer 1


This one and this one cover $(1)$ and $(2)$. The argument that I gave in the second paragraph of my answer to the second question actually shows that the set in $(3)$ is also dense in $\Bbb R$: in it I found an $m\in\Bbb Z$ and an $n\in\Bbb Z^+$ such that $m+n\alpha\in(a,b)$, where $(a,b)$ was an arbitrary non-empty open interval in $\Bbb R$.

The interesting one is $(4)$. Suppose that $x\in\Bbb R$ is a limit point of $A=\{m+n\alpha:m,n\in\Bbb N\}$; then there are sequences $\langle m_k:k\in\Bbb N\rangle$ and $\langle n_k:k\in\Bbb N\rangle$ such that the sequence $\langle m_k+n_k\alpha:k\in\Bbb N\rangle$ in $A\setminus\{x\}$ converges to $x$.

  • Show that without loss of generality we may assume that $m_k\le m_{k+1}$ and $n_k\le n_{k+1}$ for each $k\in\Bbb N$.
  • Conclude that the sequence $\langle x-m_k:k\in\Bbb N\rangle$ is non-increasing: $x-m_{k+1}\le x-m_k$ for all $k\in\Bbb N$.

Suppose first that $\alpha>0$; clearly the sequence $\langle n_k\alpha:k\in\Bbb N\rangle$ is non-decreasing: $n_k\alpha\le n_{k+1}\alpha$ for each $k\in\Bbb N$. It’s also clear that $m+n\alpha\ge 0$ for all $m,n\in\Bbb N$, so we must have $x\ge 0$.

  • Show that there are $k_0,m\in\Bbb N$ such that $m_k=m$ for all $k\ge k_0$, and conclude that $\langle n_k\alpha:k\in\Bbb N\rangle$ converges to $x-m$.
  • Show that there are a $k_1\ge k_0$ and an $n\in\Bbb N$ such that $n_k=n$ for each $k\ge k_1$, and conclude that $m_k+n_k\alpha=x$ for each $k\ge k_1$, contradicting the hypothesis that $\langle m_k+n_k\alpha:k\in\Bbb N\rangle$ is in $A\setminus\{x\}$.

This shows that $A$ has no limit points if $\alpha>0$.

Matters are very different, though, if $\alpha<0$. The argument in the third paragraph of my answer to the second question linked above actually shows in this case that if $(a,b)$ is any non-empty open interval in $\Bbb R$, there are $n\in\Bbb Z^+$ and $m\in\Bbb N$ such that $m+n\alpha\in(a,b)$, so that in this case $A$ is dense in $\Bbb R$.

  • $\begingroup$ Thanking you...you finally give your precise time for my question....thanks a lot again... $\endgroup$
    – neelkanth
    Sep 17, 2015 at 7:13
  • $\begingroup$ @neela: You’re very welcome; I’m sorry that I had to be away so long. $\endgroup$ Sep 17, 2015 at 7:15

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