Find all accumulation points of the following set in R^2 with the usual metric.
{ (1/n+m , 1/n + 1/m) : n, m ∈ N }
If anyone helps me, I will be thankful.
Here is a starting point.
Is $\displaystyle (0,0)$ an accumulation point of that set? Why or why not?
Is $\displaystyle (0, \frac{1}{2})$ an accumulation point of that set? Why or why not?
Is $\displaystyle (1,0)$ an accumulation point of that set? Why or why not?
I think that both poster and Mathstud28 need to consider the makeup of the set
$\displaystyle S = \left\{ {\left( {\frac{1}{{n + m}},\frac{1}{n} + \frac{1}{m}} \right):\left\{ {n,m} \right\} \subseteq \mathbb{Z}^ + } \right\}$.
Note that the set $\displaystyle S$ is in the quarter closed disk, centered at (0,0) with radius $\displaystyle
\frac{{\sqrt {17} }}{2}$.
The point furthest from the origin is $\displaystyle \left( {\frac{1}{2},2} \right)$ happens when $\displaystyle n = 1\,\& \,m = 1$.
If we let $\displaystyle n = 1\,\& \,m = k$ then we get the points $\displaystyle \left( {\frac{1}{1+k},\frac{k+1}{k}} \right)$.
Now, we must remind ourselves of the definition of accumulation point.
If $\displaystyle p$ is an accumulation point of $\displaystyle S$ then $\displaystyle p$ must be the limit of a sequence of distinct points in $\displaystyle S$.
Again think about letting $\displaystyle n = 1\,\& \,m = k$ then we get the points $\displaystyle \left( {\frac{1}{1+k},\frac{k+1}{k}} \right)$ as $\displaystyle k \to \infty $.
Does that help you to think about this set?
Take a good look at the definition of $\displaystyle S$.
The sums are symmetric. So holding n constant and varying m yields the same limit as holding m constant and varying n for a given constant.
The set of accumulation points of $\displaystyle S$ is $\displaystyle \left\{ {\left( {0,0} \right)} \right\} \cup \left\{ {\left( {0,\frac{1}{J}} \right):J \in \mathbb{Z}^ + } \right\}$.