1. Accumlation Points

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.

2. 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?

3. Is the answer (in white) all the numbers in [0,2]X[0,2] since this is the set of all rationals in (0,2]X(0,2]and the rationals are dense in the reals?

4. Originally Posted by Mathstud28
Is the answer (in white) all the numbers in [0,2]X[0,2] since this is the set of all rationals in (0,2]X(0,2]and the rationals are dense in the reals?
Find a pair of positive integers n,m such that (1/n)+(1/m) is in (3/2,2)

CB

5. I think the answer should be [0,1/2] X [0,2] . Am I right?

Since when n=m=1, (1/2, 2)

when n=m=2, (1/4, 1)

when n and m go to infinity, (0,0)

6. Originally Posted by selinunan
I think the answer should be [0,1/2]X [0,2] . Am I right?
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 $\displaystyle k \to \infty$.

7. Then, the only accumulation point is (0,0), isn't it?

Because as n and m goes to infinity, the set goes to (0,0) and D( (0,0), r) contains at least one point of the set.

Is it true now?

8. Originally Posted by selinunan
the only accumulation point is (0,0), isn't it? NO
Because as n and m goes to infinity,
If you hold n fixed, $\displaystyle n = J\,\& \,m = k$ then we get the points $\displaystyle \left( {\frac{1}{J+k},\frac{k+J}{Jk}} \right)$ as $\displaystyle k \to \infty$
you get the limit $\displaystyle \left( 0 ,\frac{1}{J} \right)$ .
Do you see that?

9. But it will change if you hold m constant.

If you hold m fixed,then the limit will be (0, 1/k).

Then is it possible to say that the accumulation points are (0,1/m) and (0,1/n)?

Sorry for asking too much but I've got to learn that.

10. Originally Posted by selinunan
But it will change if you hold m constant. NO!
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 $\displaystyle \left\{ {\left( {0,0} \right)} \right\} \cup \left\{ {\left( {0,\frac{1}{J}} \right):J \in \mathbb{Z}^ + } \right\}$.