Accumlation Points

**selinunan** · December 22nd 2008, 08:57 AM

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.

**Plato** · December 22nd 2008, 10:14 AM

Here is a starting point.
Is $(0,0)$ an accumulation point of that set? Why or why not?

Is $(0, \frac{1}{2})$ an accumulation point of that set? Why or why not?

Is $(1,0)$ an accumulation point of that set? Why or why not?

**Mathstud28** · December 22nd 2008, 08:30 PM

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?

**CaptainBlack** · December 22nd 2008, 10:42 PM

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

**selinunan** · December 23rd 2008, 11:10 AM

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)

**Plato** · December 23rd 2008, 12:43 PM

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
$S = \left\{ {\left( {\frac{1}{{n + m}},\frac{1}{n} + \frac{1}{m}} \right):\left\{ {n,m} \right\} \subseteq \mathbb{Z}^ + } \right\}$ .
Note that the set $S$ is in the quarter closed disk, centered at (0,0) with radius $<br /> \frac{{\sqrt {17} }}{2}$ .
The point furthest from the origin is $\left( {\frac{1}{2},2} \right)$ happens when $n = 1\,\& \,m = 1$ .
If we let $n = 1\,\& \,m = k$ then we get the points $\left( {\frac{1}{1+k},\frac{k+1}{k}} \right)$ .

Now, we must remind ourselves of the definition of accumulation point.
If $p$ is an accumulation point of $S$ then $p$ must be the limit of a sequence of distinct points in $S$ .

Again think about letting $n = 1\,\& \,m = k$ then we get the points $\left( {\frac{1}{1+k},\frac{k+1}{k}} \right)$ as $k \to \infty$ .

Does that help you to think about this set?

**selinunan** · December 23rd 2008, 01:06 PM

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?

**Plato** · December 23rd 2008, 01:26 PM

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, $n = J\,\& \,m = k$ then we get the points $\left( {\frac{1}{J+k},\frac{k+J}{Jk}} \right)$ as $k \to \infty$
you get the limit $\left( 0 ,\frac{1}{J} \right)$ .
Do you see that?

**selinunan** · December 23rd 2008, 01:38 PM

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.

**Plato** · December 23rd 2008, 02:14 PM

Originally Posted by selinunan

But it will change if you hold m constant. NO!

Take a good look at the definition of $S$ .
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 $S$ is $\left\{ {\left( {0,0} \right)} \right\} \cup \left\{ {\left( {0,\frac{1}{J}} \right):J \in \mathbb{Z}^ + } \right\}$ .

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