MHB Bounded Set with Two Limit Points

[OhMyMarkov]

Hello everyone!

I'm asked to find a set that is bounded and that has exactly two limit points, now this is how I am thinking.

Consider the set $A_n = [0,\frac{1}{n}) \cup(2-\frac{1}{n},2]$, if $A_1 = [0,1)\cup(1,2]$, $A_2=[0,1/2)\cup (3/2,2]$. If I let $n$ grow indefinitely, I will have only two limit points, 0 and 2, right?

Any help is appreciated!

 

[Plato2]

I'm asked to find a set that is bounded and that has exactly two limit points, now this is how I am thinking.
Consider the set $A_n = [0,\frac{1}{n}) \cup(2-\frac{1}{n},2]$, if $A_1 = [0,1)\cup(1,2]$, $A_2=[0,1/2)\cup (3/2,2]$. If I let $n$ grow indefinitely, I will have only two limit points, 0 and 2, right?

No the set $A_1 = \left[ {0,1} \right)$ alone has infinitely many limit points.

$a_n = \left\{ {\begin{array}{rl} {1/n,} & n\text{ is odd} \\ {1 - 1/n,} & n\text{ is even} \\\end{array}} \right.$

 

[OhMyMarkov]

Hi Plato,

Thanks for your reply. I didn't quite understand. I realize [0,1) has infinitely many limit points, I didn't say otherwise. Can you please explain.

 

[Plato2]

Thanks for your reply. I didn't quite understand. I realize [0,1) has infinitely many limit points, I didn't say otherwise. Can you please explain.

Perhaps the problem here is your using interval notation when you mean set notation.
Maybe you meant let $A_n = \left\{ {0,\frac{1}{n}} \right\} \cup \left\{ {2 - \frac{1}{n},2} \right\}$ then let $A = \bigcup\limits_n {A_n } $.
Now $A$ is a bounded infinite set having exactly two limit points.