Exercise 2.17.b

Exercise 2.17.b: Consider the metric space $$\mathbb{R}$$ (with the usual Euclidean metric). Show that every point in $$K = \{k_n = \frac{1}{n} : n\text{ is a positive integer }\}$$ is NOT a limit point of $$K$$.

Olomana
To show that a point $$k_n$$ is not a limit point of $$K$$, all we have to do is construct one open set which contains $$k_n$$ but no other points in $$K$$. The sequence is strictly decreasing ($$k_n > k_{n+1}$$), so if our set does not contain $$k_{n-1}$$ or $$k_{n+1}$$, it won't contain any other points in $$K$$ either. Let:

$$\epsilon = \frac{1}{2}min(d(k_{n-1}, k_n), d(k_n, k_{n+1}))$$

(For $$n = 1$$, the first term in the $$min$$ function disappears, but the construction still works.)

The ball $$B(k_n, \epsilon)$$ is an open set which contains $$k_n$$ but no other points in $$K$$, therefore every point $$k_n$$ in $$K$$ is not a limit point of $$K$$.

Olomana 07:02, March 17, 2011 (UTC)

'''Yep. Note that $$d(k_n, k_{n+1})=\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n^2+n}$$ < $$ \frac{1}{n^2-n}=\frac{1}{n}-\frac{1}{n-1} = d(k_{n-1},k_n)$$, so the whole $$min$$ function was unnecessary, but good for emphasis (as I'm sure you intended). I'd only leave it off to avoid the nonexistent $$k_0$$. --Steven.clontz 20:29, March 17, 2011 (UTC)'''

Actually, I started out like you did, but switched to the $$min$$ notation when I realized that I could prove a more general result, one that does not depend on the particular function defining the sequence. All I need is $$k_n > k_{n+1}$$. The nonexistent $$k_0$$ is awkward, but your inequality has a $$\frac{1}{n-1}$$ in it. Maybe we can't avoid treating $$k_1$$ as a special case.

On second thought, why don't we just define $$k_0 = 2$$? $$k_0$$ is not in $$K$$, but so what? The argument still works.

Olomana 21:38, March 17, 2011 (UTC)