I(A) is the set of all isolated points of A

Question

(isolated point means point $x$ such that $x \in A$ but $x$ is not a limit point of A)

${S_k}$ is a sequnce such that For all $k\in N$, $I(S_k)=\varnothing$

$S=\cap S_k$

Then $I(S)=\varnothing$?

I couldn't find a counterexample, so is this correct?

score 0 · Answer 1 · answered Apr 07 '22 at 13:55

0

Hint:

When looking for a counterexample here, choose an $S$ first, preferably as simple as possible, with $I(S)\neq\varnothing$. Then build a suitable sequence $S_k$.

answered Apr 07 '22 at 13:55

Arthur

199,419

score 0 · Accepted Answer · answered Apr 07 '22 at 14:06

$(\Bbb{R}, \tau_{euclidean}) $

Choose, $S_k=(-\frac{1}{k}, {\frac{1}{k}}) \subset \Bbb{R}$

Prove :

$I(S_k) =\emptyset$
$S= \cap_{k\in\Bbb{N}} S_k =\{0\}$

(Hint : Use Archimedean property of real)

$I(S) =\{0\}$

Done. Let's try to find different examples , probably different topological spaces.

I(A) is the set of all isolated points of A

2 Answers2