Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the set $(-\infty, 0]\cup\{1/n:n\in\mathbb{N}\}$ with the subspace topology. Then

  1. $0$ is an isolated point
  2. $(–2, 0]$ is an open set
  3. $0$ is a limit point of the subset $\{1/n:n\in\mathbb{N}\}$
  4. $(–2, 0)$ is an open set.

I think 3 and 4 are I right?

share|cite|improve this question
Yes. Can you answer why you're right? – Henrique Tyrrell Dec 18 '12 at 2:51

1 Answer 1

$3$. $0$ is a limit point of the subset $\{1/n : n \in \mathbb{N}\}$, Proof: Let $U_0$ be an arbitrary open neighborhood of $0$ of the form $(-a, \epsilon) \cap (-\infty,0]\cup\{1/n : n \in \mathbb{N}\}$, with $\epsilon > 0 $, $a>0$ subject to choice. Then for $N>\frac{1}{\epsilon}$, we have $\frac{1}{N} < \epsilon$. So then:
$$ \{\frac{1}{N},0\} \subset (-a, \epsilon) \cap (-\infty,0]\cup\{1/n : n \in \mathbb{N}\} \cap \{1/n : n \in \mathbb{N}\} $$ i.e. the intersection contains something other than $0$, namely: $\{\frac{1}{N}\}$. So $0$ is a limit point for the set: $\{1/n : n \in \mathbb{N}\}$.

$4.$ $(-2,0)$ is open because $(-2,0)$ is open in $\mathbb{R}$ and $$ (-2,0)\cap (-\infty,0]\cup \{1/n : n \in \mathbb{N}\} = (-2,0) $$ is open in the subspace topology by definition.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.