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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?

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Yes. Can you answer why you're right? – Henrique Tyrrell Dec 18 '12 at 2:51

$3$. $0$ 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.

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