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  • If $D \subseteq \mathbb{R}^n$ contains an isolated point then $D$ is not open.

  • Let $D \subseteq \mathbb{R}^n$ and $a \in D$. If $a$ is a limit point of $D$ then there is some $\epsilon > 0$ such that $B_{\epsilon}(a) \subseteq D$.

For the first, by definition of isolated point, it must be in $D$ but not be a limit point of $D$. I have a feeling if $D = \mathbb{R}^n$ this could be a counter example, as it is a clopen set.

For the second, the consequent is the definition of open if I'm not mistaken, but I'm not sure how this relates to $a$ being a limit point of $D$.

Any help would be appreciated?

qwertyburty
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1 Answers1

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If $x$ is an isolated point of $D$, there is an $\epsilon_0>0$ such that $B(x,\epsilon_0)\cap D=\{x\}$. If $D$ were open, there would be an $\epsilon_1>0$ such that $B(x,\epsilon_1)\subseteq D$. Let $\epsilon=\min\{\epsilon_0,\epsilon_1)$; then on the one hand $$x\in B(x,\epsilon)\cap D\subseteq B(x,\epsilon_0)\cap D=\{x\}\,,$$ so $B(x,\epsilon)\cap D=\{x\}$, but on the other hand

$$B(x,\epsilon)\subseteq B(x,\epsilon_1)\subseteq D\,,$$

so $B(x,\epsilon)\cap D=B(x,\epsilon)$. Thus, $B(x,\epsilon)=\{x\}$, which is false in $\Bbb R^n$. This shows that $D$ cannot be open.

For the second question, let $n=1$, and let $D=(0,1)$; $0$ is a limit point of $D$, but there is no $\epsilon>0$ such that $(-\epsilon,\epsilon)\subseteq(0,1)$.

Brian M. Scott
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