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Find the interior points of the set $\,A=\left\{\sqrt2+n\,:\,n\in\Bbb N\right\}$ on the usual topology on $\,\Bbb R\,.$

I think it is the empty set because if $\,p\in A\,$ and $\,p\in U\,,\,$ where $\,U\,$ is an open set, then $\,U\,$ is not a subset of $\,A\,.$

Angelo
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Arghya Santra
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    Arghya, you are right, there does not exist any interior point of the set $A.$ – Angelo Jul 03 '23 at 15:55
  • @Angelo thanks for confirmation – Arghya Santra Jul 03 '23 at 16:00
  • $A$ is s discrete closed set of $\mathbb{R}$, the same way that $\mathbb{N}$ is also a discrete and closed subset of $\mathbb{R}$. – Mittens Jul 03 '23 at 21:14
  • Maybe I misunderstood the question but with usual topology on $\mathbb{R}$, $A$ I think gains the discrete topology with the subspace topology and the interior isn't empty but $A$, @Angelo am I wrong? – Turquoise Tilt Jul 04 '23 at 17:57
  • @TurquoiseTilt, the interior of the subset $A$ of the topological space $\Bbb R$ with the usual topology is empty, indeed for any $a\in A$ and for any open subset $U(a)$ of $\Bbb R$ which contains $a$, it results that $U(a)\not\subseteq A$. – Angelo Jul 04 '23 at 18:38

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