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Let $D:= \{1/n: n \in\mathbb N_+\} \cup\{0\}$ and $f: D \to \mathbb R$ be any function. Show that $f$ is continuous at $1/n$ for every $n\in \mathbb N$.

I didn't understand what it means by saying continuous at $1/n$. can you please explain it to me?

Ricky
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SANDRA P M
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Sep 29 '23 at 08:44
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    The concept of continuity at a point is introduced in every book on Calculus. – Kavi Rama Murthy Sep 29 '23 at 08:47
  • How will I show that the function is continuous at 1/n for every natural number n – SANDRA P M Sep 29 '23 at 08:54
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    $|x-\frac 1 n| <\delta\equiv \frac 1 {n(n+1)}, x \in D$ implies $x=\frac 1n$. – Kavi Rama Murthy Sep 29 '23 at 08:57
  • @geetha290krm Thank you – SANDRA P M Sep 29 '23 at 11:34
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    $f$ is continuous at $x\in D$ if for all open $U$ with $f(x)\in U$ there is open $V\subseteq D$, $x\in V$, with $f[V]\subseteq U$. For your specific choice of $D$, the points $1/n$ for $n\in\mathbb{N}$ are isolated, that is ${1/n}$ is open, so $V = {1/n}$ always works. So, more generally, any function is continuous at an isolated point. – Jakobian Sep 29 '23 at 11:36
  • You may find this article helpful even if it's long. https://mathforums.com/t/doubt-about-continuity-and-accumulation-points-with-value-of-limit.355714/ – nickalh Sep 29 '23 at 13:32

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