$A=\left\{n+\frac{1}{2n}|n\in\mathbb{N}\right\}$
Is this set closed?
I was reading answer form Example to show the distance between two closed sets can be 0 even if the two sets are disjoint
where Nishrito mentioned above set is closed .
But On that account then
$B=\left\{\frac{1}{2n}|n\in\mathbb{N}\right\}$
This set also become closed but I know which is not as $0$ is limit point of it which is not belong to that set.
Any Help in this regard will be appreciated .
Thanks a lot
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Gibbs
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Curious student
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0 is not a limit point of A. – Christian Sykes Apr 29 '18 at 17:12
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but 1 will be limit point then – Curious student Apr 29 '18 at 17:13
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$A$ is a discrete subset of $\Bbb R$, just like $\Bbb N$ is. – Angina Seng Apr 29 '18 at 17:13
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1 isn't either. In fact, A has no limit points at all. – Christian Sykes Apr 29 '18 at 17:13
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Since $\lim_n \frac{1}{2n} = 0$ and $0$ is not in $B$, $B$ is not closed. – Gibbs Apr 29 '18 at 17:14
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Please can you explain ? – Curious student Apr 29 '18 at 17:14
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@MathStudent $A={3/2,9/4,19/6,\cdots}$. Observe that $1<3/2<2<9/4<3<19/6<4<\cdots$, – Angina Seng Apr 29 '18 at 17:15
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Can you find a convergent sequence of distinct points in A? – Christian Sykes Apr 29 '18 at 17:15
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There is no convergent sub sequence in A – Curious student Apr 29 '18 at 17:16
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Exactly. Thus A has no limit points. – Christian Sykes Apr 29 '18 at 17:18
2 Answers
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Yes, $A$ is a closed set. It follows from the fact that there is a $r>0$ such that the distance between any two distinct elements of $A$ is greater than $r$ ($\frac15$ will do, for instance). But $B$ is not closed and therefore there is no way of deducing that $B$ is closed from the fact that $A$ is closed.
José Carlos Santos
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1@MathStudent The set ${1/n:n\in\mathbb N}$ is discrete (every singleton is closed) but is not closed in $\mathbb R$. – Wojowu Apr 29 '18 at 17:26
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The complement of $A$ is an infinite union of open intervals, so the complement is open which makes $A$ an open set.
Also $0$ is not a limit point of $A$, on the contrary it is an isolated point of $A.$
In fact every point of $A$ is an isolated point.
Mohammad Riazi-Kermani
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