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let $n \in \mathbb{N}$ . Set $A_n = \{ n , n+1, n+2,....\} $ and $ T = \{ \emptyset , A_n\}_{ n\in \mathbb{N} }$

$1.$Is $T $is separated ?

$2.$ what are dense set in $T$ ?

$3.$ what is the interior and the closure of $\{4\}$ and of $\{2,4,6,8.....\}$?

My attempt : for $ 1,$ I know that $ T$ is not separated because take $A_1 = \{ 1, 2,3,...\}$ and $A_2 = \{ 2, 3,4,,,,,,,\}$ here both $A_1 \cap A_2 \ne \emptyset$

For $2)$ i thinks $\mathbb{N} $

For $3)$ im confused

Any hints/solution will be appreciated

thanks u

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    $\begingroup$ Please use \emptyset to denote the empty set, not $\phi$! $\endgroup$ – B.Swan May 1 at 10:47
  • $\begingroup$ @B.Swan okkss.. $\endgroup$ – jasmine May 1 at 10:48
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$1)$ Actually, $A_n\cap A_m\neq \emptyset$ for every $n,m\in\Bbb{N}$. That should help you prove that no pair of points can be separated.

$2)$ A subspace $S$ is dense when its closure $\overline{S}$ is the whole space. A point $x\in\overline{S}$ is a point for which every neighbourhood contains a point of $S$. It is not hard to prove that $S$ is dense if and only if $S$ is infinite. If $S$ is finite, then it has a maximum element $s_0$, so any point $s>s_0$ has a neighbourhood which does not intersect $S$. Similarly, if for every $n\in \Bbb{N}$ there is a neighbourhood containing a point in $S$, then $S$ must be infinite.

$3)$ Their interior is empty because no open set is contained in $\{4\}$ nor in $\{2,4,6,\dots\}$.

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    $\begingroup$ thanks u javi,, $\endgroup$ – jasmine May 1 at 11:09
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Your answers:

1) Incorrect. The fact that two particular open sets intersect does not imply at all that the topology is not separated. You should pick two natural numbers $m,n$ (call $m$ the lesser one, for example) and try to show that this points can not be separated.

2) Incorrect. Of course $\Bbb N$ is dense. But how are the closed sets? What would be the closure of the set of, say, even numbers?

For the third one, perhaps you have a clearer idea of how the topology is. Good try!

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  • $\begingroup$ .okss @ajotatxe for 2) i thinks ur assumption is wrong why are r u taking even number ? for third i think interior of$ \{4\}= \emptyset$ and closure $\{4\} = \mathbb{N}$ $\endgroup$ – jasmine May 1 at 11:01
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    $\begingroup$ @jasmine The closure of $\{4\}$ is $\{1, 2, 3, 4\}$. Try showing that, if an open set contains $1$, $2$, or $3$, then it must contain $4$. Also, if $n > 4$, then try finding an open set that contains $n$ but not $4$. (Or just write out all the closed sets, and find the smallest one that contains $4$.) $\endgroup$ – Theo Bendit May 1 at 11:06
  • $\begingroup$ @TheoBendit thanks u $\endgroup$ – jasmine May 1 at 11:09

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