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$X=[0,+\infty) \subset \mathbb{R}$, and $B={[0,n): n \in \mathbb{Z^+} } $.

And let $\tau(B)$ be the topology generated from $B$.

Find the closure and the interior of the following subsets:

$A_1=(1,3)$, $A_2=[0,5]$, $A_3=[0,\sqrt(5)]$, $A_4=\mathbb{Q^+}$

Sol.:

I call $I(A)$ the interior and $Cl(A)$ the closure. And $C_X(A)$ is the complementar set of $A$ in $X$

I tried to understand how are the close sets of $\tau(B)$ and I think they are $[n,+\infty)$ since $C_X([n,+\infty))=[0,n) \in \tau(B)$

So:

$I(A_1)=\emptyset$, because there's no way to get an open set with the extreme left different from $0$. $Cl(A_1)=[1,+\infty)$

$I(A_2)=[0,5)$

$Cl(A_2)=?$

$I(A_3)=[0,2)$

$Cl(A_3)=?$

$I(A_4)=[0,+\infty]$ $Cl(A)=\emptyset$

I really can't find the closure, and I'm not sure also about the interior. Thanks to everyone will give me a hint.

(sorry for my bad english, I'm italian).

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  • $\begingroup$ interpret closure as the intersection of all closed sets $[n,+\infty)$ containing some $A_i$ $\endgroup$ – serg_1 Jun 26 '17 at 12:28
  • $\begingroup$ Which $A_i$ do I have to take to get the closure of $(1,3)$? $\endgroup$ – VoB Jun 26 '17 at 12:31
  • $\begingroup$ only two close sets contain $(1,3)$ it's $X=[0,+\infty)$ and $[1,+\infty)$, so their intersections give you $[1,+\infty)$ $\endgroup$ – serg_1 Jun 26 '17 at 12:48
  • $\begingroup$ the closure of $A_i$ is the SMALLEST closed set containing some $A_i$ $\endgroup$ – serg_1 Jun 26 '17 at 13:10
  • $\begingroup$ the interior of $A_i$ is the LARGEST open set contained in some $A_i$ $\endgroup$ – serg_1 Jun 26 '17 at 13:11
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Closure of a set is the smallest closed set that contains the set.So the compliment of closure is an open set which does not intersects that set.Clearly of $A_2$ and $A_3$ is contained by all the nontrivial open sets,So we have to choose empty set as the open set and it's complement $[0,\infty) $ becomes the closure of $A_2$ and $A_3$. $A_4$,...If you take an arbitrary point say,r then the smallest open set containing r is [0, [r]+1); [r] is the nearest positive integer less than r,It contains irrational points also and hence not contained in Q,All the open sets containing r clearly not contained in Q,So r is not interior point.Since r is arbitrary ,no point of $A_4$ is interior point.So $I(A)$ is empty set;It's closure is $[0,\infty)$,It comes from the same logic that $Q$ has closure $R$.But you are correct in $A_1$.

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  • $\begingroup$ Thanks so much! So $I(A_4)=\emptyset$ because of the density of $Q$ in $R$ and $Cl(A_4)=X$. Is it correct? $\endgroup$ – VoB Jun 26 '17 at 23:06
  • $\begingroup$ Yes,the same idea.If you take any point in $X$then all the open sets containing it intersects $A_4$. $\endgroup$ – Subhajit Saha Jun 27 '17 at 3:53

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