Containment of Sets Implies Containment of Closures [duplicate]

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  • If $Asubset B$ then $barAsubset barB$

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If $A subset B$, does it imply $overlineA subset overlineB$?






general-topology elementary-set-theory







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marked as duplicate by Martin R, Saucy O'Path, amWhy elementary-set-theory
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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • I'd be curious to know which book does not mention this topic, pedantic as general topology books tend to be.
    – Saucy O'Path
    Sep 9 at 12:57











  • Actually its implicitly used in a proof in Munkres on page 104. Now I'm stuck because I'm wondering if I can have a huge limiting set that doesn't make this true.
    – yoshi
    Sep 9 at 12:59






  • 1




    Also: If A is a subset of B, then the closure of A is contained in the closure of B.
    – Martin R
    Sep 9 at 13:05







  • 1




    @yoshi Ah, ok. It should be exercise 6 page 101, at the end of §17.
    – Saucy O'Path
    Sep 9 at 13:07










  • I may be wrong here but take the affine line over $mathbbC$. Then we have $(0) subset (x-a)$ but $x-a$ is a maximal ideal hence a closed point while the closure of $(0)$ is everything.
    – Sheel Stueber
    Sep 9 at 13:10














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  • If $Asubset B$ then $barAsubset barB$

    3 answers



If $A subset B$, does it imply $overlineA subset overlineB$?






general-topology elementary-set-theory







share|cite|improve this question













marked as duplicate by Martin R, Saucy O'Path, amWhy elementary-set-theory
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 Sep 9 at 13:10


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  • I'd be curious to know which book does not mention this topic, pedantic as general topology books tend to be.
    – Saucy O'Path
    Sep 9 at 12:57











  • Actually its implicitly used in a proof in Munkres on page 104. Now I'm stuck because I'm wondering if I can have a huge limiting set that doesn't make this true.
    – yoshi
    Sep 9 at 12:59






  • 1




    Also: If A is a subset of B, then the closure of A is contained in the closure of B.
    – Martin R
    Sep 9 at 13:05







  • 1




    @yoshi Ah, ok. It should be exercise 6 page 101, at the end of §17.
    – Saucy O'Path
    Sep 9 at 13:07










  • I may be wrong here but take the affine line over $mathbbC$. Then we have $(0) subset (x-a)$ but $x-a$ is a maximal ideal hence a closed point while the closure of $(0)$ is everything.
    – Sheel Stueber
    Sep 9 at 13:10












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This question already has an answer here:



  • If $Asubset B$ then $barAsubset barB$

    3 answers



If $A subset B$, does it imply $overlineA subset overlineB$?






general-topology elementary-set-theory







share|cite|improve this question














This question already has an answer here:



  • If $Asubset B$ then $barAsubset barB$

    3 answers



If $A subset B$, does it imply $overlineA subset overlineB$?





This question already has an answer here:



  • If $Asubset B$ then $barAsubset barB$

    3 answers






general-topology elementary-set-theory




general-topology elementary-set-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 9 at 12:53









yoshi

912716




912716




marked as duplicate by Martin R, Saucy O'Path, amWhy elementary-set-theory
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marked as duplicate by Martin R, Saucy O'Path, amWhy elementary-set-theory
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 Sep 9 at 13:10


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • I'd be curious to know which book does not mention this topic, pedantic as general topology books tend to be.
    – Saucy O'Path
    Sep 9 at 12:57











  • Actually its implicitly used in a proof in Munkres on page 104. Now I'm stuck because I'm wondering if I can have a huge limiting set that doesn't make this true.
    – yoshi
    Sep 9 at 12:59






  • 1




    Also: If A is a subset of B, then the closure of A is contained in the closure of B.
    – Martin R
    Sep 9 at 13:05







  • 1




    @yoshi Ah, ok. It should be exercise 6 page 101, at the end of §17.
    – Saucy O'Path
    Sep 9 at 13:07










  • I may be wrong here but take the affine line over $mathbbC$. Then we have $(0) subset (x-a)$ but $x-a$ is a maximal ideal hence a closed point while the closure of $(0)$ is everything.
    – Sheel Stueber
    Sep 9 at 13:10
















  • I'd be curious to know which book does not mention this topic, pedantic as general topology books tend to be.
    – Saucy O'Path
    Sep 9 at 12:57











  • Actually its implicitly used in a proof in Munkres on page 104. Now I'm stuck because I'm wondering if I can have a huge limiting set that doesn't make this true.
    – yoshi
    Sep 9 at 12:59






  • 1




    Also: If A is a subset of B, then the closure of A is contained in the closure of B.
    – Martin R
    Sep 9 at 13:05







  • 1




    @yoshi Ah, ok. It should be exercise 6 page 101, at the end of §17.
    – Saucy O'Path
    Sep 9 at 13:07










  • I may be wrong here but take the affine line over $mathbbC$. Then we have $(0) subset (x-a)$ but $x-a$ is a maximal ideal hence a closed point while the closure of $(0)$ is everything.
    – Sheel Stueber
    Sep 9 at 13:10















I'd be curious to know which book does not mention this topic, pedantic as general topology books tend to be.
– Saucy O'Path
Sep 9 at 12:57





I'd be curious to know which book does not mention this topic, pedantic as general topology books tend to be.
– Saucy O'Path
Sep 9 at 12:57













Actually its implicitly used in a proof in Munkres on page 104. Now I'm stuck because I'm wondering if I can have a huge limiting set that doesn't make this true.
– yoshi
Sep 9 at 12:59




Actually its implicitly used in a proof in Munkres on page 104. Now I'm stuck because I'm wondering if I can have a huge limiting set that doesn't make this true.
– yoshi
Sep 9 at 12:59




1




1




Also: If A is a subset of B, then the closure of A is contained in the closure of B.
– Martin R
Sep 9 at 13:05





Also: If A is a subset of B, then the closure of A is contained in the closure of B.
– Martin R
Sep 9 at 13:05





1




1




@yoshi Ah, ok. It should be exercise 6 page 101, at the end of §17.
– Saucy O'Path
Sep 9 at 13:07




@yoshi Ah, ok. It should be exercise 6 page 101, at the end of §17.
– Saucy O'Path
Sep 9 at 13:07












I may be wrong here but take the affine line over $mathbbC$. Then we have $(0) subset (x-a)$ but $x-a$ is a maximal ideal hence a closed point while the closure of $(0)$ is everything.
– Sheel Stueber
Sep 9 at 13:10




I may be wrong here but take the affine line over $mathbbC$. Then we have $(0) subset (x-a)$ but $x-a$ is a maximal ideal hence a closed point while the closure of $(0)$ is everything.
– Sheel Stueber
Sep 9 at 13:10










1 Answer
1






active

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up vote
1
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By definition $barA = bigcaplimits_A subseteq C space and space C space closed C$ and the same holds for $barB$.



Now, if $C$ is closed and $B subseteq C$ then $A subseteq C$, so $barA subseteq C$. Considering the intersection over all $C$ closed such that $B subseteq C$ we have $barA subseteq barB$.






share|cite|improve this answer




















  • Simpler version of this approach: math.stackexchange.com/a/1620306/42969.
    – Martin R
    Sep 9 at 13:08










  • Oh, you're right! :)
    – LucaMac
    Sep 9 at 13:08

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote













By definition $barA = bigcaplimits_A subseteq C space and space C space closed C$ and the same holds for $barB$.



Now, if $C$ is closed and $B subseteq C$ then $A subseteq C$, so $barA subseteq C$. Considering the intersection over all $C$ closed such that $B subseteq C$ we have $barA subseteq barB$.






share|cite|improve this answer




















  • Simpler version of this approach: math.stackexchange.com/a/1620306/42969.
    – Martin R
    Sep 9 at 13:08










  • Oh, you're right! :)
    – LucaMac
    Sep 9 at 13:08














up vote
1
down vote













By definition $barA = bigcaplimits_A subseteq C space and space C space closed C$ and the same holds for $barB$.



Now, if $C$ is closed and $B subseteq C$ then $A subseteq C$, so $barA subseteq C$. Considering the intersection over all $C$ closed such that $B subseteq C$ we have $barA subseteq barB$.






share|cite|improve this answer




















  • Simpler version of this approach: math.stackexchange.com/a/1620306/42969.
    – Martin R
    Sep 9 at 13:08










  • Oh, you're right! :)
    – LucaMac
    Sep 9 at 13:08












up vote
1
down vote










up vote
1
down vote









By definition $barA = bigcaplimits_A subseteq C space and space C space closed C$ and the same holds for $barB$.



Now, if $C$ is closed and $B subseteq C$ then $A subseteq C$, so $barA subseteq C$. Considering the intersection over all $C$ closed such that $B subseteq C$ we have $barA subseteq barB$.






share|cite|improve this answer












By definition $barA = bigcaplimits_A subseteq C space and space C space closed C$ and the same holds for $barB$.



Now, if $C$ is closed and $B subseteq C$ then $A subseteq C$, so $barA subseteq C$. Considering the intersection over all $C$ closed such that $B subseteq C$ we have $barA subseteq barB$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 9 at 13:05









LucaMac

1,694116




1,694116











  • Simpler version of this approach: math.stackexchange.com/a/1620306/42969.
    – Martin R
    Sep 9 at 13:08










  • Oh, you're right! :)
    – LucaMac
    Sep 9 at 13:08
















  • Simpler version of this approach: math.stackexchange.com/a/1620306/42969.
    – Martin R
    Sep 9 at 13:08










  • Oh, you're right! :)
    – LucaMac
    Sep 9 at 13:08















Simpler version of this approach: math.stackexchange.com/a/1620306/42969.
– Martin R
Sep 9 at 13:08




Simpler version of this approach: math.stackexchange.com/a/1620306/42969.
– Martin R
Sep 9 at 13:08












Oh, you're right! :)
– LucaMac
Sep 9 at 13:08




Oh, you're right! :)
– LucaMac
Sep 9 at 13:08


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