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
marked as duplicate by Martin R, Saucy O'Path, amWhy
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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.
add a comment |Â
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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
marked as duplicate by Martin R, Saucy O'Path, amWhy
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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.
â 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.
â 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.
â Martin R
Sep 9 at 13:05
1
@yoshi Ah, ok. It should be exercise 6 page 101, at the end of §17.
â 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.
â Sheel Stueber
Sep 9 at 13:10
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up vote
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favorite
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
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
asked Sep 9 at 12:53
yoshi
912716
912716
marked as duplicate by Martin R, Saucy O'Path, amWhy
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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.
marked as duplicate by Martin R, Saucy O'Path, amWhy
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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.
â 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.
â 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.
â Martin R
Sep 9 at 13:05
1
@yoshi Ah, ok. It should be exercise 6 page 101, at the end of §17.
â 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.
â Sheel Stueber
Sep 9 at 13:10
add a comment |Â
I'd be curious to know which book does not mention this topic, pedantic as general topology books tend to be.
â 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.
â 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.
â Martin R
Sep 9 at 13:05
1
@yoshi Ah, ok. It should be exercise 6 page 101, at the end of §17.
â 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.
â 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.
â 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.
â 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.
â 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.
â 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.
â 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.
â Martin R
Sep 9 at 13:05
1
1
@yoshi Ah, ok. It should be exercise 6 page 101, at the end of §17.
â Saucy O'Path
Sep 9 at 13:07
@yoshi Ah, ok. It should be exercise 6 page 101, at the end of §17.
â 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.
â 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.
â Sheel Stueber
Sep 9 at 13:10
add a comment |Â
1 Answer
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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 $barA subseteq barB$.
Simpler version of this approach: math.stackexchange.com/a/1620306/42969.
â Martin R
Sep 9 at 13:08
Oh, you're right! :)
â LucaMac
Sep 9 at 13:08
add a comment |Â
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 $barA subseteq barB$.
Simpler version of this approach: math.stackexchange.com/a/1620306/42969.
â Martin R
Sep 9 at 13:08
Oh, you're right! :)
â LucaMac
Sep 9 at 13:08
add a comment |Â
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 $barA subseteq barB$.
Simpler version of this approach: math.stackexchange.com/a/1620306/42969.
â Martin R
Sep 9 at 13:08
Oh, you're right! :)
â LucaMac
Sep 9 at 13:08
add a comment |Â
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 $barA subseteq barB$.
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 $barA subseteq barB$.
answered Sep 9 at 13:05
LucaMac
1,694116
1,694116
Simpler version of this approach: math.stackexchange.com/a/1620306/42969.
â Martin R
Sep 9 at 13:08
Oh, you're right! :)
â LucaMac
Sep 9 at 13:08
add a comment |Â
Simpler version of this approach: math.stackexchange.com/a/1620306/42969.
â Martin R
Sep 9 at 13:08
Oh, you're right! :)
â LucaMac
Sep 9 at 13:08
Simpler version of this approach: math.stackexchange.com/a/1620306/42969.
â Martin R
Sep 9 at 13:08
Simpler version of this approach: math.stackexchange.com/a/1620306/42969.
â Martin R
Sep 9 at 13:08
Oh, you're right! :)
â LucaMac
Sep 9 at 13:08
Oh, you're right! :)
â LucaMac
Sep 9 at 13:08
add a comment |Â
I'd be curious to know which book does not mention this topic, pedantic as general topology books tend to be.
â 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.
â 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.
â Martin R
Sep 9 at 13:05
1
@yoshi Ah, ok. It should be exercise 6 page 101, at the end of §17.
â 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.
â Sheel Stueber
Sep 9 at 13:10