Prove that $A$ is open when $A cap bd(A) = emptyset$

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Before I saw the solution to the problem below, I was attempting to prove it using De Morgan's laws.



$bd(A)$ is the boundary of $A$, and $overline A$ is the complement of $A$



My attempt



$overlineAcap bd(A)=X$



and by De Morgan's laws



$overlineA cup overline bd(A)=X$



and by the definition of $bd(A)$ and another application of De Morgan



$overlineA cup overline cl(A) cap overlineA=X$



$overlineA cup overline cl(A) cup A=X$



where $cl(A)$ is the closure of $A$



I was unsure how to proceed from this point. I was considering isolating $A$ by writing $A=X$ $overline A cup overlinecl(A)$, but I wasn't sure if I could do this, and even if I could, I still don't see how I'd get that A is open.



Thanks, and here's the problem along with another solution for reference purposes



enter image description here



enter image description here




real-analysis general-topology analysis proof-writing




share|cite|improve this question





















  • What is $bd(A)$? Do write $overlineA$ for the complement of $A$ in $X$? Hence $overlineA=Xsetminus A$?
    – Cornman
    Jul 24 at 0:19










  • $bd(A)$ is the boundary of $A$ and yes, $overline A$ is the complement as well as $X setminus A$. I've edited my question
    – john fowles
    Jul 24 at 0:24











  • @Chickenmancer but the conclusion is that $A$ is open. Where did I assume this?
    – john fowles
    Jul 24 at 0:33











  • The very first line of your proof is that $overlineAcap bd(A)=X.$
    – Chickenmancer
    Jul 24 at 0:34










  • That follows directly from the condition given, as you pointed out in your previous comment
    – john fowles
    Jul 24 at 0:36














up vote
1
down vote

favorite












Before I saw the solution to the problem below, I was attempting to prove it using De Morgan's laws.



$bd(A)$ is the boundary of $A$, and $overline A$ is the complement of $A$



My attempt



$overlineAcap bd(A)=X$



and by De Morgan's laws



$overlineA cup overline bd(A)=X$



and by the definition of $bd(A)$ and another application of De Morgan



$overlineA cup overline cl(A) cap overlineA=X$



$overlineA cup overline cl(A) cup A=X$



where $cl(A)$ is the closure of $A$



I was unsure how to proceed from this point. I was considering isolating $A$ by writing $A=X$ $overline A cup overlinecl(A)$, but I wasn't sure if I could do this, and even if I could, I still don't see how I'd get that A is open.



Thanks, and here's the problem along with another solution for reference purposes



enter image description here



enter image description here




real-analysis general-topology analysis proof-writing




share|cite|improve this question





















  • What is $bd(A)$? Do write $overlineA$ for the complement of $A$ in $X$? Hence $overlineA=Xsetminus A$?
    – Cornman
    Jul 24 at 0:19










  • $bd(A)$ is the boundary of $A$ and yes, $overline A$ is the complement as well as $X setminus A$. I've edited my question
    – john fowles
    Jul 24 at 0:24











  • @Chickenmancer but the conclusion is that $A$ is open. Where did I assume this?
    – john fowles
    Jul 24 at 0:33











  • The very first line of your proof is that $overlineAcap bd(A)=X.$
    – Chickenmancer
    Jul 24 at 0:34










  • That follows directly from the condition given, as you pointed out in your previous comment
    – john fowles
    Jul 24 at 0:36












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Before I saw the solution to the problem below, I was attempting to prove it using De Morgan's laws.



$bd(A)$ is the boundary of $A$, and $overline A$ is the complement of $A$



My attempt



$overlineAcap bd(A)=X$



and by De Morgan's laws



$overlineA cup overline bd(A)=X$



and by the definition of $bd(A)$ and another application of De Morgan



$overlineA cup overline cl(A) cap overlineA=X$



$overlineA cup overline cl(A) cup A=X$



where $cl(A)$ is the closure of $A$



I was unsure how to proceed from this point. I was considering isolating $A$ by writing $A=X$ $overline A cup overlinecl(A)$, but I wasn't sure if I could do this, and even if I could, I still don't see how I'd get that A is open.



Thanks, and here's the problem along with another solution for reference purposes



enter image description here



enter image description here




real-analysis general-topology analysis proof-writing




share|cite|improve this question













Before I saw the solution to the problem below, I was attempting to prove it using De Morgan's laws.



$bd(A)$ is the boundary of $A$, and $overline A$ is the complement of $A$



My attempt



$overlineAcap bd(A)=X$



and by De Morgan's laws



$overlineA cup overline bd(A)=X$



and by the definition of $bd(A)$ and another application of De Morgan



$overlineA cup overline cl(A) cap overlineA=X$



$overlineA cup overline cl(A) cup A=X$



where $cl(A)$ is the closure of $A$



I was unsure how to proceed from this point. I was considering isolating $A$ by writing $A=X$ $overline A cup overlinecl(A)$, but I wasn't sure if I could do this, and even if I could, I still don't see how I'd get that A is open.



Thanks, and here's the problem along with another solution for reference purposes



enter image description here



enter image description here





real-analysis general-topology analysis proof-writing





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 24 at 4:13









William Elliot

5,0942416




5,0942416









asked Jul 24 at 0:04









john fowles

1,088817




1,088817











  • What is $bd(A)$? Do write $overlineA$ for the complement of $A$ in $X$? Hence $overlineA=Xsetminus A$?
    – Cornman
    Jul 24 at 0:19










  • $bd(A)$ is the boundary of $A$ and yes, $overline A$ is the complement as well as $X setminus A$. I've edited my question
    – john fowles
    Jul 24 at 0:24











  • @Chickenmancer but the conclusion is that $A$ is open. Where did I assume this?
    – john fowles
    Jul 24 at 0:33











  • The very first line of your proof is that $overlineAcap bd(A)=X.$
    – Chickenmancer
    Jul 24 at 0:34










  • That follows directly from the condition given, as you pointed out in your previous comment
    – john fowles
    Jul 24 at 0:36
















  • What is $bd(A)$? Do write $overlineA$ for the complement of $A$ in $X$? Hence $overlineA=Xsetminus A$?
    – Cornman
    Jul 24 at 0:19










  • $bd(A)$ is the boundary of $A$ and yes, $overline A$ is the complement as well as $X setminus A$. I've edited my question
    – john fowles
    Jul 24 at 0:24











  • @Chickenmancer but the conclusion is that $A$ is open. Where did I assume this?
    – john fowles
    Jul 24 at 0:33











  • The very first line of your proof is that $overlineAcap bd(A)=X.$
    – Chickenmancer
    Jul 24 at 0:34










  • That follows directly from the condition given, as you pointed out in your previous comment
    – john fowles
    Jul 24 at 0:36















What is $bd(A)$? Do write $overlineA$ for the complement of $A$ in $X$? Hence $overlineA=Xsetminus A$?
– Cornman
Jul 24 at 0:19




What is $bd(A)$? Do write $overlineA$ for the complement of $A$ in $X$? Hence $overlineA=Xsetminus A$?
– Cornman
Jul 24 at 0:19












$bd(A)$ is the boundary of $A$ and yes, $overline A$ is the complement as well as $X setminus A$. I've edited my question
– john fowles
Jul 24 at 0:24





$bd(A)$ is the boundary of $A$ and yes, $overline A$ is the complement as well as $X setminus A$. I've edited my question
– john fowles
Jul 24 at 0:24













@Chickenmancer but the conclusion is that $A$ is open. Where did I assume this?
– john fowles
Jul 24 at 0:33





@Chickenmancer but the conclusion is that $A$ is open. Where did I assume this?
– john fowles
Jul 24 at 0:33













The very first line of your proof is that $overlineAcap bd(A)=X.$
– Chickenmancer
Jul 24 at 0:34




The very first line of your proof is that $overlineAcap bd(A)=X.$
– Chickenmancer
Jul 24 at 0:34












That follows directly from the condition given, as you pointed out in your previous comment
– john fowles
Jul 24 at 0:36




That follows directly from the condition given, as you pointed out in your previous comment
– john fowles
Jul 24 at 0:36










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










I don't think you should use deMorgan's laws in this case. (At least, i don't see how to.)



It's easy just to note that any $ain A$ has to have a nbhd that doesn't meet $Xsetminus A$ by definition of boundary (as in the solution you included).



But, here's an alternate approach:
$partial A=partial A^c$. Hence $Acappartial A=emptysetimplies Acappartial A^c=emptysetimplies partial A^csubset A^cimplies A^c$ is closed. (This uses the sort of complementary fact that a set is closed iff it contains its boundary...)






share|cite|improve this answer























  • Both approaches make sense, thanks! I was also wondering if I wanted to isolate $A$, for example in the expression $overlineA cup overline cl(A) cup A=X$ then would I write $A=X setminus overline A cup overline cl(A)$? I'm not trying to make that proof work, just wanting to know how to perform operations on unions and intersections
    – john fowles
    Jul 26 at 0:54











  • Well, of course $Asubset text cl Aimplies overlinetext cl Asubset bar A$. Try using this to simplify matters.
    – Chris Custer
    Jul 26 at 1:16











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oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










I don't think you should use deMorgan's laws in this case. (At least, i don't see how to.)



It's easy just to note that any $ain A$ has to have a nbhd that doesn't meet $Xsetminus A$ by definition of boundary (as in the solution you included).



But, here's an alternate approach:
$partial A=partial A^c$. Hence $Acappartial A=emptysetimplies Acappartial A^c=emptysetimplies partial A^csubset A^cimplies A^c$ is closed. (This uses the sort of complementary fact that a set is closed iff it contains its boundary...)






share|cite|improve this answer























  • Both approaches make sense, thanks! I was also wondering if I wanted to isolate $A$, for example in the expression $overlineA cup overline cl(A) cup A=X$ then would I write $A=X setminus overline A cup overline cl(A)$? I'm not trying to make that proof work, just wanting to know how to perform operations on unions and intersections
    – john fowles
    Jul 26 at 0:54











  • Well, of course $Asubset text cl Aimplies overlinetext cl Asubset bar A$. Try using this to simplify matters.
    – Chris Custer
    Jul 26 at 1:16















up vote
1
down vote



accepted










I don't think you should use deMorgan's laws in this case. (At least, i don't see how to.)



It's easy just to note that any $ain A$ has to have a nbhd that doesn't meet $Xsetminus A$ by definition of boundary (as in the solution you included).



But, here's an alternate approach:
$partial A=partial A^c$. Hence $Acappartial A=emptysetimplies Acappartial A^c=emptysetimplies partial A^csubset A^cimplies A^c$ is closed. (This uses the sort of complementary fact that a set is closed iff it contains its boundary...)






share|cite|improve this answer























  • Both approaches make sense, thanks! I was also wondering if I wanted to isolate $A$, for example in the expression $overlineA cup overline cl(A) cup A=X$ then would I write $A=X setminus overline A cup overline cl(A)$? I'm not trying to make that proof work, just wanting to know how to perform operations on unions and intersections
    – john fowles
    Jul 26 at 0:54











  • Well, of course $Asubset text cl Aimplies overlinetext cl Asubset bar A$. Try using this to simplify matters.
    – Chris Custer
    Jul 26 at 1:16













up vote
1
down vote



accepted







up vote
1
down vote



accepted






I don't think you should use deMorgan's laws in this case. (At least, i don't see how to.)



It's easy just to note that any $ain A$ has to have a nbhd that doesn't meet $Xsetminus A$ by definition of boundary (as in the solution you included).



But, here's an alternate approach:
$partial A=partial A^c$. Hence $Acappartial A=emptysetimplies Acappartial A^c=emptysetimplies partial A^csubset A^cimplies A^c$ is closed. (This uses the sort of complementary fact that a set is closed iff it contains its boundary...)






share|cite|improve this answer















I don't think you should use deMorgan's laws in this case. (At least, i don't see how to.)



It's easy just to note that any $ain A$ has to have a nbhd that doesn't meet $Xsetminus A$ by definition of boundary (as in the solution you included).



But, here's an alternate approach:
$partial A=partial A^c$. Hence $Acappartial A=emptysetimplies Acappartial A^c=emptysetimplies partial A^csubset A^cimplies A^c$ is closed. (This uses the sort of complementary fact that a set is closed iff it contains its boundary...)







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 25 at 23:56


























answered Jul 24 at 0:46









Chris Custer

5,4082622




5,4082622











  • Both approaches make sense, thanks! I was also wondering if I wanted to isolate $A$, for example in the expression $overlineA cup overline cl(A) cup A=X$ then would I write $A=X setminus overline A cup overline cl(A)$? I'm not trying to make that proof work, just wanting to know how to perform operations on unions and intersections
    – john fowles
    Jul 26 at 0:54











  • Well, of course $Asubset text cl Aimplies overlinetext cl Asubset bar A$. Try using this to simplify matters.
    – Chris Custer
    Jul 26 at 1:16

















  • Both approaches make sense, thanks! I was also wondering if I wanted to isolate $A$, for example in the expression $overlineA cup overline cl(A) cup A=X$ then would I write $A=X setminus overline A cup overline cl(A)$? I'm not trying to make that proof work, just wanting to know how to perform operations on unions and intersections
    – john fowles
    Jul 26 at 0:54











  • Well, of course $Asubset text cl Aimplies overlinetext cl Asubset bar A$. Try using this to simplify matters.
    – Chris Custer
    Jul 26 at 1:16
















Both approaches make sense, thanks! I was also wondering if I wanted to isolate $A$, for example in the expression $overlineA cup overline cl(A) cup A=X$ then would I write $A=X setminus overline A cup overline cl(A)$? I'm not trying to make that proof work, just wanting to know how to perform operations on unions and intersections
– john fowles
Jul 26 at 0:54





Both approaches make sense, thanks! I was also wondering if I wanted to isolate $A$, for example in the expression $overlineA cup overline cl(A) cup A=X$ then would I write $A=X setminus overline A cup overline cl(A)$? I'm not trying to make that proof work, just wanting to know how to perform operations on unions and intersections
– john fowles
Jul 26 at 0:54













Well, of course $Asubset text cl Aimplies overlinetext cl Asubset bar A$. Try using this to simplify matters.
– Chris Custer
Jul 26 at 1:16





Well, of course $Asubset text cl Aimplies overlinetext cl Asubset bar A$. Try using this to simplify matters.
– Chris Custer
Jul 26 at 1:16













 

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