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Prove that $∂A$ is closed given $∂A = textCl(A) − textInt(A)$

“Prove that ∂A ∪ Int(A) = Cl(A)”Prove that $partial A$ is a cutset of connected $X$ if $operatornameInt(A)$ and $operatornameInt(X - A)$ are nonemptyQuestions on the counterexample for “connectedness doesn't imply path-connectedness”Understanding a proof of Klein Bottle being embeddable in $mathbb R^4$“Prove that a topology Ƭ on X is the discrete topology if and only if x ∈ Ƭ for all x ∈ X”“Let Ƭ consist of ∅, ℝ, and all intervals (−∞, p) for p ∈ ℝ. Prove that Ƭ is a topology on ℝ.”Determine which of the following are open sets in $mathbbRl$. In each case, prove your assertion“Show that a single-point set n is closed in the digital line topology if and only if n is even.”“Prove that Cl(ℚ) = ℝ in the standard topology on ℝ.”Prove that $partial A = mathrmCl(A) cap mathrmCl(X − A)$“Prove that ∂A ∪ Int(A) = Cl(A)”

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

3

$begingroup$

Similar questions have been asked, but none with the given information. My textbook doesn't give me the fact that $∂A = textCl(A) − textInt(A)$. If that were the case, I could just state that definition and note that it's the intersection of closed sets.

I have very little knowledge of set theory and proofs, so I'm not sure how else to prove this. As always, I appreciate any help.

The question comes from “Introduction to Topology: Pure and Applied” by Colin Adams and Robert Franzosa.

"Let $A$ be a subset of a topological space $X$. Prove that $∂A$ is closed given $∂A = textCl(A) − textInt(A)$"

general-topology

share|cite|improve this question

edited Jun 17 at 9:42

YuiTo Cheng

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asked Jun 16 at 18:07

Math_Student_1Math_Student_1

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$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Use the fact that set subtraction is the same as intersecting the complement.
  $endgroup$
  – Tony S.F.
  Jun 16 at 18:12
  

  
  
  
  

add a comment | 

3

$begingroup$

Similar questions have been asked, but none with the given information. My textbook doesn't give me the fact that $∂A = textCl(A) − textInt(A)$. If that were the case, I could just state that definition and note that it's the intersection of closed sets.

I have very little knowledge of set theory and proofs, so I'm not sure how else to prove this. As always, I appreciate any help.

The question comes from “Introduction to Topology: Pure and Applied” by Colin Adams and Robert Franzosa.

"Let $A$ be a subset of a topological space $X$. Prove that $∂A$ is closed given $∂A = textCl(A) − textInt(A)$"

general-topology

share|cite|improve this question

edited Jun 17 at 9:42

YuiTo Cheng

3,38377 gold badges1515 silver badges4646 bronze badges

asked Jun 16 at 18:07

Math_Student_1Math_Student_1

4766 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Use the fact that set subtraction is the same as intersecting the complement.
  $endgroup$
  – Tony S.F.
  Jun 16 at 18:12
  

  
  
  
  

add a comment | 

$begingroup$

Similar questions have been asked, but none with the given information. My textbook doesn't give me the fact that $∂A = textCl(A) − textInt(A)$. If that were the case, I could just state that definition and note that it's the intersection of closed sets.

I have very little knowledge of set theory and proofs, so I'm not sure how else to prove this. As always, I appreciate any help.

The question comes from “Introduction to Topology: Pure and Applied” by Colin Adams and Robert Franzosa.

"Let $A$ be a subset of a topological space $X$. Prove that $∂A$ is closed given $∂A = textCl(A) − textInt(A)$"

general-topology

share|cite|improve this question

edited Jun 17 at 9:42

YuiTo Cheng

3,38377 gold badges1515 silver badges4646 bronze badges

asked Jun 16 at 18:07

Math_Student_1Math_Student_1

4766 bronze badges

$endgroup$

share|cite|improve this question

edited Jun 17 at 9:42

YuiTo Cheng

3,38377 gold badges1515 silver badges4646 bronze badges

asked Jun 16 at 18:07

Math_Student_1Math_Student_1

4766 bronze badges

share|cite|improve this question

edited Jun 17 at 9:42

YuiTo Cheng

3,38377 gold badges1515 silver badges4646 bronze badges

asked Jun 16 at 18:07

Math_Student_1Math_Student_1

4766 bronze badges

edited Jun 17 at 9:42

YuiTo Cheng

3,38377 gold badges1515 silver badges4646 bronze badges

edited Jun 17 at 9:42

YuiTo Cheng

3,38377 gold badges1515 silver badges4646 bronze badges

asked Jun 16 at 18:07

Math_Student_1Math_Student_1

4766 bronze badges

asked Jun 16 at 18:07

Math_Student_1Math_Student_1

4766 bronze badges

2 Answers
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9

$begingroup$

$partial A= mathrmCl(A)-mathrmInt(A)=mathrmCl(A) cap (X - mathrmInt(A))$. Now $mathrmCl(A)$ is closed by definition, $X-mathrmInt(A)$ is closed since it's the complement of an open, and intersection of closed is closed, so $partial A$ is closed.

share|cite|improve this answer

edited Jun 16 at 18:21

J. W. Tanner

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answered Jun 16 at 18:12

user289143user289143

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$endgroup$

add a comment | 

0

$begingroup$

Since, $partial A=overline Asetminusmathring A$, $(partial A)^complement=overline A^complementcupmathring A$, which is open, since it's the union of two open sets.

share|cite|improve this answer

edited Jun 16 at 18:59

answered Jun 16 at 18:13

José Carlos SantosJosé Carlos Santos

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$endgroup$

add a comment | 

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9

$begingroup$

$partial A= mathrmCl(A)-mathrmInt(A)=mathrmCl(A) cap (X - mathrmInt(A))$. Now $mathrmCl(A)$ is closed by definition, $X-mathrmInt(A)$ is closed since it's the complement of an open, and intersection of closed is closed, so $partial A$ is closed.

share|cite|improve this answer

edited Jun 16 at 18:21

J. W. Tanner

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answered Jun 16 at 18:12

user289143user289143

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$endgroup$

add a comment | 

9

$begingroup$

$partial A= mathrmCl(A)-mathrmInt(A)=mathrmCl(A) cap (X - mathrmInt(A))$. Now $mathrmCl(A)$ is closed by definition, $X-mathrmInt(A)$ is closed since it's the complement of an open, and intersection of closed is closed, so $partial A$ is closed.

share|cite|improve this answer

edited Jun 16 at 18:21

J. W. Tanner

8,76011 gold badge77 silver badges2323 bronze badges

answered Jun 16 at 18:12

user289143user289143

1,75733 silver badges1414 bronze badges

$endgroup$

add a comment | 

$begingroup$

$partial A= mathrmCl(A)-mathrmInt(A)=mathrmCl(A) cap (X - mathrmInt(A))$. Now $mathrmCl(A)$ is closed by definition, $X-mathrmInt(A)$ is closed since it's the complement of an open, and intersection of closed is closed, so $partial A$ is closed.

share|cite|improve this answer

edited Jun 16 at 18:21

J. W. Tanner

8,76011 gold badge77 silver badges2323 bronze badges

answered Jun 16 at 18:12

user289143user289143

1,75733 silver badges1414 bronze badges

$endgroup$

share|cite|improve this answer

edited Jun 16 at 18:21

J. W. Tanner

8,76011 gold badge77 silver badges2323 bronze badges

answered Jun 16 at 18:12

user289143user289143

1,75733 silver badges1414 bronze badges

edited Jun 16 at 18:21

J. W. Tanner

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edited Jun 16 at 18:21

J. W. Tanner

8,76011 gold badge77 silver badges2323 bronze badges

answered Jun 16 at 18:12

user289143user289143

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answered Jun 16 at 18:12

user289143user289143

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0

$begingroup$

Since, $partial A=overline Asetminusmathring A$, $(partial A)^complement=overline A^complementcupmathring A$, which is open, since it's the union of two open sets.

share|cite|improve this answer

edited Jun 16 at 18:59

answered Jun 16 at 18:13

José Carlos SantosJosé Carlos Santos

194k2424 gold badges152152 silver badges269269 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

Since, $partial A=overline Asetminusmathring A$, $(partial A)^complement=overline A^complementcupmathring A$, which is open, since it's the union of two open sets.

share|cite|improve this answer

edited Jun 16 at 18:59

answered Jun 16 at 18:13

José Carlos SantosJosé Carlos Santos

194k2424 gold badges152152 silver badges269269 bronze badges

$endgroup$

add a comment | 

$begingroup$

Since, $partial A=overline Asetminusmathring A$, $(partial A)^complement=overline A^complementcupmathring A$, which is open, since it's the union of two open sets.

share|cite|improve this answer

edited Jun 16 at 18:59

answered Jun 16 at 18:13

José Carlos SantosJosé Carlos Santos

194k2424 gold badges152152 silver badges269269 bronze badges

$endgroup$

share|cite|improve this answer

edited Jun 16 at 18:59

answered Jun 16 at 18:13

José Carlos SantosJosé Carlos Santos

194k2424 gold badges152152 silver badges269269 bronze badges

answered Jun 16 at 18:13

José Carlos SantosJosé Carlos Santos

194k2424 gold badges152152 silver badges269269 bronze badges

answered Jun 16 at 18:13

José Carlos SantosJosé Carlos Santos

194k2424 gold badges152152 silver badges269269 bronze badges