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Boundary of $A times B$

0

$begingroup$

Let $X$ and $Y$ be topological spaces. Let $A subset X$ and $B subset Y$. Find the boundary of $A times B$ in term of boundary of $A$ and boundary of $B$.

general-topology

share|cite|improve this question

edited Mar 25 at 17:56

Andrés E. Caicedo

66.1k8160252

asked Mar 25 at 16:32

SUDIP SINGHASUDIP SINGHA

84

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math.SE! What do you think should be the correct approach to solving the problem? Where are you getting stuck? HINT If a point is on the boundary of $A$ and in the interior of $B$, is it on the boundary of $A times B$?
  $endgroup$
  – gt6989b
  Mar 25 at 16:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are your thoughts on this question? Have you made any progress? Are you stuck in any particular place? Please note that questions which simply demand a solution, without any evidence of your work, are often closed here on math.stackexchange. Here's some information about how to ask a good question, with particular emphasis on how to provide context.
  $endgroup$
  – Lee Mosher
  Mar 25 at 16:41
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. Yes it is on the boundary of A×B.
  $endgroup$
  – SUDIP SINGHA
  Mar 25 at 16:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Similarly a point (x,y)in X×Y lies interior of A(x lies in intA)and boundary of B(y lies boundary of B) lies on boundary of A×B. Also a point (x,y) where x lies on boundary of A and y lies on boundary of B .Then it also lies on boundary of A×B. Are there another points not belonging type of those but in boundary of A×B?
  $endgroup$
  – SUDIP SINGHA
  Mar 25 at 16:56
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $X$ and $Y$ be topological spaces. Let $A subset X$ and $B subset Y$. Find the boundary of $A times B$ in term of boundary of $A$ and boundary of $B$.

general-topology

share|cite|improve this question

edited Mar 25 at 17:56

Andrés E. Caicedo

66.1k8160252

asked Mar 25 at 16:32

SUDIP SINGHASUDIP SINGHA

84

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math.SE! What do you think should be the correct approach to solving the problem? Where are you getting stuck? HINT If a point is on the boundary of $A$ and in the interior of $B$, is it on the boundary of $A times B$?
  $endgroup$
  – gt6989b
  Mar 25 at 16:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are your thoughts on this question? Have you made any progress? Are you stuck in any particular place? Please note that questions which simply demand a solution, without any evidence of your work, are often closed here on math.stackexchange. Here's some information about how to ask a good question, with particular emphasis on how to provide context.
  $endgroup$
  – Lee Mosher
  Mar 25 at 16:41
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. Yes it is on the boundary of A×B.
  $endgroup$
  – SUDIP SINGHA
  Mar 25 at 16:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Similarly a point (x,y)in X×Y lies interior of A(x lies in intA)and boundary of B(y lies boundary of B) lies on boundary of A×B. Also a point (x,y) where x lies on boundary of A and y lies on boundary of B .Then it also lies on boundary of A×B. Are there another points not belonging type of those but in boundary of A×B?
  $endgroup$
  – SUDIP SINGHA
  Mar 25 at 16:56
  

  
  
  
  

add a comment | 

$begingroup$

Let $X$ and $Y$ be topological spaces. Let $A subset X$ and $B subset Y$. Find the boundary of $A times B$ in term of boundary of $A$ and boundary of $B$.

general-topology

share|cite|improve this question

edited Mar 25 at 17:56

Andrés E. Caicedo

66.1k8160252

asked Mar 25 at 16:32

SUDIP SINGHASUDIP SINGHA

84

$endgroup$

share|cite|improve this question

edited Mar 25 at 17:56

Andrés E. Caicedo

66.1k8160252

asked Mar 25 at 16:32

SUDIP SINGHASUDIP SINGHA

84

share|cite|improve this question

edited Mar 25 at 17:56

Andrés E. Caicedo

66.1k8160252

asked Mar 25 at 16:32

SUDIP SINGHASUDIP SINGHA

84

edited Mar 25 at 17:56

Andrés E. Caicedo

66.1k8160252

edited Mar 25 at 17:56

Andrés E. Caicedo

66.1k8160252

asked Mar 25 at 16:32

SUDIP SINGHASUDIP SINGHA

84

asked Mar 25 at 16:32

SUDIP SINGHASUDIP SINGHA

84

1 Answer
1

active

oldest

votes

1

$begingroup$

Denoting by $partial A$ the boundary of $A$, we have

$$partial (A times B) = (partial A times overline{B}) cup (overline{A} times partial B)tag{*}$$

and we can expand that if you like, using $overline{C} = C cup partial C$ (for all $C$) as

$$(partial A times B)cup (partial A times partial B) cup (A times partial B) cup (partial A times partial B)= (partial A times B)cup (partial A times partial B) cup (A times partial B)$$

eliminating the double term, if you want everything in terms of the original sets and their boundaries.

The first identity $(ast)$ is a straightforward application of the definitions of boundary and the product topology, showing two inclusions. Try to prove it. You can attempt a more algebraic proof if you already know

$$overline{A times B}= overline{A} times overline{B}$$

$$operatorname{int}(A times B) = operatorname{int}(A) times operatorname{int}(B)$$

and $$partial C = overline{C}setminus operatorname{int}(C)$$

applied to $C=A times B$.

share|cite|improve this answer

edited Mar 25 at 16:58

answered Mar 25 at 16:50

Henno BrandsmaHenno Brandsma

117k350128

$endgroup$

add a comment | 

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1

$begingroup$

Denoting by $partial A$ the boundary of $A$, we have

$$partial (A times B) = (partial A times overline{B}) cup (overline{A} times partial B)tag{*}$$

and we can expand that if you like, using $overline{C} = C cup partial C$ (for all $C$) as

$$(partial A times B)cup (partial A times partial B) cup (A times partial B) cup (partial A times partial B)= (partial A times B)cup (partial A times partial B) cup (A times partial B)$$

eliminating the double term, if you want everything in terms of the original sets and their boundaries.

The first identity $(ast)$ is a straightforward application of the definitions of boundary and the product topology, showing two inclusions. Try to prove it. You can attempt a more algebraic proof if you already know

$$overline{A times B}= overline{A} times overline{B}$$

$$operatorname{int}(A times B) = operatorname{int}(A) times operatorname{int}(B)$$

and $$partial C = overline{C}setminus operatorname{int}(C)$$

applied to $C=A times B$.

share|cite|improve this answer

edited Mar 25 at 16:58

answered Mar 25 at 16:50

Henno BrandsmaHenno Brandsma

117k350128

$endgroup$

add a comment | 

1

$begingroup$

Denoting by $partial A$ the boundary of $A$, we have

$$partial (A times B) = (partial A times overline{B}) cup (overline{A} times partial B)tag{*}$$

and we can expand that if you like, using $overline{C} = C cup partial C$ (for all $C$) as

$$(partial A times B)cup (partial A times partial B) cup (A times partial B) cup (partial A times partial B)= (partial A times B)cup (partial A times partial B) cup (A times partial B)$$

eliminating the double term, if you want everything in terms of the original sets and their boundaries.

The first identity $(ast)$ is a straightforward application of the definitions of boundary and the product topology, showing two inclusions. Try to prove it. You can attempt a more algebraic proof if you already know

$$overline{A times B}= overline{A} times overline{B}$$

$$operatorname{int}(A times B) = operatorname{int}(A) times operatorname{int}(B)$$

and $$partial C = overline{C}setminus operatorname{int}(C)$$

applied to $C=A times B$.

share|cite|improve this answer

edited Mar 25 at 16:58

answered Mar 25 at 16:50

Henno BrandsmaHenno Brandsma

117k350128

$endgroup$

add a comment | 

$begingroup$

Denoting by $partial A$ the boundary of $A$, we have

$$partial (A times B) = (partial A times overline{B}) cup (overline{A} times partial B)tag{*}$$

and we can expand that if you like, using $overline{C} = C cup partial C$ (for all $C$) as

$$(partial A times B)cup (partial A times partial B) cup (A times partial B) cup (partial A times partial B)= (partial A times B)cup (partial A times partial B) cup (A times partial B)$$

eliminating the double term, if you want everything in terms of the original sets and their boundaries.

The first identity $(ast)$ is a straightforward application of the definitions of boundary and the product topology, showing two inclusions. Try to prove it. You can attempt a more algebraic proof if you already know

$$overline{A times B}= overline{A} times overline{B}$$

$$operatorname{int}(A times B) = operatorname{int}(A) times operatorname{int}(B)$$

and $$partial C = overline{C}setminus operatorname{int}(C)$$

applied to $C=A times B$.

share|cite|improve this answer

edited Mar 25 at 16:58

answered Mar 25 at 16:50

Henno BrandsmaHenno Brandsma

117k350128

$endgroup$

share|cite|improve this answer

edited Mar 25 at 16:58

answered Mar 25 at 16:50

Henno BrandsmaHenno Brandsma

117k350128

answered Mar 25 at 16:50

Henno BrandsmaHenno Brandsma

117k350128

answered Mar 25 at 16:50

Henno BrandsmaHenno Brandsma

117k350128