Determine interior and boundary of $Atimes B$
$begingroup$
Let $(X,mathcal{O}_X)$ and $(Y,mathcal{O}_Y)$ be topological spaces and $A,B$ subset of $X,Y$ respectively. I have to find the interior $(Atimes B)^circ$ and the boundary $partial(Atimes B)$. Furthermore, it is that the open sets are given by the product topology on $Atimes B$.
This sounds pretty easy, nevertheless I am not quite sure what exactly I should show. The interior and boundary both depend on the topology itself. Is this context to general to actually "show" something?
general-topology
asked Dec 14 '17 at 17:39
EpsilonDeltaEpsilonDelta
6921615
$endgroup$
add a comment |
$begingroup$
Let $(X,mathcal{O}_X)$ and $(Y,mathcal{O}_Y)$ be topological spaces and $A,B$ subset of $X,Y$ respectively. I have to find the interior $(Atimes B)^circ$ and the boundary $partial(Atimes B)$. Furthermore, it is that the open sets are given by the product topology on $Atimes B$.
This sounds pretty easy, nevertheless I am not quite sure what exactly I should show. The interior and boundary both depend on the topology itself. Is this context to general to actually "show" something?
general-topology
asked Dec 14 '17 at 17:39
EpsilonDeltaEpsilonDelta
6921615
$endgroup$
1
$begingroup$
I suspect it wants you to find the interior and boundary in terms of the interiors and boundaries of $A$ and $B$ in their respective topologies.
$endgroup$
– gj255
Dec 14 '17 at 17:47
2
$begingroup$
For instance, wouldn't it be nice if $(Atimes B)^circ=A^circtimes B^circ$ and $partial(Atimes B)=partial Atimespartial B$? Sadly, only one of these is true (which one?) The other requires some modification. But this is the kind of thing they're after.
$endgroup$
– Arthur
Dec 14 '17 at 17:54
$begingroup$
I could show the first relation. Not sure about the second one though...
$endgroup$
– EpsilonDelta
Dec 14 '17 at 18:24
1
$begingroup$
Draw a picture in the plane using open intervals for $A$ and $B$ to get an feel for the boundary case. Or search the site.
$endgroup$
– Henno Brandsma
Dec 14 '17 at 23:01
add a comment |
$begingroup$
Let $(X,mathcal{O}_X)$ and $(Y,mathcal{O}_Y)$ be topological spaces and $A,B$ subset of $X,Y$ respectively. I have to find the interior $(Atimes B)^circ$ and the boundary $partial(Atimes B)$. Furthermore, it is that the open sets are given by the product topology on $Atimes B$.
This sounds pretty easy, nevertheless I am not quite sure what exactly I should show. The interior and boundary both depend on the topology itself. Is this context to general to actually "show" something?
general-topology
asked Dec 14 '17 at 17:39
EpsilonDeltaEpsilonDelta
6921615
$endgroup$
Let $(X,mathcal{O}_X)$ and $(Y,mathcal{O}_Y)$ be topological spaces and $A,B$ subset of $X,Y$ respectively. I have to find the interior $(Atimes B)^circ$ and the boundary $partial(Atimes B)$. Furthermore, it is that the open sets are given by the product topology on $Atimes B$.
This sounds pretty easy, nevertheless I am not quite sure what exactly I should show. The interior and boundary both depend on the topology itself. Is this context to general to actually "show" something?
general-topology
general-topology
asked Dec 14 '17 at 17:39
EpsilonDeltaEpsilonDelta
6921615
asked Dec 14 '17 at 17:39
EpsilonDeltaEpsilonDelta
6921615
asked Dec 14 '17 at 17:39
EpsilonDeltaEpsilonDelta
6921615
asked Dec 14 '17 at 17:39
EpsilonDeltaEpsilonDelta
6921615
asked Dec 14 '17 at 17:39
EpsilonDeltaEpsilonDelta
6921615
6921615
1
$begingroup$
I suspect it wants you to find the interior and boundary in terms of the interiors and boundaries of $A$ and $B$ in their respective topologies.
$endgroup$
– gj255
Dec 14 '17 at 17:47
2
$begingroup$
For instance, wouldn't it be nice if $(Atimes B)^circ=A^circtimes B^circ$ and $partial(Atimes B)=partial Atimespartial B$? Sadly, only one of these is true (which one?) The other requires some modification. But this is the kind of thing they're after.
$endgroup$
– Arthur
Dec 14 '17 at 17:54
$begingroup$
I could show the first relation. Not sure about the second one though...
$endgroup$
– EpsilonDelta
Dec 14 '17 at 18:24
1
$begingroup$
Draw a picture in the plane using open intervals for $A$ and $B$ to get an feel for the boundary case. Or search the site.
$endgroup$
– Henno Brandsma
Dec 14 '17 at 23:01
add a comment |
1
$begingroup$
I suspect it wants you to find the interior and boundary in terms of the interiors and boundaries of $A$ and $B$ in their respective topologies.
$endgroup$
– gj255
Dec 14 '17 at 17:47
2
$begingroup$
For instance, wouldn't it be nice if $(Atimes B)^circ=A^circtimes B^circ$ and $partial(Atimes B)=partial Atimespartial B$? Sadly, only one of these is true (which one?) The other requires some modification. But this is the kind of thing they're after.
$endgroup$
– Arthur
Dec 14 '17 at 17:54
$begingroup$
I could show the first relation. Not sure about the second one though...
$endgroup$
– EpsilonDelta
Dec 14 '17 at 18:24
1
$begingroup$
Draw a picture in the plane using open intervals for $A$ and $B$ to get an feel for the boundary case. Or search the site.
$endgroup$
– Henno Brandsma
Dec 14 '17 at 23:01
1
1
$begingroup$
I suspect it wants you to find the interior and boundary in terms of the interiors and boundaries of $A$ and $B$ in their respective topologies.
$endgroup$
– gj255
Dec 14 '17 at 17:47
$begingroup$
I suspect it wants you to find the interior and boundary in terms of the interiors and boundaries of $A$ and $B$ in their respective topologies.
$endgroup$
– gj255
Dec 14 '17 at 17:47
2
2
$begingroup$
For instance, wouldn't it be nice if $(Atimes B)^circ=A^circtimes B^circ$ and $partial(Atimes B)=partial Atimespartial B$? Sadly, only one of these is true (which one?) The other requires some modification. But this is the kind of thing they're after.
$endgroup$
– Arthur
Dec 14 '17 at 17:54
$begingroup$
For instance, wouldn't it be nice if $(Atimes B)^circ=A^circtimes B^circ$ and $partial(Atimes B)=partial Atimespartial B$? Sadly, only one of these is true (which one?) The other requires some modification. But this is the kind of thing they're after.
$endgroup$
– Arthur
Dec 14 '17 at 17:54
$begingroup$
I could show the first relation. Not sure about the second one though...
$endgroup$
– EpsilonDelta
Dec 14 '17 at 18:24
$begingroup$
I could show the first relation. Not sure about the second one though...
$endgroup$
– EpsilonDelta
Dec 14 '17 at 18:24
1
1
$begingroup$
Draw a picture in the plane using open intervals for $A$ and $B$ to get an feel for the boundary case. Or search the site.
$endgroup$
– Henno Brandsma
Dec 14 '17 at 23:01
$begingroup$
Draw a picture in the plane using open intervals for $A$ and $B$ to get an feel for the boundary case. Or search the site.
$endgroup$
– Henno Brandsma
Dec 14 '17 at 23:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
(To remove the question from unanswered).
According to Exercise 2.3.B from [Eng], $(Atimes B)^circ=A^circtimes B^circ$ and $partial(Atimes B)=partial Atimes overline{B}cup overline{A}times partial B$. Prove this.
By Proposition 2.3.1 from [Eng], the set $A^circtimes B^circ$ is open, so $A^circtimes B^circsubset (Atimes B)^circ$. On the other hand, let $(x,y)in (Atimes B)^circ$ be any point. Then there exists an element $Utimes V$ of the canonical base at $Xtimes Y$ such that $(x,y)in Utimes Vsubset Atimes B$. Then $Usubset A$ and $Vsubset B$. Since $U$ and $V$ are open in $X$ and $Y$, respectively, we have $Usubset A^circ$ and $Vsubset B^circ$. Then $(x,y)in A^circtimes B^circ$.
By Proposition 2.3.3 from [Eng], $overline{Atimes B}=overline Atimes overline B$, so
$$partial(Atimes B)=$$ $$overline{Atimes B}setminus (Atimes B)^circ=$$
$$overline Atimes overline B setminus A^circtimes B^circ =$$ $$
((partial Acup A^circ)times (partial Bcup B^circ)) setminus A^circtimes B^circ=$$ $$partial Atimes partial Bcup partial Atimes B^circ cup A^circ times partial B=$$ $$(partial Atimes partial Bcup partial Atimes B^circ)cup
(partial Atimes partial Bcup A^circ times partial B)=$$ $$ partial Atimes overline{B}cup overline{A}times partial B.$$
References
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
answered Jan 4 at 2:38
Alex RavskyAlex Ravsky
42.7k32383
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add a comment |
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$begingroup$
(To remove the question from unanswered).
According to Exercise 2.3.B from [Eng], $(Atimes B)^circ=A^circtimes B^circ$ and $partial(Atimes B)=partial Atimes overline{B}cup overline{A}times partial B$. Prove this.
By Proposition 2.3.1 from [Eng], the set $A^circtimes B^circ$ is open, so $A^circtimes B^circsubset (Atimes B)^circ$. On the other hand, let $(x,y)in (Atimes B)^circ$ be any point. Then there exists an element $Utimes V$ of the canonical base at $Xtimes Y$ such that $(x,y)in Utimes Vsubset Atimes B$. Then $Usubset A$ and $Vsubset B$. Since $U$ and $V$ are open in $X$ and $Y$, respectively, we have $Usubset A^circ$ and $Vsubset B^circ$. Then $(x,y)in A^circtimes B^circ$.
By Proposition 2.3.3 from [Eng], $overline{Atimes B}=overline Atimes overline B$, so
$$partial(Atimes B)=$$ $$overline{Atimes B}setminus (Atimes B)^circ=$$
$$overline Atimes overline B setminus A^circtimes B^circ =$$ $$
((partial Acup A^circ)times (partial Bcup B^circ)) setminus A^circtimes B^circ=$$ $$partial Atimes partial Bcup partial Atimes B^circ cup A^circ times partial B=$$ $$(partial Atimes partial Bcup partial Atimes B^circ)cup
(partial Atimes partial Bcup A^circ times partial B)=$$ $$ partial Atimes overline{B}cup overline{A}times partial B.$$
References
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
answered Jan 4 at 2:38
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
add a comment |
$begingroup$
(To remove the question from unanswered).
According to Exercise 2.3.B from [Eng], $(Atimes B)^circ=A^circtimes B^circ$ and $partial(Atimes B)=partial Atimes overline{B}cup overline{A}times partial B$. Prove this.
By Proposition 2.3.1 from [Eng], the set $A^circtimes B^circ$ is open, so $A^circtimes B^circsubset (Atimes B)^circ$. On the other hand, let $(x,y)in (Atimes B)^circ$ be any point. Then there exists an element $Utimes V$ of the canonical base at $Xtimes Y$ such that $(x,y)in Utimes Vsubset Atimes B$. Then $Usubset A$ and $Vsubset B$. Since $U$ and $V$ are open in $X$ and $Y$, respectively, we have $Usubset A^circ$ and $Vsubset B^circ$. Then $(x,y)in A^circtimes B^circ$.
By Proposition 2.3.3 from [Eng], $overline{Atimes B}=overline Atimes overline B$, so
$$partial(Atimes B)=$$ $$overline{Atimes B}setminus (Atimes B)^circ=$$
$$overline Atimes overline B setminus A^circtimes B^circ =$$ $$
((partial Acup A^circ)times (partial Bcup B^circ)) setminus A^circtimes B^circ=$$ $$partial Atimes partial Bcup partial Atimes B^circ cup A^circ times partial B=$$ $$(partial Atimes partial Bcup partial Atimes B^circ)cup
(partial Atimes partial Bcup A^circ times partial B)=$$ $$ partial Atimes overline{B}cup overline{A}times partial B.$$
References
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
answered Jan 4 at 2:38
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
add a comment |
$begingroup$
(To remove the question from unanswered).
According to Exercise 2.3.B from [Eng], $(Atimes B)^circ=A^circtimes B^circ$ and $partial(Atimes B)=partial Atimes overline{B}cup overline{A}times partial B$. Prove this.
By Proposition 2.3.1 from [Eng], the set $A^circtimes B^circ$ is open, so $A^circtimes B^circsubset (Atimes B)^circ$. On the other hand, let $(x,y)in (Atimes B)^circ$ be any point. Then there exists an element $Utimes V$ of the canonical base at $Xtimes Y$ such that $(x,y)in Utimes Vsubset Atimes B$. Then $Usubset A$ and $Vsubset B$. Since $U$ and $V$ are open in $X$ and $Y$, respectively, we have $Usubset A^circ$ and $Vsubset B^circ$. Then $(x,y)in A^circtimes B^circ$.
By Proposition 2.3.3 from [Eng], $overline{Atimes B}=overline Atimes overline B$, so
$$partial(Atimes B)=$$ $$overline{Atimes B}setminus (Atimes B)^circ=$$
$$overline Atimes overline B setminus A^circtimes B^circ =$$ $$
((partial Acup A^circ)times (partial Bcup B^circ)) setminus A^circtimes B^circ=$$ $$partial Atimes partial Bcup partial Atimes B^circ cup A^circ times partial B=$$ $$(partial Atimes partial Bcup partial Atimes B^circ)cup
(partial Atimes partial Bcup A^circ times partial B)=$$ $$ partial Atimes overline{B}cup overline{A}times partial B.$$
References
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
answered Jan 4 at 2:38
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
(To remove the question from unanswered).
According to Exercise 2.3.B from [Eng], $(Atimes B)^circ=A^circtimes B^circ$ and $partial(Atimes B)=partial Atimes overline{B}cup overline{A}times partial B$. Prove this.
By Proposition 2.3.1 from [Eng], the set $A^circtimes B^circ$ is open, so $A^circtimes B^circsubset (Atimes B)^circ$. On the other hand, let $(x,y)in (Atimes B)^circ$ be any point. Then there exists an element $Utimes V$ of the canonical base at $Xtimes Y$ such that $(x,y)in Utimes Vsubset Atimes B$. Then $Usubset A$ and $Vsubset B$. Since $U$ and $V$ are open in $X$ and $Y$, respectively, we have $Usubset A^circ$ and $Vsubset B^circ$. Then $(x,y)in A^circtimes B^circ$.
By Proposition 2.3.3 from [Eng], $overline{Atimes B}=overline Atimes overline B$, so
$$partial(Atimes B)=$$ $$overline{Atimes B}setminus (Atimes B)^circ=$$
$$overline Atimes overline B setminus A^circtimes B^circ =$$ $$
((partial Acup A^circ)times (partial Bcup B^circ)) setminus A^circtimes B^circ=$$ $$partial Atimes partial Bcup partial Atimes B^circ cup A^circ times partial B=$$ $$(partial Atimes partial Bcup partial Atimes B^circ)cup
(partial Atimes partial Bcup A^circ times partial B)=$$ $$ partial Atimes overline{B}cup overline{A}times partial B.$$
References
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
answered Jan 4 at 2:38
Alex RavskyAlex Ravsky
42.7k32383
answered Jan 4 at 2:38
Alex RavskyAlex Ravsky
42.7k32383
answered Jan 4 at 2:38
Alex RavskyAlex Ravsky
42.7k32383
answered Jan 4 at 2:38
Alex RavskyAlex Ravsky
42.7k32383
42.7k32383
add a comment |
add a comment |
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1
$begingroup$
I suspect it wants you to find the interior and boundary in terms of the interiors and boundaries of $A$ and $B$ in their respective topologies.
$endgroup$
– gj255
Dec 14 '17 at 17:47
2
$begingroup$
For instance, wouldn't it be nice if $(Atimes B)^circ=A^circtimes B^circ$ and $partial(Atimes B)=partial Atimespartial B$? Sadly, only one of these is true (which one?) The other requires some modification. But this is the kind of thing they're after.
$endgroup$
– Arthur
Dec 14 '17 at 17:54
$begingroup$
I could show the first relation. Not sure about the second one though...
$endgroup$
– EpsilonDelta
Dec 14 '17 at 18:24
1
$begingroup$
Draw a picture in the plane using open intervals for $A$ and $B$ to get an feel for the boundary case. Or search the site.
$endgroup$
– Henno Brandsma
Dec 14 '17 at 23:01