Mixing
Is this interior-closure identity true? $(Ccap mathrm{Cl}(A))cup(Ccap mathrm{Cl}(Xsetminus A))=Ccup...
$begingroup$
Let $X$ be a topological space. Let $A,Csubseteq X$. Then, is it true that
$$(Ccap mathrm{Cl}(A))cup(Ccap mathrm{Cl}(Xsetminus A))=Ccup (Ccappartial A)$$
I've shown (if I'm not wrong) that
$$Ccap mathrm{Cl}(Xsetminus A)=Csetminus mathrm{Int}(A)$$ but I don't know what else to do. Any help would be appreciated.
general-topology
edited Jan 21 at 13:34
Andrés E. Caicedo
65.6k8159250
asked Jan 21 at 11:33
Ray BernRay Bern
17110
$endgroup$
add a comment |
$begingroup$
Let $X$ be a topological space. Let $A,Csubseteq X$. Then, is it true that
$$(Ccap mathrm{Cl}(A))cup(Ccap mathrm{Cl}(Xsetminus A))=Ccup (Ccappartial A)$$
I've shown (if I'm not wrong) that
$$Ccap mathrm{Cl}(Xsetminus A)=Csetminus mathrm{Int}(A)$$ but I don't know what else to do. Any help would be appreciated.
general-topology
edited Jan 21 at 13:34
Andrés E. Caicedo
65.6k8159250
asked Jan 21 at 11:33
Ray BernRay Bern
17110
$endgroup$
1
$begingroup$
Note that any set $Asubset X$ partitions $X$ into three sets: the interior points of $A$, the boundary points of $A$, and the exterior points of $A$. The closure is interior + boundary. This idea helps for a lot of identities involving interiors, boundaries and closures.
$endgroup$
– Christoph
Jan 21 at 11:49
add a comment |
$begingroup$
Let $X$ be a topological space. Let $A,Csubseteq X$. Then, is it true that
$$(Ccap mathrm{Cl}(A))cup(Ccap mathrm{Cl}(Xsetminus A))=Ccup (Ccappartial A)$$
I've shown (if I'm not wrong) that
$$Ccap mathrm{Cl}(Xsetminus A)=Csetminus mathrm{Int}(A)$$ but I don't know what else to do. Any help would be appreciated.
general-topology
edited Jan 21 at 13:34
Andrés E. Caicedo
65.6k8159250
asked Jan 21 at 11:33
Ray BernRay Bern
17110
$endgroup$
Let $X$ be a topological space. Let $A,Csubseteq X$. Then, is it true that
$$(Ccap mathrm{Cl}(A))cup(Ccap mathrm{Cl}(Xsetminus A))=Ccup (Ccappartial A)$$
I've shown (if I'm not wrong) that
$$Ccap mathrm{Cl}(Xsetminus A)=Csetminus mathrm{Int}(A)$$ but I don't know what else to do. Any help would be appreciated.
general-topology
general-topology
edited Jan 21 at 13:34
Andrés E. Caicedo
65.6k8159250
asked Jan 21 at 11:33
Ray BernRay Bern
17110
edited Jan 21 at 13:34
Andrés E. Caicedo
65.6k8159250
asked Jan 21 at 11:33
Ray BernRay Bern
17110
edited Jan 21 at 13:34
Andrés E. Caicedo
65.6k8159250
edited Jan 21 at 13:34
Andrés E. Caicedo
65.6k8159250
edited Jan 21 at 13:34
Andrés E. Caicedo
65.6k8159250
65.6k8159250
asked Jan 21 at 11:33
Ray BernRay Bern
17110
asked Jan 21 at 11:33
Ray BernRay Bern
17110
asked Jan 21 at 11:33
Ray BernRay Bern
17110
17110
1
$begingroup$
Note that any set $Asubset X$ partitions $X$ into three sets: the interior points of $A$, the boundary points of $A$, and the exterior points of $A$. The closure is interior + boundary. This idea helps for a lot of identities involving interiors, boundaries and closures.
$endgroup$
– Christoph
Jan 21 at 11:49
add a comment |
1
$begingroup$
Note that any set $Asubset X$ partitions $X$ into three sets: the interior points of $A$, the boundary points of $A$, and the exterior points of $A$. The closure is interior + boundary. This idea helps for a lot of identities involving interiors, boundaries and closures.
$endgroup$
– Christoph
Jan 21 at 11:49
1
1
$begingroup$
Note that any set $Asubset X$ partitions $X$ into three sets: the interior points of $A$, the boundary points of $A$, and the exterior points of $A$. The closure is interior + boundary. This idea helps for a lot of identities involving interiors, boundaries and closures.
$endgroup$
– Christoph
Jan 21 at 11:49
$begingroup$
Note that any set $Asubset X$ partitions $X$ into three sets: the interior points of $A$, the boundary points of $A$, and the exterior points of $A$. The closure is interior + boundary. This idea helps for a lot of identities involving interiors, boundaries and closures.
$endgroup$
– Christoph
Jan 21 at 11:49
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
By the distributive law:
$$(Ccap overline{A}) cup (C cap overline{(Xsetminus A)})= C cap (overline{A}cup overline{Xsetminus A})= C cap X = C$$
Your statement is not wrong since $C cap delta A$ is a subset of C, thus $C cup (C cap delta A)= C$.
answered Jan 21 at 12:01
Leander Tilsted KristensenLeander Tilsted Kristensen
213
$endgroup$
$begingroup$
My pedantic style would be $ Csupseteq (Ccap overline A)cup (Ccap overline {(Xsetminus A)},)$ $=Ccap (overline Acup overline {(Xsetminus A)},)supseteq$ $ supseteq Ccap (Acup (Xsetminus A))=$ $Ccap X=C$........+1.
$endgroup$
– DanielWainfleet
Jan 21 at 20:18
add a comment |
$begingroup$
Yes, correct on both. Derive a complicated formula
involving $partial$(A $cup$ B) and $partial$(A $cap$ B).
answered Jan 21 at 11:46
William ElliotWilliam Elliot
8,5022720
$endgroup$
add a comment |
$begingroup$
Since $Ccap partial A$ is a subset of $C$, the RHS of the identity if just $C$. The LHS simplifies to
$$
C cap left( operatorname{Cl}(A) cup operatorname{Cl}(Xsetminus A) right).
$$
Since $operatorname{Cl}(A) cup operatorname{Cl}(Xsetminus A)$ contains all the points of $A$ and all the points of $Xsetminus A$ it is equal to $X$. Hence the LHS is $Ccap X = C$ and agrees with the RHS.
answered Jan 21 at 11:53
ChristophChristoph
12.5k1642
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By the distributive law:
$$(Ccap overline{A}) cup (C cap overline{(Xsetminus A)})= C cap (overline{A}cup overline{Xsetminus A})= C cap X = C$$
Your statement is not wrong since $C cap delta A$ is a subset of C, thus $C cup (C cap delta A)= C$.
answered Jan 21 at 12:01
Leander Tilsted KristensenLeander Tilsted Kristensen
213
$endgroup$
$begingroup$
My pedantic style would be $ Csupseteq (Ccap overline A)cup (Ccap overline {(Xsetminus A)},)$ $=Ccap (overline Acup overline {(Xsetminus A)},)supseteq$ $ supseteq Ccap (Acup (Xsetminus A))=$ $Ccap X=C$........+1.
$endgroup$
– DanielWainfleet
Jan 21 at 20:18
add a comment |
$begingroup$
By the distributive law:
$$(Ccap overline{A}) cup (C cap overline{(Xsetminus A)})= C cap (overline{A}cup overline{Xsetminus A})= C cap X = C$$
Your statement is not wrong since $C cap delta A$ is a subset of C, thus $C cup (C cap delta A)= C$.
answered Jan 21 at 12:01
Leander Tilsted KristensenLeander Tilsted Kristensen
213
$endgroup$
$begingroup$
My pedantic style would be $ Csupseteq (Ccap overline A)cup (Ccap overline {(Xsetminus A)},)$ $=Ccap (overline Acup overline {(Xsetminus A)},)supseteq$ $ supseteq Ccap (Acup (Xsetminus A))=$ $Ccap X=C$........+1.
$endgroup$
– DanielWainfleet
Jan 21 at 20:18
add a comment |
$begingroup$
By the distributive law:
$$(Ccap overline{A}) cup (C cap overline{(Xsetminus A)})= C cap (overline{A}cup overline{Xsetminus A})= C cap X = C$$
Your statement is not wrong since $C cap delta A$ is a subset of C, thus $C cup (C cap delta A)= C$.
answered Jan 21 at 12:01
Leander Tilsted KristensenLeander Tilsted Kristensen
213
$endgroup$
By the distributive law:
$$(Ccap overline{A}) cup (C cap overline{(Xsetminus A)})= C cap (overline{A}cup overline{Xsetminus A})= C cap X = C$$
Your statement is not wrong since $C cap delta A$ is a subset of C, thus $C cup (C cap delta A)= C$.
answered Jan 21 at 12:01
Leander Tilsted KristensenLeander Tilsted Kristensen
213
answered Jan 21 at 12:01
Leander Tilsted KristensenLeander Tilsted Kristensen
213
answered Jan 21 at 12:01
Leander Tilsted KristensenLeander Tilsted Kristensen
213
answered Jan 21 at 12:01
Leander Tilsted KristensenLeander Tilsted Kristensen
213
213
$begingroup$
My pedantic style would be $ Csupseteq (Ccap overline A)cup (Ccap overline {(Xsetminus A)},)$ $=Ccap (overline Acup overline {(Xsetminus A)},)supseteq$ $ supseteq Ccap (Acup (Xsetminus A))=$ $Ccap X=C$........+1.
$endgroup$
– DanielWainfleet
Jan 21 at 20:18
add a comment |
$begingroup$
My pedantic style would be $ Csupseteq (Ccap overline A)cup (Ccap overline {(Xsetminus A)},)$ $=Ccap (overline Acup overline {(Xsetminus A)},)supseteq$ $ supseteq Ccap (Acup (Xsetminus A))=$ $Ccap X=C$........+1.
$endgroup$
– DanielWainfleet
Jan 21 at 20:18
$begingroup$
My pedantic style would be $ Csupseteq (Ccap overline A)cup (Ccap overline {(Xsetminus A)},)$ $=Ccap (overline Acup overline {(Xsetminus A)},)supseteq$ $ supseteq Ccap (Acup (Xsetminus A))=$ $Ccap X=C$........+1.
$endgroup$
– DanielWainfleet
Jan 21 at 20:18
$begingroup$
My pedantic style would be $ Csupseteq (Ccap overline A)cup (Ccap overline {(Xsetminus A)},)$ $=Ccap (overline Acup overline {(Xsetminus A)},)supseteq$ $ supseteq Ccap (Acup (Xsetminus A))=$ $Ccap X=C$........+1.
$endgroup$
– DanielWainfleet
Jan 21 at 20:18
add a comment |
$begingroup$
Yes, correct on both. Derive a complicated formula
involving $partial$(A $cup$ B) and $partial$(A $cap$ B).
answered Jan 21 at 11:46
William ElliotWilliam Elliot
8,5022720
$endgroup$
add a comment |
$begingroup$
Yes, correct on both. Derive a complicated formula
involving $partial$(A $cup$ B) and $partial$(A $cap$ B).
answered Jan 21 at 11:46
William ElliotWilliam Elliot
8,5022720
$endgroup$
add a comment |
$begingroup$
Yes, correct on both. Derive a complicated formula
involving $partial$(A $cup$ B) and $partial$(A $cap$ B).
answered Jan 21 at 11:46
William ElliotWilliam Elliot
8,5022720
$endgroup$
Yes, correct on both. Derive a complicated formula
involving $partial$(A $cup$ B) and $partial$(A $cap$ B).
answered Jan 21 at 11:46
William ElliotWilliam Elliot
8,5022720
answered Jan 21 at 11:46
William ElliotWilliam Elliot
8,5022720
answered Jan 21 at 11:46
William ElliotWilliam Elliot
8,5022720
answered Jan 21 at 11:46
William ElliotWilliam Elliot
8,5022720
8,5022720
add a comment |
add a comment |
$begingroup$
Since $Ccap partial A$ is a subset of $C$, the RHS of the identity if just $C$. The LHS simplifies to
$$
C cap left( operatorname{Cl}(A) cup operatorname{Cl}(Xsetminus A) right).
$$
Since $operatorname{Cl}(A) cup operatorname{Cl}(Xsetminus A)$ contains all the points of $A$ and all the points of $Xsetminus A$ it is equal to $X$. Hence the LHS is $Ccap X = C$ and agrees with the RHS.
answered Jan 21 at 11:53
ChristophChristoph
12.5k1642
$endgroup$
add a comment |
$begingroup$
Since $Ccap partial A$ is a subset of $C$, the RHS of the identity if just $C$. The LHS simplifies to
$$
C cap left( operatorname{Cl}(A) cup operatorname{Cl}(Xsetminus A) right).
$$
Since $operatorname{Cl}(A) cup operatorname{Cl}(Xsetminus A)$ contains all the points of $A$ and all the points of $Xsetminus A$ it is equal to $X$. Hence the LHS is $Ccap X = C$ and agrees with the RHS.
answered Jan 21 at 11:53
ChristophChristoph
12.5k1642
$endgroup$
add a comment |
$begingroup$
Since $Ccap partial A$ is a subset of $C$, the RHS of the identity if just $C$. The LHS simplifies to
$$
C cap left( operatorname{Cl}(A) cup operatorname{Cl}(Xsetminus A) right).
$$
Since $operatorname{Cl}(A) cup operatorname{Cl}(Xsetminus A)$ contains all the points of $A$ and all the points of $Xsetminus A$ it is equal to $X$. Hence the LHS is $Ccap X = C$ and agrees with the RHS.
answered Jan 21 at 11:53
ChristophChristoph
12.5k1642
$endgroup$
Since $Ccap partial A$ is a subset of $C$, the RHS of the identity if just $C$. The LHS simplifies to
$$
C cap left( operatorname{Cl}(A) cup operatorname{Cl}(Xsetminus A) right).
$$
Since $operatorname{Cl}(A) cup operatorname{Cl}(Xsetminus A)$ contains all the points of $A$ and all the points of $Xsetminus A$ it is equal to $X$. Hence the LHS is $Ccap X = C$ and agrees with the RHS.
answered Jan 21 at 11:53
ChristophChristoph
12.5k1642
answered Jan 21 at 11:53
ChristophChristoph
12.5k1642
answered Jan 21 at 11:53
ChristophChristoph
12.5k1642
answered Jan 21 at 11:53
ChristophChristoph
12.5k1642
12.5k1642
add a comment |
add a comment |
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1
$begingroup$
Note that any set $Asubset X$ partitions $X$ into three sets: the interior points of $A$, the boundary points of $A$, and the exterior points of $A$. The closure is interior + boundary. This idea helps for a lot of identities involving interiors, boundaries and closures.
$endgroup$
– Christoph
Jan 21 at 11:49