Mixing
$A, B$ be two subsets of a metric space such that $bar A cap bar B = emptyset$, then prove that $partial(Acup B)=partial A cup partial B$
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To prove the above-mentioned statement, I am just able to show $partial(Acup B)subseteq partial A cup partial B$ in the following manner
$partial A cup partial B=$
$(bar A backslash A^circ)cup (bar B backslash B^circ)=(bar A cup bar B)backslash (A^circ cup B^circ)=(overline A cup B)backslash (A^circ cup B^circ)supseteq (overline A cup B)backslash (A cup B)^circ= partial (Acup B)$
$implies partial (A cup B) subseteq partial A cup partial B tag 1labeleq1$
But it is general implication, I've not used the fact $bar A cap bar B = emptyset$ to prove $eqrefeq1$.
How to prove that $partial A cup partial B subseteq partial(Acup B) $?
Can anybody help me find the wayout? Thank for the assisatance in advance.
N.B. Here $bar A = A cup A^d$, where $A^d$ denotes the set of all limit points of $A$; $A^circ$ denotes the set of all interior points of $A$; $partial A$ denotes the set of all boundary points of A i.e. set of those points which are not interior nor exterior points of $A$.
metric-spaces
asked Jul 18 at 6:59
Biswarup Saha
2318
add a comment |Â
up vote
2
down vote
favorite
To prove the above-mentioned statement, I am just able to show $partial(Acup B)subseteq partial A cup partial B$ in the following manner
$partial A cup partial B=$
$(bar A backslash A^circ)cup (bar B backslash B^circ)=(bar A cup bar B)backslash (A^circ cup B^circ)=(overline A cup B)backslash (A^circ cup B^circ)supseteq (overline A cup B)backslash (A cup B)^circ= partial (Acup B)$
$implies partial (A cup B) subseteq partial A cup partial B tag 1labeleq1$
But it is general implication, I've not used the fact $bar A cap bar B = emptyset$ to prove $eqrefeq1$.
How to prove that $partial A cup partial B subseteq partial(Acup B) $?
Can anybody help me find the wayout? Thank for the assisatance in advance.
N.B. Here $bar A = A cup A^d$, where $A^d$ denotes the set of all limit points of $A$; $A^circ$ denotes the set of all interior points of $A$; $partial A$ denotes the set of all boundary points of A i.e. set of those points which are not interior nor exterior points of $A$.
metric-spaces
asked Jul 18 at 6:59
Biswarup Saha
2318
You just edited all your inclusions to go the other way. I don't think that that was right. Considering $A = (0,1)$ and $B = (1, 2)$ in $Bbb R$, the "easy" and generally true conclusion is that $partial(Acup B)subseteq partial Acup partial B$ (which in my example here would be $0, 2subseteq 0,1,2$), while the one where you need $bar Acapbar B=varnothing$ is $partial Acuppartial Bsubseteq partial(Acup B)$.
– Arthur
Jul 18 at 7:43
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
To prove the above-mentioned statement, I am just able to show $partial(Acup B)subseteq partial A cup partial B$ in the following manner
$partial A cup partial B=$
$(bar A backslash A^circ)cup (bar B backslash B^circ)=(bar A cup bar B)backslash (A^circ cup B^circ)=(overline A cup B)backslash (A^circ cup B^circ)supseteq (overline A cup B)backslash (A cup B)^circ= partial (Acup B)$
$implies partial (A cup B) subseteq partial A cup partial B tag 1labeleq1$
But it is general implication, I've not used the fact $bar A cap bar B = emptyset$ to prove $eqrefeq1$.
How to prove that $partial A cup partial B subseteq partial(Acup B) $?
Can anybody help me find the wayout? Thank for the assisatance in advance.
N.B. Here $bar A = A cup A^d$, where $A^d$ denotes the set of all limit points of $A$; $A^circ$ denotes the set of all interior points of $A$; $partial A$ denotes the set of all boundary points of A i.e. set of those points which are not interior nor exterior points of $A$.
metric-spaces
asked Jul 18 at 6:59
Biswarup Saha
2318
To prove the above-mentioned statement, I am just able to show $partial(Acup B)subseteq partial A cup partial B$ in the following manner
$partial A cup partial B=$
$(bar A backslash A^circ)cup (bar B backslash B^circ)=(bar A cup bar B)backslash (A^circ cup B^circ)=(overline A cup B)backslash (A^circ cup B^circ)supseteq (overline A cup B)backslash (A cup B)^circ= partial (Acup B)$
$implies partial (A cup B) subseteq partial A cup partial B tag 1labeleq1$
But it is general implication, I've not used the fact $bar A cap bar B = emptyset$ to prove $eqrefeq1$.
How to prove that $partial A cup partial B subseteq partial(Acup B) $?
Can anybody help me find the wayout? Thank for the assisatance in advance.
N.B. Here $bar A = A cup A^d$, where $A^d$ denotes the set of all limit points of $A$; $A^circ$ denotes the set of all interior points of $A$; $partial A$ denotes the set of all boundary points of A i.e. set of those points which are not interior nor exterior points of $A$.
metric-spaces
asked Jul 18 at 6:59
Biswarup Saha
2318
edited Jul 18 at 12:38
asked Jul 18 at 6:59
Biswarup Saha
2318
asked Jul 18 at 6:59
Biswarup Saha
2318
asked Jul 18 at 6:59
Biswarup Saha
2318
2318
You just edited all your inclusions to go the other way. I don't think that that was right. Considering $A = (0,1)$ and $B = (1, 2)$ in $Bbb R$, the "easy" and generally true conclusion is that $partial(Acup B)subseteq partial Acup partial B$ (which in my example here would be $0, 2subseteq 0,1,2$), while the one where you need $bar Acapbar B=varnothing$ is $partial Acuppartial Bsubseteq partial(Acup B)$.
– Arthur
Jul 18 at 7:43
add a comment |Â
You just edited all your inclusions to go the other way. I don't think that that was right. Considering $A = (0,1)$ and $B = (1, 2)$ in $Bbb R$, the "easy" and generally true conclusion is that $partial(Acup B)subseteq partial Acup partial B$ (which in my example here would be $0, 2subseteq 0,1,2$), while the one where you need $bar Acapbar B=varnothing$ is $partial Acuppartial Bsubseteq partial(Acup B)$.
– Arthur
Jul 18 at 7:43
You just edited all your inclusions to go the other way. I don't think that that was right. Considering $A = (0,1)$ and $B = (1, 2)$ in $Bbb R$, the "easy" and generally true conclusion is that $partial(Acup B)subseteq partial Acup partial B$ (which in my example here would be $0, 2subseteq 0,1,2$), while the one where you need $bar Acapbar B=varnothing$ is $partial Acuppartial Bsubseteq partial(Acup B)$.
– Arthur
Jul 18 at 7:43
You just edited all your inclusions to go the other way. I don't think that that was right. Considering $A = (0,1)$ and $B = (1, 2)$ in $Bbb R$, the "easy" and generally true conclusion is that $partial(Acup B)subseteq partial Acup partial B$ (which in my example here would be $0, 2subseteq 0,1,2$), while the one where you need $bar Acapbar B=varnothing$ is $partial Acuppartial Bsubseteq partial(Acup B)$.
– Arthur
Jul 18 at 7:43
add a comment |Â
1 Answer
1
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oldest
votes
up vote
2
down vote
I've not used the fact $bar A cap bar B = emptyset$ to prove $eqrefeq1$. How to prove that $partial A cup partial B subseteq partial(Acup B)$?
Well, you're kindof answering your own question there. The fact $bar A cap bar B = emptyset$ is exactly what makes the $supseteq$ in $overline(Acup B)setminus(A^circ cup B^circ)supseteq (overline A cup B)setminus(A cup B)^circ$ into $=$ because it implies
$$A^circ cup B^circ=(A cup B)^circ$$
answered Jul 18 at 7:10
Arthur
98.8k793175
1
you are saying that $bar A cap bar B=emptyset$ implies $(Acup B)^circ = A^circ cup B^circ$. Am I right?
– Biswarup Saha
Jul 18 at 7:15
@BiswarupSaha That is what I'm saying, yes.
– Arthur
Jul 18 at 7:21
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
I've not used the fact $bar A cap bar B = emptyset$ to prove $eqrefeq1$. How to prove that $partial A cup partial B subseteq partial(Acup B)$?
Well, you're kindof answering your own question there. The fact $bar A cap bar B = emptyset$ is exactly what makes the $supseteq$ in $overline(Acup B)setminus(A^circ cup B^circ)supseteq (overline A cup B)setminus(A cup B)^circ$ into $=$ because it implies
$$A^circ cup B^circ=(A cup B)^circ$$
answered Jul 18 at 7:10
Arthur
98.8k793175
1
you are saying that $bar A cap bar B=emptyset$ implies $(Acup B)^circ = A^circ cup B^circ$. Am I right?
– Biswarup Saha
Jul 18 at 7:15
@BiswarupSaha That is what I'm saying, yes.
– Arthur
Jul 18 at 7:21
add a comment |Â
up vote
2
down vote
I've not used the fact $bar A cap bar B = emptyset$ to prove $eqrefeq1$. How to prove that $partial A cup partial B subseteq partial(Acup B)$?
Well, you're kindof answering your own question there. The fact $bar A cap bar B = emptyset$ is exactly what makes the $supseteq$ in $overline(Acup B)setminus(A^circ cup B^circ)supseteq (overline A cup B)setminus(A cup B)^circ$ into $=$ because it implies
$$A^circ cup B^circ=(A cup B)^circ$$
answered Jul 18 at 7:10
Arthur
98.8k793175
1
you are saying that $bar A cap bar B=emptyset$ implies $(Acup B)^circ = A^circ cup B^circ$. Am I right?
– Biswarup Saha
Jul 18 at 7:15
@BiswarupSaha That is what I'm saying, yes.
– Arthur
Jul 18 at 7:21
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I've not used the fact $bar A cap bar B = emptyset$ to prove $eqrefeq1$. How to prove that $partial A cup partial B subseteq partial(Acup B)$?
Well, you're kindof answering your own question there. The fact $bar A cap bar B = emptyset$ is exactly what makes the $supseteq$ in $overline(Acup B)setminus(A^circ cup B^circ)supseteq (overline A cup B)setminus(A cup B)^circ$ into $=$ because it implies
$$A^circ cup B^circ=(A cup B)^circ$$
answered Jul 18 at 7:10
Arthur
98.8k793175
I've not used the fact $bar A cap bar B = emptyset$ to prove $eqrefeq1$. How to prove that $partial A cup partial B subseteq partial(Acup B)$?
Well, you're kindof answering your own question there. The fact $bar A cap bar B = emptyset$ is exactly what makes the $supseteq$ in $overline(Acup B)setminus(A^circ cup B^circ)supseteq (overline A cup B)setminus(A cup B)^circ$ into $=$ because it implies
$$A^circ cup B^circ=(A cup B)^circ$$
answered Jul 18 at 7:10
Arthur
98.8k793175
answered Jul 18 at 7:10
Arthur
98.8k793175
answered Jul 18 at 7:10
Arthur
98.8k793175
answered Jul 18 at 7:10
Arthur
98.8k793175
98.8k793175
1
you are saying that $bar A cap bar B=emptyset$ implies $(Acup B)^circ = A^circ cup B^circ$. Am I right?
– Biswarup Saha
Jul 18 at 7:15
@BiswarupSaha That is what I'm saying, yes.
– Arthur
Jul 18 at 7:21
add a comment |Â
1
you are saying that $bar A cap bar B=emptyset$ implies $(Acup B)^circ = A^circ cup B^circ$. Am I right?
– Biswarup Saha
Jul 18 at 7:15
@BiswarupSaha That is what I'm saying, yes.
– Arthur
Jul 18 at 7:21
1
1
you are saying that $bar A cap bar B=emptyset$ implies $(Acup B)^circ = A^circ cup B^circ$. Am I right?
– Biswarup Saha
Jul 18 at 7:15
you are saying that $bar A cap bar B=emptyset$ implies $(Acup B)^circ = A^circ cup B^circ$. Am I right?
– Biswarup Saha
Jul 18 at 7:15
@BiswarupSaha That is what I'm saying, yes.
– Arthur
Jul 18 at 7:21
@BiswarupSaha That is what I'm saying, yes.
– Arthur
Jul 18 at 7:21
add a comment |Â
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You just edited all your inclusions to go the other way. I don't think that that was right. Considering $A = (0,1)$ and $B = (1, 2)$ in $Bbb R$, the "easy" and generally true conclusion is that $partial(Acup B)subseteq partial Acup partial B$ (which in my example here would be $0, 2subseteq 0,1,2$), while the one where you need $bar Acapbar B=varnothing$ is $partial Acuppartial Bsubseteq partial(Acup B)$.
– Arthur
Jul 18 at 7:43