Boundary of union equal union of boundaries
$begingroup$
I would like to show that
$$tag{*}
partial ( A cup B ) = partial A cup partial B
$$
under the condition
$$tag{**}
overline{A} cap B = varnothing = A cap overline{B}
$$
This question has already been asked and correctly answered in, for instance, here.
What I'm looking for is a proof in a direct form as here, without proving the double inclusion. Is it possible? My attempts gave no positive outcome.
Something like
$$
begin{align}
partial ( A cup B )&=overline{Acup B}-(Acup B)^o\
&= overline{A}cup overline{B}-(Acup B)^C{}^C{}^o\
&= overline{A}cup overline{B}-(A^Ccap B^C){}^C{}^o\
&= (overline{A}cup overline{B})cap (A^Ccap B^C){}^C{}^o{}^C\
&= (overline{A}cup overline{B})cap overline{A^Ccap B^C}\
end{align}
$$
or starting from the bottom
$$
begin{align}
partial A cup partial B
&=(overline{A}-A^o)cup(overline{B}-B^o)\
&=(overline{A}cap A^o{}^C)cup(overline{B}cap B^o{}^C)\
&=(overline{A}cap overline{A^C})cup(overline{B}cap overline{B^C})
end{align}
$$
and somewhere apply (**) to get (*).
general-topology
asked Dec 29 '18 at 13:06
PeptideChainPeptideChain
464311
$endgroup$
add a comment |
$begingroup$
I would like to show that
$$tag{*}
partial ( A cup B ) = partial A cup partial B
$$
under the condition
$$tag{**}
overline{A} cap B = varnothing = A cap overline{B}
$$
This question has already been asked and correctly answered in, for instance, here.
What I'm looking for is a proof in a direct form as here, without proving the double inclusion. Is it possible? My attempts gave no positive outcome.
Something like
$$
begin{align}
partial ( A cup B )&=overline{Acup B}-(Acup B)^o\
&= overline{A}cup overline{B}-(Acup B)^C{}^C{}^o\
&= overline{A}cup overline{B}-(A^Ccap B^C){}^C{}^o\
&= (overline{A}cup overline{B})cap (A^Ccap B^C){}^C{}^o{}^C\
&= (overline{A}cup overline{B})cap overline{A^Ccap B^C}\
end{align}
$$
or starting from the bottom
$$
begin{align}
partial A cup partial B
&=(overline{A}-A^o)cup(overline{B}-B^o)\
&=(overline{A}cap A^o{}^C)cup(overline{B}cap B^o{}^C)\
&=(overline{A}cap overline{A^C})cup(overline{B}cap overline{B^C})
end{align}
$$
and somewhere apply (**) to get (*).
general-topology
asked Dec 29 '18 at 13:06
PeptideChainPeptideChain
464311
$endgroup$
$begingroup$
If you provide your attempt it will help me understand what you mean by "a proof in a direct form".
$endgroup$
– Yanko
Dec 29 '18 at 13:10
$begingroup$
@Yanko I would like a sequence of equalities; I edited the question
$endgroup$
– PeptideChain
Dec 29 '18 at 13:27
$begingroup$
A proof is a proof. Why does it have to be a chain of equalities? Maybe you're too stuck on set algebra?
$endgroup$
– Henno Brandsma
Dec 29 '18 at 17:32
$begingroup$
@HennoBrandsma In my option would be a nicer proof. I can do also without it. However the condition (**) is so simple that I imagined that it could be replaced in some expansion of one of the two sides of the equation. If someone find the proof is good, otherwise it is also good.
$endgroup$
– PeptideChain
Dec 30 '18 at 10:41
add a comment |
$begingroup$
I would like to show that
$$tag{*}
partial ( A cup B ) = partial A cup partial B
$$
under the condition
$$tag{**}
overline{A} cap B = varnothing = A cap overline{B}
$$
This question has already been asked and correctly answered in, for instance, here.
What I'm looking for is a proof in a direct form as here, without proving the double inclusion. Is it possible? My attempts gave no positive outcome.
Something like
$$
begin{align}
partial ( A cup B )&=overline{Acup B}-(Acup B)^o\
&= overline{A}cup overline{B}-(Acup B)^C{}^C{}^o\
&= overline{A}cup overline{B}-(A^Ccap B^C){}^C{}^o\
&= (overline{A}cup overline{B})cap (A^Ccap B^C){}^C{}^o{}^C\
&= (overline{A}cup overline{B})cap overline{A^Ccap B^C}\
end{align}
$$
or starting from the bottom
$$
begin{align}
partial A cup partial B
&=(overline{A}-A^o)cup(overline{B}-B^o)\
&=(overline{A}cap A^o{}^C)cup(overline{B}cap B^o{}^C)\
&=(overline{A}cap overline{A^C})cup(overline{B}cap overline{B^C})
end{align}
$$
and somewhere apply (**) to get (*).
general-topology
asked Dec 29 '18 at 13:06
PeptideChainPeptideChain
464311
$endgroup$
I would like to show that
$$tag{*}
partial ( A cup B ) = partial A cup partial B
$$
under the condition
$$tag{**}
overline{A} cap B = varnothing = A cap overline{B}
$$
This question has already been asked and correctly answered in, for instance, here.
What I'm looking for is a proof in a direct form as here, without proving the double inclusion. Is it possible? My attempts gave no positive outcome.
Something like
$$
begin{align}
partial ( A cup B )&=overline{Acup B}-(Acup B)^o\
&= overline{A}cup overline{B}-(Acup B)^C{}^C{}^o\
&= overline{A}cup overline{B}-(A^Ccap B^C){}^C{}^o\
&= (overline{A}cup overline{B})cap (A^Ccap B^C){}^C{}^o{}^C\
&= (overline{A}cup overline{B})cap overline{A^Ccap B^C}\
end{align}
$$
or starting from the bottom
$$
begin{align}
partial A cup partial B
&=(overline{A}-A^o)cup(overline{B}-B^o)\
&=(overline{A}cap A^o{}^C)cup(overline{B}cap B^o{}^C)\
&=(overline{A}cap overline{A^C})cup(overline{B}cap overline{B^C})
end{align}
$$
and somewhere apply (**) to get (*).
general-topology
general-topology
asked Dec 29 '18 at 13:06
PeptideChainPeptideChain
464311
asked Dec 29 '18 at 13:06
PeptideChainPeptideChain
464311
edited Dec 29 '18 at 13:25
PeptideChain
asked Dec 29 '18 at 13:06
PeptideChainPeptideChain
464311
asked Dec 29 '18 at 13:06
PeptideChainPeptideChain
464311
asked Dec 29 '18 at 13:06
PeptideChainPeptideChain
464311
464311
$begingroup$
If you provide your attempt it will help me understand what you mean by "a proof in a direct form".
$endgroup$
– Yanko
Dec 29 '18 at 13:10
$begingroup$
@Yanko I would like a sequence of equalities; I edited the question
$endgroup$
– PeptideChain
Dec 29 '18 at 13:27
$begingroup$
A proof is a proof. Why does it have to be a chain of equalities? Maybe you're too stuck on set algebra?
$endgroup$
– Henno Brandsma
Dec 29 '18 at 17:32
$begingroup$
@HennoBrandsma In my option would be a nicer proof. I can do also without it. However the condition (**) is so simple that I imagined that it could be replaced in some expansion of one of the two sides of the equation. If someone find the proof is good, otherwise it is also good.
$endgroup$
– PeptideChain
Dec 30 '18 at 10:41
add a comment |
$begingroup$
If you provide your attempt it will help me understand what you mean by "a proof in a direct form".
$endgroup$
– Yanko
Dec 29 '18 at 13:10
$begingroup$
@Yanko I would like a sequence of equalities; I edited the question
$endgroup$
– PeptideChain
Dec 29 '18 at 13:27
$begingroup$
A proof is a proof. Why does it have to be a chain of equalities? Maybe you're too stuck on set algebra?
$endgroup$
– Henno Brandsma
Dec 29 '18 at 17:32
$begingroup$
@HennoBrandsma In my option would be a nicer proof. I can do also without it. However the condition (**) is so simple that I imagined that it could be replaced in some expansion of one of the two sides of the equation. If someone find the proof is good, otherwise it is also good.
$endgroup$
– PeptideChain
Dec 30 '18 at 10:41
$begingroup$
If you provide your attempt it will help me understand what you mean by "a proof in a direct form".
$endgroup$
– Yanko
Dec 29 '18 at 13:10
$begingroup$
If you provide your attempt it will help me understand what you mean by "a proof in a direct form".
$endgroup$
– Yanko
Dec 29 '18 at 13:10
$begingroup$
@Yanko I would like a sequence of equalities; I edited the question
$endgroup$
– PeptideChain
Dec 29 '18 at 13:27
$begingroup$
@Yanko I would like a sequence of equalities; I edited the question
$endgroup$
– PeptideChain
Dec 29 '18 at 13:27
$begingroup$
A proof is a proof. Why does it have to be a chain of equalities? Maybe you're too stuck on set algebra?
$endgroup$
– Henno Brandsma
Dec 29 '18 at 17:32
$begingroup$
A proof is a proof. Why does it have to be a chain of equalities? Maybe you're too stuck on set algebra?
$endgroup$
– Henno Brandsma
Dec 29 '18 at 17:32
$begingroup$
@HennoBrandsma In my option would be a nicer proof. I can do also without it. However the condition (**) is so simple that I imagined that it could be replaced in some expansion of one of the two sides of the equation. If someone find the proof is good, otherwise it is also good.
$endgroup$
– PeptideChain
Dec 30 '18 at 10:41
$begingroup$
@HennoBrandsma In my option would be a nicer proof. I can do also without it. However the condition (**) is so simple that I imagined that it could be replaced in some expansion of one of the two sides of the equation. If someone find the proof is good, otherwise it is also good.
$endgroup$
– PeptideChain
Dec 30 '18 at 10:41
add a comment |
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$begingroup$
If you provide your attempt it will help me understand what you mean by "a proof in a direct form".
$endgroup$
– Yanko
Dec 29 '18 at 13:10
$begingroup$
@Yanko I would like a sequence of equalities; I edited the question
$endgroup$
– PeptideChain
Dec 29 '18 at 13:27
$begingroup$
A proof is a proof. Why does it have to be a chain of equalities? Maybe you're too stuck on set algebra?
$endgroup$
– Henno Brandsma
Dec 29 '18 at 17:32
$begingroup$
@HennoBrandsma In my option would be a nicer proof. I can do also without it. However the condition (**) is so simple that I imagined that it could be replaced in some expansion of one of the two sides of the equation. If someone find the proof is good, otherwise it is also good.
$endgroup$
– PeptideChain
Dec 30 '18 at 10:41