Mixing
$partial(A)=cl(A)Longleftrightarrow int(A)=emptyset$
$begingroup$
Is it write to say:
begin{align}
int(A)=emptyset& Longleftrightarrow C_{E}(int(A))=E\
&Longleftrightarrow cl(A)cap C_{E}(int(A))=cl({A})cap E=cl({A})\
&Longleftrightarrow partial(A)= cl(A)
end{align}
If I start from $partial(A)=cl(A)$ I don't know how to get $int({A})=emptyset$??
general-topology
asked Jan 1 at 16:08
VrouvrouVrouvrou
1,9191722
$endgroup$
|
show 3 more comments
$begingroup$
Is it write to say:
begin{align}
int(A)=emptyset& Longleftrightarrow C_{E}(int(A))=E\
&Longleftrightarrow cl(A)cap C_{E}(int(A))=cl({A})cap E=cl({A})\
&Longleftrightarrow partial(A)= cl(A)
end{align}
If I start from $partial(A)=cl(A)$ I don't know how to get $int({A})=emptyset$??
general-topology
asked Jan 1 at 16:08
VrouvrouVrouvrou
1,9191722
$endgroup$
$begingroup$
What is $adh(A)$?
$endgroup$
– Thomas Shelby
Jan 1 at 16:16
$begingroup$
the adherence of A,
$endgroup$
– Vrouvrou
Jan 1 at 16:17
$begingroup$
Could you clarify your notation a little better? I think $adh(A)$ are the adherent points of the set $A$ (aka closure of set) but I have no idea what $Fr(A)$ is then.
$endgroup$
– DreaDk
Jan 1 at 16:18
$begingroup$
@DreaDk $Fr= partial$
$endgroup$
– Vrouvrou
Jan 1 at 16:24
$begingroup$
I don't understand why -2 ????
$endgroup$
– Vrouvrou
Jan 1 at 16:38
|
show 3 more comments
$begingroup$
Is it write to say:
begin{align}
int(A)=emptyset& Longleftrightarrow C_{E}(int(A))=E\
&Longleftrightarrow cl(A)cap C_{E}(int(A))=cl({A})cap E=cl({A})\
&Longleftrightarrow partial(A)= cl(A)
end{align}
If I start from $partial(A)=cl(A)$ I don't know how to get $int({A})=emptyset$??
general-topology
asked Jan 1 at 16:08
VrouvrouVrouvrou
1,9191722
$endgroup$
Is it write to say:
begin{align}
int(A)=emptyset& Longleftrightarrow C_{E}(int(A))=E\
&Longleftrightarrow cl(A)cap C_{E}(int(A))=cl({A})cap E=cl({A})\
&Longleftrightarrow partial(A)= cl(A)
end{align}
If I start from $partial(A)=cl(A)$ I don't know how to get $int({A})=emptyset$??
general-topology
general-topology
asked Jan 1 at 16:08
VrouvrouVrouvrou
1,9191722
asked Jan 1 at 16:08
VrouvrouVrouvrou
1,9191722
edited Jan 1 at 16:39
Vrouvrou
asked Jan 1 at 16:08
VrouvrouVrouvrou
1,9191722
asked Jan 1 at 16:08
VrouvrouVrouvrou
1,9191722
asked Jan 1 at 16:08
VrouvrouVrouvrou
1,9191722
1,9191722
$begingroup$
What is $adh(A)$?
$endgroup$
– Thomas Shelby
Jan 1 at 16:16
$begingroup$
the adherence of A,
$endgroup$
– Vrouvrou
Jan 1 at 16:17
$begingroup$
Could you clarify your notation a little better? I think $adh(A)$ are the adherent points of the set $A$ (aka closure of set) but I have no idea what $Fr(A)$ is then.
$endgroup$
– DreaDk
Jan 1 at 16:18
$begingroup$
@DreaDk $Fr= partial$
$endgroup$
– Vrouvrou
Jan 1 at 16:24
$begingroup$
I don't understand why -2 ????
$endgroup$
– Vrouvrou
Jan 1 at 16:38
|
show 3 more comments
$begingroup$
What is $adh(A)$?
$endgroup$
– Thomas Shelby
Jan 1 at 16:16
$begingroup$
the adherence of A,
$endgroup$
– Vrouvrou
Jan 1 at 16:17
$begingroup$
Could you clarify your notation a little better? I think $adh(A)$ are the adherent points of the set $A$ (aka closure of set) but I have no idea what $Fr(A)$ is then.
$endgroup$
– DreaDk
Jan 1 at 16:18
$begingroup$
@DreaDk $Fr= partial$
$endgroup$
– Vrouvrou
Jan 1 at 16:24
$begingroup$
I don't understand why -2 ????
$endgroup$
– Vrouvrou
Jan 1 at 16:38
$begingroup$
What is $adh(A)$?
$endgroup$
– Thomas Shelby
Jan 1 at 16:16
$begingroup$
What is $adh(A)$?
$endgroup$
– Thomas Shelby
Jan 1 at 16:16
$begingroup$
the adherence of A,
$endgroup$
– Vrouvrou
Jan 1 at 16:17
$begingroup$
the adherence of A,
$endgroup$
– Vrouvrou
Jan 1 at 16:17
$begingroup$
Could you clarify your notation a little better? I think $adh(A)$ are the adherent points of the set $A$ (aka closure of set) but I have no idea what $Fr(A)$ is then.
$endgroup$
– DreaDk
Jan 1 at 16:18
$begingroup$
Could you clarify your notation a little better? I think $adh(A)$ are the adherent points of the set $A$ (aka closure of set) but I have no idea what $Fr(A)$ is then.
$endgroup$
– DreaDk
Jan 1 at 16:18
$begingroup$
@DreaDk $Fr= partial$
$endgroup$
– Vrouvrou
Jan 1 at 16:24
$begingroup$
@DreaDk $Fr= partial$
$endgroup$
– Vrouvrou
Jan 1 at 16:24
$begingroup$
I don't understand why -2 ????
$endgroup$
– Vrouvrou
Jan 1 at 16:38
$begingroup$
I don't understand why -2 ????
$endgroup$
– Vrouvrou
Jan 1 at 16:38
|
show 3 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Note that $mbox{Fr}(A)=overline{A}setminus A^°$. Now, if $A^°=emptyset$, $mbox{Fr}(A)=overline{A}$ trivially. If, $mbox{Fr}(A)=overline{A}$, it means that $overline{A}cap mbox{int}(A)^c=overline{A}$, which is equivalent to, $overline{A}subset mbox{int}(A)^c$, i.e, $mbox{int}(A)^csubsetoverline{A}^c$, now intersect with $A$, and then $A^°=A^°cap Asubset overline{A}^ccap A=emptyset$ since $Asubsetoverline{A}$.
answered Jan 1 at 16:18
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
$endgroup$
add a comment |
$begingroup$
Another way to see this, $Fr(A)=adh(A)cap C_E(int(A))$ by definition, $Fr(A)=adh(A)$ implies that $int(A)subset adh(A)=C_E(int(A)$ this implies that $int(A)$ is empty since it is contained in its complementary space.
answered Jan 1 at 16:19
Tsemo AristideTsemo Aristide
56.6k11444
$endgroup$
add a comment |
$begingroup$
$partial A = overline{A} setminus A^circ$. (closure minus interior). Also $A^circ subseteq A subseteq overline{A}$.
From this it's immedediate that $partial A = overline{A}$ iff $A^circ = emptyset$.
As the OP asks left to right: suppose $x in A^circ$ existed. Then $x in overline{A}$ but $x notin partial A$, contradicting that $partial A = overline{A}$. So $A^circ = emptyset$; the reverse is just substituting in the formula for $partial A$.
answered Jan 1 at 22:59
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
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3 Answers
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active
oldest
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3 Answers
3
active
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votes
active
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oldest
votes
$begingroup$
Note that $mbox{Fr}(A)=overline{A}setminus A^°$. Now, if $A^°=emptyset$, $mbox{Fr}(A)=overline{A}$ trivially. If, $mbox{Fr}(A)=overline{A}$, it means that $overline{A}cap mbox{int}(A)^c=overline{A}$, which is equivalent to, $overline{A}subset mbox{int}(A)^c$, i.e, $mbox{int}(A)^csubsetoverline{A}^c$, now intersect with $A$, and then $A^°=A^°cap Asubset overline{A}^ccap A=emptyset$ since $Asubsetoverline{A}$.
answered Jan 1 at 16:18
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
$endgroup$
add a comment |
$begingroup$
Note that $mbox{Fr}(A)=overline{A}setminus A^°$. Now, if $A^°=emptyset$, $mbox{Fr}(A)=overline{A}$ trivially. If, $mbox{Fr}(A)=overline{A}$, it means that $overline{A}cap mbox{int}(A)^c=overline{A}$, which is equivalent to, $overline{A}subset mbox{int}(A)^c$, i.e, $mbox{int}(A)^csubsetoverline{A}^c$, now intersect with $A$, and then $A^°=A^°cap Asubset overline{A}^ccap A=emptyset$ since $Asubsetoverline{A}$.
answered Jan 1 at 16:18
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
$endgroup$
add a comment |
$begingroup$
Note that $mbox{Fr}(A)=overline{A}setminus A^°$. Now, if $A^°=emptyset$, $mbox{Fr}(A)=overline{A}$ trivially. If, $mbox{Fr}(A)=overline{A}$, it means that $overline{A}cap mbox{int}(A)^c=overline{A}$, which is equivalent to, $overline{A}subset mbox{int}(A)^c$, i.e, $mbox{int}(A)^csubsetoverline{A}^c$, now intersect with $A$, and then $A^°=A^°cap Asubset overline{A}^ccap A=emptyset$ since $Asubsetoverline{A}$.
answered Jan 1 at 16:18
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
$endgroup$
Note that $mbox{Fr}(A)=overline{A}setminus A^°$. Now, if $A^°=emptyset$, $mbox{Fr}(A)=overline{A}$ trivially. If, $mbox{Fr}(A)=overline{A}$, it means that $overline{A}cap mbox{int}(A)^c=overline{A}$, which is equivalent to, $overline{A}subset mbox{int}(A)^c$, i.e, $mbox{int}(A)^csubsetoverline{A}^c$, now intersect with $A$, and then $A^°=A^°cap Asubset overline{A}^ccap A=emptyset$ since $Asubsetoverline{A}$.
answered Jan 1 at 16:18
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
answered Jan 1 at 16:18
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
answered Jan 1 at 16:18
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
answered Jan 1 at 16:18
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
825110
add a comment |
add a comment |
$begingroup$
Another way to see this, $Fr(A)=adh(A)cap C_E(int(A))$ by definition, $Fr(A)=adh(A)$ implies that $int(A)subset adh(A)=C_E(int(A)$ this implies that $int(A)$ is empty since it is contained in its complementary space.
answered Jan 1 at 16:19
Tsemo AristideTsemo Aristide
56.6k11444
$endgroup$
add a comment |
$begingroup$
Another way to see this, $Fr(A)=adh(A)cap C_E(int(A))$ by definition, $Fr(A)=adh(A)$ implies that $int(A)subset adh(A)=C_E(int(A)$ this implies that $int(A)$ is empty since it is contained in its complementary space.
answered Jan 1 at 16:19
Tsemo AristideTsemo Aristide
56.6k11444
$endgroup$
add a comment |
$begingroup$
Another way to see this, $Fr(A)=adh(A)cap C_E(int(A))$ by definition, $Fr(A)=adh(A)$ implies that $int(A)subset adh(A)=C_E(int(A)$ this implies that $int(A)$ is empty since it is contained in its complementary space.
answered Jan 1 at 16:19
Tsemo AristideTsemo Aristide
56.6k11444
$endgroup$
Another way to see this, $Fr(A)=adh(A)cap C_E(int(A))$ by definition, $Fr(A)=adh(A)$ implies that $int(A)subset adh(A)=C_E(int(A)$ this implies that $int(A)$ is empty since it is contained in its complementary space.
answered Jan 1 at 16:19
Tsemo AristideTsemo Aristide
56.6k11444
answered Jan 1 at 16:19
Tsemo AristideTsemo Aristide
56.6k11444
answered Jan 1 at 16:19
Tsemo AristideTsemo Aristide
56.6k11444
answered Jan 1 at 16:19
Tsemo AristideTsemo Aristide
56.6k11444
56.6k11444
add a comment |
add a comment |
$begingroup$
$partial A = overline{A} setminus A^circ$. (closure minus interior). Also $A^circ subseteq A subseteq overline{A}$.
From this it's immedediate that $partial A = overline{A}$ iff $A^circ = emptyset$.
As the OP asks left to right: suppose $x in A^circ$ existed. Then $x in overline{A}$ but $x notin partial A$, contradicting that $partial A = overline{A}$. So $A^circ = emptyset$; the reverse is just substituting in the formula for $partial A$.
answered Jan 1 at 22:59
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
$partial A = overline{A} setminus A^circ$. (closure minus interior). Also $A^circ subseteq A subseteq overline{A}$.
From this it's immedediate that $partial A = overline{A}$ iff $A^circ = emptyset$.
As the OP asks left to right: suppose $x in A^circ$ existed. Then $x in overline{A}$ but $x notin partial A$, contradicting that $partial A = overline{A}$. So $A^circ = emptyset$; the reverse is just substituting in the formula for $partial A$.
answered Jan 1 at 22:59
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
$partial A = overline{A} setminus A^circ$. (closure minus interior). Also $A^circ subseteq A subseteq overline{A}$.
From this it's immedediate that $partial A = overline{A}$ iff $A^circ = emptyset$.
As the OP asks left to right: suppose $x in A^circ$ existed. Then $x in overline{A}$ but $x notin partial A$, contradicting that $partial A = overline{A}$. So $A^circ = emptyset$; the reverse is just substituting in the formula for $partial A$.
answered Jan 1 at 22:59
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
$partial A = overline{A} setminus A^circ$. (closure minus interior). Also $A^circ subseteq A subseteq overline{A}$.
From this it's immedediate that $partial A = overline{A}$ iff $A^circ = emptyset$.
As the OP asks left to right: suppose $x in A^circ$ existed. Then $x in overline{A}$ but $x notin partial A$, contradicting that $partial A = overline{A}$. So $A^circ = emptyset$; the reverse is just substituting in the formula for $partial A$.
answered Jan 1 at 22:59
Henno BrandsmaHenno Brandsma
106k347114
answered Jan 1 at 22:59
Henno BrandsmaHenno Brandsma
106k347114
answered Jan 1 at 22:59
Henno BrandsmaHenno Brandsma
106k347114
answered Jan 1 at 22:59
Henno BrandsmaHenno Brandsma
106k347114
106k347114
add a comment |
add a comment |
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$begingroup$
What is $adh(A)$?
$endgroup$
– Thomas Shelby
Jan 1 at 16:16
$begingroup$
the adherence of A,
$endgroup$
– Vrouvrou
Jan 1 at 16:17
$begingroup$
Could you clarify your notation a little better? I think $adh(A)$ are the adherent points of the set $A$ (aka closure of set) but I have no idea what $Fr(A)$ is then.
$endgroup$
– DreaDk
Jan 1 at 16:18
$begingroup$
@DreaDk $Fr= partial$
$endgroup$
– Vrouvrou
Jan 1 at 16:24
$begingroup$
I don't understand why -2 ????
$endgroup$
– Vrouvrou
Jan 1 at 16:38