To prove or disprove $operatornameint(operatornamecl(A))subseteq operatornamecl(operatornameintA))$.Show the $operatornameint(A)$ is open.Give an example of a metric space $(X,d)$ and $Asubseteq X$ such that $textint(overlineA)notsubseteqoverlinetextint(A)$ and vice versaBoundary of Boundary of a set?Proving that $operatornamecl(operatornameint(Y))=operatornamecl(Y)$ for subsets $Y$ of a topological space $X$If $operatornameBdA cap operatornameBdB= emptyset$, show that $operatornameint(A)cupoperatornameint(B)=operatornameint(Acup B)$Does $operatornameint(f^-1[A]) subseteq f^−1 [operatornameint(A)]$ hold for continuous $f$?prove or disprove on topological space the following:Prove interior point is a topological invariantIf $A$ is open then $textInt(A)=A$Prove that $A$ is open if and only if $A=operatornameintA$
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To prove or disprove $operatornameint(operatornamecl(A))subseteq operatornamecl(operatornameintA))$.
Show the $operatornameint(A)$ is open.Give an example of a metric space $(X,d)$ and $Asubseteq X$ such that $textint(overlineA)notsubseteqoverlinetextint(A)$ and vice versaBoundary of Boundary of a set?Proving that $operatornamecl(operatornameint(Y))=operatornamecl(Y)$ for subsets $Y$ of a topological space $X$If $operatornameBdA cap operatornameBdB= emptyset$, show that $operatornameint(A)cupoperatornameint(B)=operatornameint(Acup B)$Does $operatornameint(f^-1[A]) subseteq f^−1 [operatornameint(A)]$ hold for continuous $f$?prove or disprove on topological space the following:Prove interior point is a topological invariantIf $A$ is open then $textInt(A)=A$Prove that $A$ is open if and only if $A=operatornameintA$
$begingroup$
Let $(X,tau)$ be a topological space, $A$ be a subset of $X$, $operatornamecl(A)$ and $operatornameint(A)$ denote the closure and interior of $A$, respectively. I want to find a relation between $operatornameint(operatornamecl(A))$ and $operatornamecl(operatornameintA))$. By an example I found that $operatornamecl(operatornameintA))nsubseteq operatornameint(operatornamecl(A))$, but I could not prove or disprove $operatornameint(operatornamecl(A))subseteq operatornamecl(operatornameintA))$. Any hint to prove or disprove it?
general-topology
edited yesterday
Henno Brandsma
112k348120
asked yesterday
InfinityInfinity
629514
$endgroup$
add a comment |
$begingroup$
Let $(X,tau)$ be a topological space, $A$ be a subset of $X$, $operatornamecl(A)$ and $operatornameint(A)$ denote the closure and interior of $A$, respectively. I want to find a relation between $operatornameint(operatornamecl(A))$ and $operatornamecl(operatornameintA))$. By an example I found that $operatornamecl(operatornameintA))nsubseteq operatornameint(operatornamecl(A))$, but I could not prove or disprove $operatornameint(operatornamecl(A))subseteq operatornamecl(operatornameintA))$. Any hint to prove or disprove it?
general-topology
edited yesterday
Henno Brandsma
112k348120
asked yesterday
InfinityInfinity
629514
$endgroup$
add a comment |
$begingroup$
Let $(X,tau)$ be a topological space, $A$ be a subset of $X$, $operatornamecl(A)$ and $operatornameint(A)$ denote the closure and interior of $A$, respectively. I want to find a relation between $operatornameint(operatornamecl(A))$ and $operatornamecl(operatornameintA))$. By an example I found that $operatornamecl(operatornameintA))nsubseteq operatornameint(operatornamecl(A))$, but I could not prove or disprove $operatornameint(operatornamecl(A))subseteq operatornamecl(operatornameintA))$. Any hint to prove or disprove it?
general-topology
edited yesterday
Henno Brandsma
112k348120
asked yesterday
InfinityInfinity
629514
$endgroup$
Let $(X,tau)$ be a topological space, $A$ be a subset of $X$, $operatornamecl(A)$ and $operatornameint(A)$ denote the closure and interior of $A$, respectively. I want to find a relation between $operatornameint(operatornamecl(A))$ and $operatornamecl(operatornameintA))$. By an example I found that $operatornamecl(operatornameintA))nsubseteq operatornameint(operatornamecl(A))$, but I could not prove or disprove $operatornameint(operatornamecl(A))subseteq operatornamecl(operatornameintA))$. Any hint to prove or disprove it?
general-topology
general-topology
edited yesterday
Henno Brandsma
112k348120
asked yesterday
InfinityInfinity
629514
edited yesterday
Henno Brandsma
112k348120
asked yesterday
InfinityInfinity
629514
edited yesterday
Henno Brandsma
112k348120
edited yesterday
Henno Brandsma
112k348120
edited yesterday
Henno Brandsma
112k348120
112k348120
asked yesterday
InfinityInfinity
629514
asked yesterday
InfinityInfinity
629514
asked yesterday
InfinityInfinity
629514
629514
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Hint: Take $X$ the set of real numbers and $A$ the subset of rational numbers.
answered yesterday
TomGrubbTomGrubb
11k11539
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Take $X$ the set of real numbers and $A$ the subset of rational numbers.
answered yesterday
TomGrubbTomGrubb
11k11539
$endgroup$
add a comment |
$begingroup$
Hint: Take $X$ the set of real numbers and $A$ the subset of rational numbers.
answered yesterday
TomGrubbTomGrubb
11k11539
$endgroup$
add a comment |
$begingroup$
Hint: Take $X$ the set of real numbers and $A$ the subset of rational numbers.
answered yesterday
TomGrubbTomGrubb
11k11539
$endgroup$
Hint: Take $X$ the set of real numbers and $A$ the subset of rational numbers.
answered yesterday
TomGrubbTomGrubb
11k11539
answered yesterday
TomGrubbTomGrubb
11k11539
answered yesterday
TomGrubbTomGrubb
11k11539
answered yesterday
TomGrubbTomGrubb
11k11539
11k11539
add a comment |
add a comment |
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