If $intoverline{X} = intoverline{Y} = emptyset$ then $int(overline{Xcup Y}) = emptyset$.
$begingroup$
Let $X$ and $Y$ be subsets of a metric space. If $intoverline{X} = intoverline{Y} = emptyset$ then $int(overline{Xcup Y}) = emptyset$.
I know that $int(overline{Xcup Y}) = int(overline{X}cup overline{Y})$ and so I'm trying to prove the more general statement
$$
int(X) = int(Y) = emptyset implies int(Xcup Y) = emptyset
$$
which seems true to me but I not getting the proof.
What I tried so far is to consider an $xin int(Xcup Y)$, so there is a open ball $B_x$ that is contained in $Xcup Y$ but it cannot be contained in $X$ or $Y$ because these have empty interior. But I don't know what to do next.
Thanks in advance for any help.
general-topology metric-spaces
asked Dec 8 '18 at 0:48
AnalyticHarmonyAnalyticHarmony
642313
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be subsets of a metric space. If $intoverline{X} = intoverline{Y} = emptyset$ then $int(overline{Xcup Y}) = emptyset$.
I know that $int(overline{Xcup Y}) = int(overline{X}cup overline{Y})$ and so I'm trying to prove the more general statement
$$
int(X) = int(Y) = emptyset implies int(Xcup Y) = emptyset
$$
which seems true to me but I not getting the proof.
What I tried so far is to consider an $xin int(Xcup Y)$, so there is a open ball $B_x$ that is contained in $Xcup Y$ but it cannot be contained in $X$ or $Y$ because these have empty interior. But I don't know what to do next.
Thanks in advance for any help.
general-topology metric-spaces
asked Dec 8 '18 at 0:48
AnalyticHarmonyAnalyticHarmony
642313
$endgroup$
4
$begingroup$
The more general statement isn't true: consider the case $X$ is the set of rational numbers and $Y$ is the set of irrational numbers in $mathbb{R}$.
$endgroup$
– Daniel Schepler
Dec 8 '18 at 0:54
add a comment |
$begingroup$
Let $X$ and $Y$ be subsets of a metric space. If $intoverline{X} = intoverline{Y} = emptyset$ then $int(overline{Xcup Y}) = emptyset$.
I know that $int(overline{Xcup Y}) = int(overline{X}cup overline{Y})$ and so I'm trying to prove the more general statement
$$
int(X) = int(Y) = emptyset implies int(Xcup Y) = emptyset
$$
which seems true to me but I not getting the proof.
What I tried so far is to consider an $xin int(Xcup Y)$, so there is a open ball $B_x$ that is contained in $Xcup Y$ but it cannot be contained in $X$ or $Y$ because these have empty interior. But I don't know what to do next.
Thanks in advance for any help.
general-topology metric-spaces
asked Dec 8 '18 at 0:48
AnalyticHarmonyAnalyticHarmony
642313
$endgroup$
Let $X$ and $Y$ be subsets of a metric space. If $intoverline{X} = intoverline{Y} = emptyset$ then $int(overline{Xcup Y}) = emptyset$.
I know that $int(overline{Xcup Y}) = int(overline{X}cup overline{Y})$ and so I'm trying to prove the more general statement
$$
int(X) = int(Y) = emptyset implies int(Xcup Y) = emptyset
$$
which seems true to me but I not getting the proof.
What I tried so far is to consider an $xin int(Xcup Y)$, so there is a open ball $B_x$ that is contained in $Xcup Y$ but it cannot be contained in $X$ or $Y$ because these have empty interior. But I don't know what to do next.
Thanks in advance for any help.
general-topology metric-spaces
general-topology metric-spaces
asked Dec 8 '18 at 0:48
AnalyticHarmonyAnalyticHarmony
642313
asked Dec 8 '18 at 0:48
AnalyticHarmonyAnalyticHarmony
642313
asked Dec 8 '18 at 0:48
AnalyticHarmonyAnalyticHarmony
642313
asked Dec 8 '18 at 0:48
AnalyticHarmonyAnalyticHarmony
642313
asked Dec 8 '18 at 0:48
AnalyticHarmonyAnalyticHarmony
642313
642313
4
$begingroup$
The more general statement isn't true: consider the case $X$ is the set of rational numbers and $Y$ is the set of irrational numbers in $mathbb{R}$.
$endgroup$
– Daniel Schepler
Dec 8 '18 at 0:54
add a comment |
4
$begingroup$
The more general statement isn't true: consider the case $X$ is the set of rational numbers and $Y$ is the set of irrational numbers in $mathbb{R}$.
$endgroup$
– Daniel Schepler
Dec 8 '18 at 0:54
4
4
$begingroup$
The more general statement isn't true: consider the case $X$ is the set of rational numbers and $Y$ is the set of irrational numbers in $mathbb{R}$.
$endgroup$
– Daniel Schepler
Dec 8 '18 at 0:54
$begingroup$
The more general statement isn't true: consider the case $X$ is the set of rational numbers and $Y$ is the set of irrational numbers in $mathbb{R}$.
$endgroup$
– Daniel Schepler
Dec 8 '18 at 0:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Unfortunately, the statement you were expecting to hold does not, as Daniel Schepler illustratively pointed out. There is however a series of elementary propositions valid over any topological space $(X, mathscr{T})$:
Proposition 1. For any $M, N subseteq X$ it holds that $overline{M cup N}=overline{M} cup overline{N}$ (finite additivity of the closure operator).
Proposition 2. For any subset $M subseteq X$ and open subset $U in mathscr{T}$ it holds that $U cap overline{M} subseteq overline{U cap M}$.
Corollary 1. If $T subseteq X$ is a dense subset (i.e. $overline{T}=X$), then $overline{U}=overline{U cap T}$.
Corollary 2. The intersection of two dense open subsets is also dense and open.
If $F, G$ are two closed sets of empty interior, then their complementaries are open dense subsets and subsequently so is their intersection (corollary 2). Hence $F cup G$ has empty interior.
edited Dec 8 '18 at 22:22
Henno Brandsma
107k347114
answered Dec 8 '18 at 8:58
ΑΘΩΑΘΩ
2463
$endgroup$
$begingroup$
And all of this is valid in any kind of topological space....+1
$endgroup$
– DanielWainfleet
Dec 9 '18 at 10:07
1
$begingroup$
Whenever possible, I like to be as general in my approach as possible (to illustrate the richness of facts which unfold from a handful of well laid-out axioms).
$endgroup$
– ΑΘΩ
Dec 9 '18 at 10:10
$begingroup$
Many Q's about metric spaces on this site are applicable to all spaces, or all Hausdorff spaces. It seems to me that many of them are from students who have had little or even no study of top. spaces in general. This may be due the order of their course material.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 18:19
add a comment |
$begingroup$
$X$ is called nowhere dense iff the interior of its closure is empty iff the complement of $X$ contains a dense and open subset iff every non-empty open subset $O$ contains a non-empty open subset $O'$ that misses $X$.
As the finite intersection of dense and open subsets is dense and open, by de Morgan, the finite union of nowhere dense sets is nowhere dense. Or using the final characterisation: first pick $O'$ to avoid $X$ and an even smaller one inside $O'$ to also avoid $Y$.
answered Dec 8 '18 at 6:21
Henno BrandsmaHenno Brandsma
107k347114
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
Unfortunately, the statement you were expecting to hold does not, as Daniel Schepler illustratively pointed out. There is however a series of elementary propositions valid over any topological space $(X, mathscr{T})$:
Proposition 1. For any $M, N subseteq X$ it holds that $overline{M cup N}=overline{M} cup overline{N}$ (finite additivity of the closure operator).
Proposition 2. For any subset $M subseteq X$ and open subset $U in mathscr{T}$ it holds that $U cap overline{M} subseteq overline{U cap M}$.
Corollary 1. If $T subseteq X$ is a dense subset (i.e. $overline{T}=X$), then $overline{U}=overline{U cap T}$.
Corollary 2. The intersection of two dense open subsets is also dense and open.
If $F, G$ are two closed sets of empty interior, then their complementaries are open dense subsets and subsequently so is their intersection (corollary 2). Hence $F cup G$ has empty interior.
edited Dec 8 '18 at 22:22
Henno Brandsma
107k347114
answered Dec 8 '18 at 8:58
ΑΘΩΑΘΩ
2463
$endgroup$
$begingroup$
And all of this is valid in any kind of topological space....+1
$endgroup$
– DanielWainfleet
Dec 9 '18 at 10:07
1
$begingroup$
Whenever possible, I like to be as general in my approach as possible (to illustrate the richness of facts which unfold from a handful of well laid-out axioms).
$endgroup$
– ΑΘΩ
Dec 9 '18 at 10:10
$begingroup$
Many Q's about metric spaces on this site are applicable to all spaces, or all Hausdorff spaces. It seems to me that many of them are from students who have had little or even no study of top. spaces in general. This may be due the order of their course material.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 18:19
add a comment |
$begingroup$
Unfortunately, the statement you were expecting to hold does not, as Daniel Schepler illustratively pointed out. There is however a series of elementary propositions valid over any topological space $(X, mathscr{T})$:
Proposition 1. For any $M, N subseteq X$ it holds that $overline{M cup N}=overline{M} cup overline{N}$ (finite additivity of the closure operator).
Proposition 2. For any subset $M subseteq X$ and open subset $U in mathscr{T}$ it holds that $U cap overline{M} subseteq overline{U cap M}$.
Corollary 1. If $T subseteq X$ is a dense subset (i.e. $overline{T}=X$), then $overline{U}=overline{U cap T}$.
Corollary 2. The intersection of two dense open subsets is also dense and open.
If $F, G$ are two closed sets of empty interior, then their complementaries are open dense subsets and subsequently so is their intersection (corollary 2). Hence $F cup G$ has empty interior.
edited Dec 8 '18 at 22:22
Henno Brandsma
107k347114
answered Dec 8 '18 at 8:58
ΑΘΩΑΘΩ
2463
$endgroup$
$begingroup$
And all of this is valid in any kind of topological space....+1
$endgroup$
– DanielWainfleet
Dec 9 '18 at 10:07
1
$begingroup$
Whenever possible, I like to be as general in my approach as possible (to illustrate the richness of facts which unfold from a handful of well laid-out axioms).
$endgroup$
– ΑΘΩ
Dec 9 '18 at 10:10
$begingroup$
Many Q's about metric spaces on this site are applicable to all spaces, or all Hausdorff spaces. It seems to me that many of them are from students who have had little or even no study of top. spaces in general. This may be due the order of their course material.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 18:19
add a comment |
$begingroup$
Unfortunately, the statement you were expecting to hold does not, as Daniel Schepler illustratively pointed out. There is however a series of elementary propositions valid over any topological space $(X, mathscr{T})$:
Proposition 1. For any $M, N subseteq X$ it holds that $overline{M cup N}=overline{M} cup overline{N}$ (finite additivity of the closure operator).
Proposition 2. For any subset $M subseteq X$ and open subset $U in mathscr{T}$ it holds that $U cap overline{M} subseteq overline{U cap M}$.
Corollary 1. If $T subseteq X$ is a dense subset (i.e. $overline{T}=X$), then $overline{U}=overline{U cap T}$.
Corollary 2. The intersection of two dense open subsets is also dense and open.
If $F, G$ are two closed sets of empty interior, then their complementaries are open dense subsets and subsequently so is their intersection (corollary 2). Hence $F cup G$ has empty interior.
edited Dec 8 '18 at 22:22
Henno Brandsma
107k347114
answered Dec 8 '18 at 8:58
ΑΘΩΑΘΩ
2463
$endgroup$
Unfortunately, the statement you were expecting to hold does not, as Daniel Schepler illustratively pointed out. There is however a series of elementary propositions valid over any topological space $(X, mathscr{T})$:
Proposition 1. For any $M, N subseteq X$ it holds that $overline{M cup N}=overline{M} cup overline{N}$ (finite additivity of the closure operator).
Proposition 2. For any subset $M subseteq X$ and open subset $U in mathscr{T}$ it holds that $U cap overline{M} subseteq overline{U cap M}$.
Corollary 1. If $T subseteq X$ is a dense subset (i.e. $overline{T}=X$), then $overline{U}=overline{U cap T}$.
Corollary 2. The intersection of two dense open subsets is also dense and open.
If $F, G$ are two closed sets of empty interior, then their complementaries are open dense subsets and subsequently so is their intersection (corollary 2). Hence $F cup G$ has empty interior.
edited Dec 8 '18 at 22:22
Henno Brandsma
107k347114
answered Dec 8 '18 at 8:58
ΑΘΩΑΘΩ
2463
edited Dec 8 '18 at 22:22
Henno Brandsma
107k347114
edited Dec 8 '18 at 22:22
Henno Brandsma
107k347114
edited Dec 8 '18 at 22:22
Henno Brandsma
107k347114
107k347114
answered Dec 8 '18 at 8:58
ΑΘΩΑΘΩ
2463
answered Dec 8 '18 at 8:58
ΑΘΩΑΘΩ
2463
answered Dec 8 '18 at 8:58
ΑΘΩΑΘΩ
2463
2463
$begingroup$
And all of this is valid in any kind of topological space....+1
$endgroup$
– DanielWainfleet
Dec 9 '18 at 10:07
1
$begingroup$
Whenever possible, I like to be as general in my approach as possible (to illustrate the richness of facts which unfold from a handful of well laid-out axioms).
$endgroup$
– ΑΘΩ
Dec 9 '18 at 10:10
$begingroup$
Many Q's about metric spaces on this site are applicable to all spaces, or all Hausdorff spaces. It seems to me that many of them are from students who have had little or even no study of top. spaces in general. This may be due the order of their course material.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 18:19
add a comment |
$begingroup$
And all of this is valid in any kind of topological space....+1
$endgroup$
– DanielWainfleet
Dec 9 '18 at 10:07
1
$begingroup$
Whenever possible, I like to be as general in my approach as possible (to illustrate the richness of facts which unfold from a handful of well laid-out axioms).
$endgroup$
– ΑΘΩ
Dec 9 '18 at 10:10
$begingroup$
Many Q's about metric spaces on this site are applicable to all spaces, or all Hausdorff spaces. It seems to me that many of them are from students who have had little or even no study of top. spaces in general. This may be due the order of their course material.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 18:19
$begingroup$
And all of this is valid in any kind of topological space....+1
$endgroup$
– DanielWainfleet
Dec 9 '18 at 10:07
$begingroup$
And all of this is valid in any kind of topological space....+1
$endgroup$
– DanielWainfleet
Dec 9 '18 at 10:07
1
1
$begingroup$
Whenever possible, I like to be as general in my approach as possible (to illustrate the richness of facts which unfold from a handful of well laid-out axioms).
$endgroup$
– ΑΘΩ
Dec 9 '18 at 10:10
$begingroup$
Whenever possible, I like to be as general in my approach as possible (to illustrate the richness of facts which unfold from a handful of well laid-out axioms).
$endgroup$
– ΑΘΩ
Dec 9 '18 at 10:10
$begingroup$
Many Q's about metric spaces on this site are applicable to all spaces, or all Hausdorff spaces. It seems to me that many of them are from students who have had little or even no study of top. spaces in general. This may be due the order of their course material.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 18:19
$begingroup$
Many Q's about metric spaces on this site are applicable to all spaces, or all Hausdorff spaces. It seems to me that many of them are from students who have had little or even no study of top. spaces in general. This may be due the order of their course material.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 18:19
add a comment |
$begingroup$
$X$ is called nowhere dense iff the interior of its closure is empty iff the complement of $X$ contains a dense and open subset iff every non-empty open subset $O$ contains a non-empty open subset $O'$ that misses $X$.
As the finite intersection of dense and open subsets is dense and open, by de Morgan, the finite union of nowhere dense sets is nowhere dense. Or using the final characterisation: first pick $O'$ to avoid $X$ and an even smaller one inside $O'$ to also avoid $Y$.
answered Dec 8 '18 at 6:21
Henno BrandsmaHenno Brandsma
107k347114
$endgroup$
add a comment |
$begingroup$
$X$ is called nowhere dense iff the interior of its closure is empty iff the complement of $X$ contains a dense and open subset iff every non-empty open subset $O$ contains a non-empty open subset $O'$ that misses $X$.
As the finite intersection of dense and open subsets is dense and open, by de Morgan, the finite union of nowhere dense sets is nowhere dense. Or using the final characterisation: first pick $O'$ to avoid $X$ and an even smaller one inside $O'$ to also avoid $Y$.
answered Dec 8 '18 at 6:21
Henno BrandsmaHenno Brandsma
107k347114
$endgroup$
add a comment |
$begingroup$
$X$ is called nowhere dense iff the interior of its closure is empty iff the complement of $X$ contains a dense and open subset iff every non-empty open subset $O$ contains a non-empty open subset $O'$ that misses $X$.
As the finite intersection of dense and open subsets is dense and open, by de Morgan, the finite union of nowhere dense sets is nowhere dense. Or using the final characterisation: first pick $O'$ to avoid $X$ and an even smaller one inside $O'$ to also avoid $Y$.
answered Dec 8 '18 at 6:21
Henno BrandsmaHenno Brandsma
107k347114
$endgroup$
$X$ is called nowhere dense iff the interior of its closure is empty iff the complement of $X$ contains a dense and open subset iff every non-empty open subset $O$ contains a non-empty open subset $O'$ that misses $X$.
As the finite intersection of dense and open subsets is dense and open, by de Morgan, the finite union of nowhere dense sets is nowhere dense. Or using the final characterisation: first pick $O'$ to avoid $X$ and an even smaller one inside $O'$ to also avoid $Y$.
answered Dec 8 '18 at 6:21
Henno BrandsmaHenno Brandsma
107k347114
edited Dec 8 '18 at 8:19
answered Dec 8 '18 at 6:21
Henno BrandsmaHenno Brandsma
107k347114
answered Dec 8 '18 at 6:21
Henno BrandsmaHenno Brandsma
107k347114
answered Dec 8 '18 at 6:21
Henno BrandsmaHenno Brandsma
107k347114
107k347114
add a comment |
add a comment |
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$begingroup$
The more general statement isn't true: consider the case $X$ is the set of rational numbers and $Y$ is the set of irrational numbers in $mathbb{R}$.
$endgroup$
– Daniel Schepler
Dec 8 '18 at 0:54