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What if metric space is replaced by arbitrary topological space, will the result still hold?
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Let $(X, d)$ be a metric space with no isolated points, and let $A$
be a relatively discrete subset of $X$. Prove that $A$ is nowhere
dense in $X$.
relatively discrete subset of $X$:= A subset $A$ of a topological space $(X,mathscr T)$ is relatively discrete provided that for each $ain A$, there exists $Uin mathscr T$ such that $U cap A=a$.
My aim is to prove $int(overlineA)=emptyset$. Let if possible $int(overlineA)neq emptyset$. Let $xin int(overlineA)$. which implies there exists $B_d(x,epsilon)subset overline A=Acup A'$.
How do I complete the proof?
What if metric space is replaced by arbitrary topological space, will the result still hold?
general-topology metric-spaces
asked 2 hours ago
Math geek
30017
add a comment |Â
up vote
1
down vote
favorite
Let $(X, d)$ be a metric space with no isolated points, and let $A$
be a relatively discrete subset of $X$. Prove that $A$ is nowhere
dense in $X$.
relatively discrete subset of $X$:= A subset $A$ of a topological space $(X,mathscr T)$ is relatively discrete provided that for each $ain A$, there exists $Uin mathscr T$ such that $U cap A=a$.
My aim is to prove $int(overlineA)=emptyset$. Let if possible $int(overlineA)neq emptyset$. Let $xin int(overlineA)$. which implies there exists $B_d(x,epsilon)subset overline A=Acup A'$.
How do I complete the proof?
What if metric space is replaced by arbitrary topological space, will the result still hold?
general-topology metric-spaces
asked 2 hours ago
Math geek
30017
1
If $X$ has indiscrete topology and $A$ is a singleton set, then $A$ is discrete, but $overline A=X$ is a nonempty open set
– Hagen von Eitzen
2 hours ago
Thank you for the counter example. How do i complete the proof? Can you please help me?
– Math geek
2 hours ago
@Mathgeek the result holds in $T_1$ spaces, I believe.
– Henno Brandsma
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $(X, d)$ be a metric space with no isolated points, and let $A$
be a relatively discrete subset of $X$. Prove that $A$ is nowhere
dense in $X$.
relatively discrete subset of $X$:= A subset $A$ of a topological space $(X,mathscr T)$ is relatively discrete provided that for each $ain A$, there exists $Uin mathscr T$ such that $U cap A=a$.
My aim is to prove $int(overlineA)=emptyset$. Let if possible $int(overlineA)neq emptyset$. Let $xin int(overlineA)$. which implies there exists $B_d(x,epsilon)subset overline A=Acup A'$.
How do I complete the proof?
What if metric space is replaced by arbitrary topological space, will the result still hold?
general-topology metric-spaces
asked 2 hours ago
Math geek
30017
Let $(X, d)$ be a metric space with no isolated points, and let $A$
be a relatively discrete subset of $X$. Prove that $A$ is nowhere
dense in $X$.
relatively discrete subset of $X$:= A subset $A$ of a topological space $(X,mathscr T)$ is relatively discrete provided that for each $ain A$, there exists $Uin mathscr T$ such that $U cap A=a$.
My aim is to prove $int(overlineA)=emptyset$. Let if possible $int(overlineA)neq emptyset$. Let $xin int(overlineA)$. which implies there exists $B_d(x,epsilon)subset overline A=Acup A'$.
How do I complete the proof?
What if metric space is replaced by arbitrary topological space, will the result still hold?
general-topology metric-spaces
general-topology metric-spaces
asked 2 hours ago
Math geek
30017
asked 2 hours ago
Math geek
30017
asked 2 hours ago
Math geek
30017
asked 2 hours ago
Math geek
30017
asked 2 hours ago
Math geek
30017
30017
1
If $X$ has indiscrete topology and $A$ is a singleton set, then $A$ is discrete, but $overline A=X$ is a nonempty open set
– Hagen von Eitzen
2 hours ago
Thank you for the counter example. How do i complete the proof? Can you please help me?
– Math geek
2 hours ago
@Mathgeek the result holds in $T_1$ spaces, I believe.
– Henno Brandsma
1 hour ago
add a comment |Â
1
If $X$ has indiscrete topology and $A$ is a singleton set, then $A$ is discrete, but $overline A=X$ is a nonempty open set
– Hagen von Eitzen
2 hours ago
Thank you for the counter example. How do i complete the proof? Can you please help me?
– Math geek
2 hours ago
@Mathgeek the result holds in $T_1$ spaces, I believe.
– Henno Brandsma
1 hour ago
1
1
If $X$ has indiscrete topology and $A$ is a singleton set, then $A$ is discrete, but $overline A=X$ is a nonempty open set
– Hagen von Eitzen
2 hours ago
If $X$ has indiscrete topology and $A$ is a singleton set, then $A$ is discrete, but $overline A=X$ is a nonempty open set
– Hagen von Eitzen
2 hours ago
Thank you for the counter example. How do i complete the proof? Can you please help me?
– Math geek
2 hours ago
Thank you for the counter example. How do i complete the proof? Can you please help me?
– Math geek
2 hours ago
@Mathgeek the result holds in $T_1$ spaces, I believe.
– Henno Brandsma
1 hour ago
@Mathgeek the result holds in $T_1$ spaces, I believe.
– Henno Brandsma
1 hour ago
add a comment |Â
3 Answers
3
active
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up vote
1
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Let $xinmathringoverline A$. Then there is a $r>0$ such that $B_r(x)subsetoverline A$. Since $B_r(x)$ is an open set which is contained in $overline A$, it contains some element $ain A$. But then, if $r'=r-d(x,a)$, $B_r'(a)subset B_r(x)$. In particular, $B_r'(a)subsetoverline A$. This is impossible, since $A$ is relatively discrete.
I will not answer the question from the title, since you already got an answer in the comments.
answered 2 hours ago
José Carlos Santos
131k17106192
add a comment |Â
up vote
1
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With your $x$ and $epsilon$, $$overline Asetminus B_d(x,epsilon/2)$$ is a strictly smaller closed set than $overline A$, hence cannot contain all of $A$. Pick $ain Acap B_d(x,epsilon/2)$, by which we achieve that $$ain B_d(a,epsilon/2)subseteq Acapoperatornameint(overline A).$$
By relative discreteness, we find $delta>0$ such that $B_d(a,delta)cap A=a$.
Wlog $deltaleepsilon/2$. Now
$$ S:=(overline Asetminus B_d(a,delta))cup a$$
is a closed set with $Asubseteq Ssubseteq overline A$, hence $S=overline A$. We still have $B_d(a,delta)subseteq overline A=S$, but $B_d(a,delta)cap S=a$, which means that $B_d(a,delta)=a$, contrary to the assumption that there are no isolated points.
answered 2 hours ago
Hagen von Eitzen
271k21264489
add a comment |Â
up vote
1
down vote
Suppose $X$ is a crowded $T_1$ space and $D$ is relatively discrete.
Suppose (for a contradiction) that there is some non-empty open set $U subseteq overlineD$
In particular, there is some $d in D cap U$ (being in the closure of $D$ means every neighbourhood intersects $D$) and as $D$ is relatively discrete, $d$ is open in $D$, so there is an open set $U_d$ of $X$ such that $U_d cap D = d$.
Now I claim that $U cap U_d = d$:
The right to left inclusion is clear, as both open sets contain $d$ and if $x neq d$ existed in $U cap U_d$, by $T_1$-ness of $X$ it follows that $U cap U_d cap (Xsetminusd)$ is an open set containing $x$ that misses $D$ entirely (clearly, as $U_d cap D = d$ and $(Xsetminus d) cap d = emptyset$) but $x in U subseteq overlineD$, so this cannot happen. This shows that indeed $U cap U_d = d$, making $d$ open, but this contradicts in turn that $X$ is crowded (has no isolated points)!
So no such $U$ can exist and $operatornameint(overlineA) = emptyset$.
answered 36 mins ago
Henno Brandsma
97.7k343105
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let $xinmathringoverline A$. Then there is a $r>0$ such that $B_r(x)subsetoverline A$. Since $B_r(x)$ is an open set which is contained in $overline A$, it contains some element $ain A$. But then, if $r'=r-d(x,a)$, $B_r'(a)subset B_r(x)$. In particular, $B_r'(a)subsetoverline A$. This is impossible, since $A$ is relatively discrete.
I will not answer the question from the title, since you already got an answer in the comments.
answered 2 hours ago
José Carlos Santos
131k17106192
add a comment |Â
up vote
1
down vote
Let $xinmathringoverline A$. Then there is a $r>0$ such that $B_r(x)subsetoverline A$. Since $B_r(x)$ is an open set which is contained in $overline A$, it contains some element $ain A$. But then, if $r'=r-d(x,a)$, $B_r'(a)subset B_r(x)$. In particular, $B_r'(a)subsetoverline A$. This is impossible, since $A$ is relatively discrete.
I will not answer the question from the title, since you already got an answer in the comments.
answered 2 hours ago
José Carlos Santos
131k17106192
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $xinmathringoverline A$. Then there is a $r>0$ such that $B_r(x)subsetoverline A$. Since $B_r(x)$ is an open set which is contained in $overline A$, it contains some element $ain A$. But then, if $r'=r-d(x,a)$, $B_r'(a)subset B_r(x)$. In particular, $B_r'(a)subsetoverline A$. This is impossible, since $A$ is relatively discrete.
I will not answer the question from the title, since you already got an answer in the comments.
answered 2 hours ago
José Carlos Santos
131k17106192
Let $xinmathringoverline A$. Then there is a $r>0$ such that $B_r(x)subsetoverline A$. Since $B_r(x)$ is an open set which is contained in $overline A$, it contains some element $ain A$. But then, if $r'=r-d(x,a)$, $B_r'(a)subset B_r(x)$. In particular, $B_r'(a)subsetoverline A$. This is impossible, since $A$ is relatively discrete.
I will not answer the question from the title, since you already got an answer in the comments.
answered 2 hours ago
José Carlos Santos
131k17106192
answered 2 hours ago
José Carlos Santos
131k17106192
answered 2 hours ago
José Carlos Santos
131k17106192
answered 2 hours ago
José Carlos Santos
131k17106192
131k17106192
add a comment |Â
add a comment |Â
up vote
1
down vote
With your $x$ and $epsilon$, $$overline Asetminus B_d(x,epsilon/2)$$ is a strictly smaller closed set than $overline A$, hence cannot contain all of $A$. Pick $ain Acap B_d(x,epsilon/2)$, by which we achieve that $$ain B_d(a,epsilon/2)subseteq Acapoperatornameint(overline A).$$
By relative discreteness, we find $delta>0$ such that $B_d(a,delta)cap A=a$.
Wlog $deltaleepsilon/2$. Now
$$ S:=(overline Asetminus B_d(a,delta))cup a$$
is a closed set with $Asubseteq Ssubseteq overline A$, hence $S=overline A$. We still have $B_d(a,delta)subseteq overline A=S$, but $B_d(a,delta)cap S=a$, which means that $B_d(a,delta)=a$, contrary to the assumption that there are no isolated points.
answered 2 hours ago
Hagen von Eitzen
271k21264489
add a comment |Â
up vote
1
down vote
With your $x$ and $epsilon$, $$overline Asetminus B_d(x,epsilon/2)$$ is a strictly smaller closed set than $overline A$, hence cannot contain all of $A$. Pick $ain Acap B_d(x,epsilon/2)$, by which we achieve that $$ain B_d(a,epsilon/2)subseteq Acapoperatornameint(overline A).$$
By relative discreteness, we find $delta>0$ such that $B_d(a,delta)cap A=a$.
Wlog $deltaleepsilon/2$. Now
$$ S:=(overline Asetminus B_d(a,delta))cup a$$
is a closed set with $Asubseteq Ssubseteq overline A$, hence $S=overline A$. We still have $B_d(a,delta)subseteq overline A=S$, but $B_d(a,delta)cap S=a$, which means that $B_d(a,delta)=a$, contrary to the assumption that there are no isolated points.
answered 2 hours ago
Hagen von Eitzen
271k21264489
add a comment |Â
up vote
1
down vote
up vote
1
down vote
With your $x$ and $epsilon$, $$overline Asetminus B_d(x,epsilon/2)$$ is a strictly smaller closed set than $overline A$, hence cannot contain all of $A$. Pick $ain Acap B_d(x,epsilon/2)$, by which we achieve that $$ain B_d(a,epsilon/2)subseteq Acapoperatornameint(overline A).$$
By relative discreteness, we find $delta>0$ such that $B_d(a,delta)cap A=a$.
Wlog $deltaleepsilon/2$. Now
$$ S:=(overline Asetminus B_d(a,delta))cup a$$
is a closed set with $Asubseteq Ssubseteq overline A$, hence $S=overline A$. We still have $B_d(a,delta)subseteq overline A=S$, but $B_d(a,delta)cap S=a$, which means that $B_d(a,delta)=a$, contrary to the assumption that there are no isolated points.
answered 2 hours ago
Hagen von Eitzen
271k21264489
With your $x$ and $epsilon$, $$overline Asetminus B_d(x,epsilon/2)$$ is a strictly smaller closed set than $overline A$, hence cannot contain all of $A$. Pick $ain Acap B_d(x,epsilon/2)$, by which we achieve that $$ain B_d(a,epsilon/2)subseteq Acapoperatornameint(overline A).$$
By relative discreteness, we find $delta>0$ such that $B_d(a,delta)cap A=a$.
Wlog $deltaleepsilon/2$. Now
$$ S:=(overline Asetminus B_d(a,delta))cup a$$
is a closed set with $Asubseteq Ssubseteq overline A$, hence $S=overline A$. We still have $B_d(a,delta)subseteq overline A=S$, but $B_d(a,delta)cap S=a$, which means that $B_d(a,delta)=a$, contrary to the assumption that there are no isolated points.
answered 2 hours ago
Hagen von Eitzen
271k21264489
answered 2 hours ago
Hagen von Eitzen
271k21264489
answered 2 hours ago
Hagen von Eitzen
271k21264489
answered 2 hours ago
Hagen von Eitzen
271k21264489
271k21264489
add a comment |Â
add a comment |Â
up vote
1
down vote
Suppose $X$ is a crowded $T_1$ space and $D$ is relatively discrete.
Suppose (for a contradiction) that there is some non-empty open set $U subseteq overlineD$
In particular, there is some $d in D cap U$ (being in the closure of $D$ means every neighbourhood intersects $D$) and as $D$ is relatively discrete, $d$ is open in $D$, so there is an open set $U_d$ of $X$ such that $U_d cap D = d$.
Now I claim that $U cap U_d = d$:
The right to left inclusion is clear, as both open sets contain $d$ and if $x neq d$ existed in $U cap U_d$, by $T_1$-ness of $X$ it follows that $U cap U_d cap (Xsetminusd)$ is an open set containing $x$ that misses $D$ entirely (clearly, as $U_d cap D = d$ and $(Xsetminus d) cap d = emptyset$) but $x in U subseteq overlineD$, so this cannot happen. This shows that indeed $U cap U_d = d$, making $d$ open, but this contradicts in turn that $X$ is crowded (has no isolated points)!
So no such $U$ can exist and $operatornameint(overlineA) = emptyset$.
answered 36 mins ago
Henno Brandsma
97.7k343105
add a comment |Â
up vote
1
down vote
Suppose $X$ is a crowded $T_1$ space and $D$ is relatively discrete.
Suppose (for a contradiction) that there is some non-empty open set $U subseteq overlineD$
In particular, there is some $d in D cap U$ (being in the closure of $D$ means every neighbourhood intersects $D$) and as $D$ is relatively discrete, $d$ is open in $D$, so there is an open set $U_d$ of $X$ such that $U_d cap D = d$.
Now I claim that $U cap U_d = d$:
The right to left inclusion is clear, as both open sets contain $d$ and if $x neq d$ existed in $U cap U_d$, by $T_1$-ness of $X$ it follows that $U cap U_d cap (Xsetminusd)$ is an open set containing $x$ that misses $D$ entirely (clearly, as $U_d cap D = d$ and $(Xsetminus d) cap d = emptyset$) but $x in U subseteq overlineD$, so this cannot happen. This shows that indeed $U cap U_d = d$, making $d$ open, but this contradicts in turn that $X$ is crowded (has no isolated points)!
So no such $U$ can exist and $operatornameint(overlineA) = emptyset$.
answered 36 mins ago
Henno Brandsma
97.7k343105
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Suppose $X$ is a crowded $T_1$ space and $D$ is relatively discrete.
Suppose (for a contradiction) that there is some non-empty open set $U subseteq overlineD$
In particular, there is some $d in D cap U$ (being in the closure of $D$ means every neighbourhood intersects $D$) and as $D$ is relatively discrete, $d$ is open in $D$, so there is an open set $U_d$ of $X$ such that $U_d cap D = d$.
Now I claim that $U cap U_d = d$:
The right to left inclusion is clear, as both open sets contain $d$ and if $x neq d$ existed in $U cap U_d$, by $T_1$-ness of $X$ it follows that $U cap U_d cap (Xsetminusd)$ is an open set containing $x$ that misses $D$ entirely (clearly, as $U_d cap D = d$ and $(Xsetminus d) cap d = emptyset$) but $x in U subseteq overlineD$, so this cannot happen. This shows that indeed $U cap U_d = d$, making $d$ open, but this contradicts in turn that $X$ is crowded (has no isolated points)!
So no such $U$ can exist and $operatornameint(overlineA) = emptyset$.
answered 36 mins ago
Henno Brandsma
97.7k343105
Suppose $X$ is a crowded $T_1$ space and $D$ is relatively discrete.
Suppose (for a contradiction) that there is some non-empty open set $U subseteq overlineD$
In particular, there is some $d in D cap U$ (being in the closure of $D$ means every neighbourhood intersects $D$) and as $D$ is relatively discrete, $d$ is open in $D$, so there is an open set $U_d$ of $X$ such that $U_d cap D = d$.
Now I claim that $U cap U_d = d$:
The right to left inclusion is clear, as both open sets contain $d$ and if $x neq d$ existed in $U cap U_d$, by $T_1$-ness of $X$ it follows that $U cap U_d cap (Xsetminusd)$ is an open set containing $x$ that misses $D$ entirely (clearly, as $U_d cap D = d$ and $(Xsetminus d) cap d = emptyset$) but $x in U subseteq overlineD$, so this cannot happen. This shows that indeed $U cap U_d = d$, making $d$ open, but this contradicts in turn that $X$ is crowded (has no isolated points)!
So no such $U$ can exist and $operatornameint(overlineA) = emptyset$.
answered 36 mins ago
Henno Brandsma
97.7k343105
answered 36 mins ago
Henno Brandsma
97.7k343105
answered 36 mins ago
Henno Brandsma
97.7k343105
answered 36 mins ago
Henno Brandsma
97.7k343105
97.7k343105
add a comment |Â
add a comment |Â
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1
If $X$ has indiscrete topology and $A$ is a singleton set, then $A$ is discrete, but $overline A=X$ is a nonempty open set
– Hagen von Eitzen
2 hours ago
Thank you for the counter example. How do i complete the proof? Can you please help me?
– Math geek
2 hours ago
@Mathgeek the result holds in $T_1$ spaces, I believe.
– Henno Brandsma
1 hour ago