Mixing
Intersection of a closed and a dense subset of a Compact Hausdorff space be empty?
Suppose that $X$ is a compact Hausdorff space. Let ${X_i}_{i in mathbb{N}}$ be a sequence of dense open subsets of $X$. Let $E$ be a closed subset of $X$. Can it happen that $left(cap_{i in mathbb{N}}X_iright)$ has empty intersection with $E$?
First by Baire Category theorem, $left(cap_{i in mathbb{N}}X_iright)$ is dense in $X$. Moreover, since $E$ is closed, it is compact. If the intersection were empty, then $E subset cup_{i in mathbb{N}}Xsetminus X_i$. But, this doesn't yield any contradiction.
A hint would be greatly appreciated.
Thanks for the help!!
real-analysis general-topology analysis compactness
asked Nov 21 '18 at 18:28
tattwamasi amrutamtattwamasi amrutam
8,19321643
add a comment |
Suppose that $X$ is a compact Hausdorff space. Let ${X_i}_{i in mathbb{N}}$ be a sequence of dense open subsets of $X$. Let $E$ be a closed subset of $X$. Can it happen that $left(cap_{i in mathbb{N}}X_iright)$ has empty intersection with $E$?
First by Baire Category theorem, $left(cap_{i in mathbb{N}}X_iright)$ is dense in $X$. Moreover, since $E$ is closed, it is compact. If the intersection were empty, then $E subset cup_{i in mathbb{N}}Xsetminus X_i$. But, this doesn't yield any contradiction.
A hint would be greatly appreciated.
Thanks for the help!!
real-analysis general-topology analysis compactness
asked Nov 21 '18 at 18:28
tattwamasi amrutamtattwamasi amrutam
8,19321643
add a comment |
Suppose that $X$ is a compact Hausdorff space. Let ${X_i}_{i in mathbb{N}}$ be a sequence of dense open subsets of $X$. Let $E$ be a closed subset of $X$. Can it happen that $left(cap_{i in mathbb{N}}X_iright)$ has empty intersection with $E$?
First by Baire Category theorem, $left(cap_{i in mathbb{N}}X_iright)$ is dense in $X$. Moreover, since $E$ is closed, it is compact. If the intersection were empty, then $E subset cup_{i in mathbb{N}}Xsetminus X_i$. But, this doesn't yield any contradiction.
A hint would be greatly appreciated.
Thanks for the help!!
real-analysis general-topology analysis compactness
asked Nov 21 '18 at 18:28
tattwamasi amrutamtattwamasi amrutam
8,19321643
Suppose that $X$ is a compact Hausdorff space. Let ${X_i}_{i in mathbb{N}}$ be a sequence of dense open subsets of $X$. Let $E$ be a closed subset of $X$. Can it happen that $left(cap_{i in mathbb{N}}X_iright)$ has empty intersection with $E$?
First by Baire Category theorem, $left(cap_{i in mathbb{N}}X_iright)$ is dense in $X$. Moreover, since $E$ is closed, it is compact. If the intersection were empty, then $E subset cup_{i in mathbb{N}}Xsetminus X_i$. But, this doesn't yield any contradiction.
A hint would be greatly appreciated.
Thanks for the help!!
real-analysis general-topology analysis compactness
real-analysis general-topology analysis compactness
asked Nov 21 '18 at 18:28
tattwamasi amrutamtattwamasi amrutam
8,19321643
asked Nov 21 '18 at 18:28
tattwamasi amrutamtattwamasi amrutam
8,19321643
asked Nov 21 '18 at 18:28
tattwamasi amrutamtattwamasi amrutam
8,19321643
asked Nov 21 '18 at 18:28
tattwamasi amrutamtattwamasi amrutam
8,19321643
asked Nov 21 '18 at 18:28
tattwamasi amrutamtattwamasi amrutam
8,19321643
8,19321643
add a comment |
add a comment |
1 Answer
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Yes, the intersection can be empty. Suppose that $X=[0,1]$, with its usual topology. Take $E={0}$ and $X_n=(0,1]$, for each natural $n$.
answered Nov 21 '18 at 18:32
José Carlos SantosJosé Carlos Santos
152k22123226
You can also take a Cantor set $C$ and $X_n = [0,1] setminus C$. Then $C$ is closed and uncountable! Since $C$ is nowhere dense, $X_n$ is dense.
– p4sch
Nov 21 '18 at 20:05
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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active
oldest
votes
active
oldest
votes
Yes, the intersection can be empty. Suppose that $X=[0,1]$, with its usual topology. Take $E={0}$ and $X_n=(0,1]$, for each natural $n$.
answered Nov 21 '18 at 18:32
José Carlos SantosJosé Carlos Santos
152k22123226
You can also take a Cantor set $C$ and $X_n = [0,1] setminus C$. Then $C$ is closed and uncountable! Since $C$ is nowhere dense, $X_n$ is dense.
– p4sch
Nov 21 '18 at 20:05
add a comment |
Yes, the intersection can be empty. Suppose that $X=[0,1]$, with its usual topology. Take $E={0}$ and $X_n=(0,1]$, for each natural $n$.
answered Nov 21 '18 at 18:32
José Carlos SantosJosé Carlos Santos
152k22123226
You can also take a Cantor set $C$ and $X_n = [0,1] setminus C$. Then $C$ is closed and uncountable! Since $C$ is nowhere dense, $X_n$ is dense.
– p4sch
Nov 21 '18 at 20:05
add a comment |
Yes, the intersection can be empty. Suppose that $X=[0,1]$, with its usual topology. Take $E={0}$ and $X_n=(0,1]$, for each natural $n$.
answered Nov 21 '18 at 18:32
José Carlos SantosJosé Carlos Santos
152k22123226
Yes, the intersection can be empty. Suppose that $X=[0,1]$, with its usual topology. Take $E={0}$ and $X_n=(0,1]$, for each natural $n$.
answered Nov 21 '18 at 18:32
José Carlos SantosJosé Carlos Santos
152k22123226
answered Nov 21 '18 at 18:32
José Carlos SantosJosé Carlos Santos
152k22123226
answered Nov 21 '18 at 18:32
José Carlos SantosJosé Carlos Santos
152k22123226
answered Nov 21 '18 at 18:32
José Carlos SantosJosé Carlos Santos
152k22123226
152k22123226
You can also take a Cantor set $C$ and $X_n = [0,1] setminus C$. Then $C$ is closed and uncountable! Since $C$ is nowhere dense, $X_n$ is dense.
– p4sch
Nov 21 '18 at 20:05
add a comment |
You can also take a Cantor set $C$ and $X_n = [0,1] setminus C$. Then $C$ is closed and uncountable! Since $C$ is nowhere dense, $X_n$ is dense.
– p4sch
Nov 21 '18 at 20:05
You can also take a Cantor set $C$ and $X_n = [0,1] setminus C$. Then $C$ is closed and uncountable! Since $C$ is nowhere dense, $X_n$ is dense.
– p4sch
Nov 21 '18 at 20:05
You can also take a Cantor set $C$ and $X_n = [0,1] setminus C$. Then $C$ is closed and uncountable! Since $C$ is nowhere dense, $X_n$ is dense.
– p4sch
Nov 21 '18 at 20:05
add a comment |
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