Does exist a compact connected $K' subset U$ such that $K subset K'$, if $U$ is an open connected and $K$ in...
$begingroup$
Let $U$ be an open and connected set in $mathbb{R}^n$.
Suppose $K subset U$ is a compact set. Is it true that there exists a compact and connected set $K' subset U$ such that $K subset K'$?
I know that there exists a compact set $K' subset U$ such that $K subset K'$.
But how I can guarantee that $K'$ is connected?
real-analysis general-topology connectedness
asked Dec 5 '18 at 10:55
SantosSantos
626514
$endgroup$
add a comment |
$begingroup$
Let $U$ be an open and connected set in $mathbb{R}^n$.
Suppose $K subset U$ is a compact set. Is it true that there exists a compact and connected set $K' subset U$ such that $K subset K'$?
I know that there exists a compact set $K' subset U$ such that $K subset K'$.
But how I can guarantee that $K'$ is connected?
real-analysis general-topology connectedness
asked Dec 5 '18 at 10:55
SantosSantos
626514
$endgroup$
add a comment |
$begingroup$
Let $U$ be an open and connected set in $mathbb{R}^n$.
Suppose $K subset U$ is a compact set. Is it true that there exists a compact and connected set $K' subset U$ such that $K subset K'$?
I know that there exists a compact set $K' subset U$ such that $K subset K'$.
But how I can guarantee that $K'$ is connected?
real-analysis general-topology connectedness
asked Dec 5 '18 at 10:55
SantosSantos
626514
$endgroup$
Let $U$ be an open and connected set in $mathbb{R}^n$.
Suppose $K subset U$ is a compact set. Is it true that there exists a compact and connected set $K' subset U$ such that $K subset K'$?
I know that there exists a compact set $K' subset U$ such that $K subset K'$.
But how I can guarantee that $K'$ is connected?
real-analysis general-topology connectedness
real-analysis general-topology connectedness
asked Dec 5 '18 at 10:55
SantosSantos
626514
asked Dec 5 '18 at 10:55
SantosSantos
626514
edited Dec 5 '18 at 11:01
Santos
asked Dec 5 '18 at 10:55
SantosSantos
626514
asked Dec 5 '18 at 10:55
SantosSantos
626514
asked Dec 5 '18 at 10:55
SantosSantos
626514
626514
add a comment |
add a comment |
2 Answers
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Since $K$ is compact, there exists $epsilon > 0$ such that $bigcup_{x in K} B(x,epsilon) subset U$.
Now, again since $K$ is compact, there exists a finite set of points $x_{1},ldots,x_{n} in K$ such that $K subset bigcup_{i} B(x_{i},epsilon/2) subset bigcup_{i} overline{B(x_{i},epsilon/2)}$.
$U$ is path connected, hence for each $i < j$ there exist a path $gamma_{i,j} : [0,1] rightarrow U$ such that $gamma(0) = x_{i}$ and $gamma(1) = x_{j}$.
Then
$K' = bigcup_{i} left( overline{B(x_{i},epsilon/2)} cup bigcup_{j > i} gamma_{i,j}([0,1]) right) subset U$
is a finite union of compact sets, hence compact, path connected by construction, hence connected, and contains $K$.
answered Dec 5 '18 at 11:32
David MorrisDavid Morris
661
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For any $xin K$ there is $epsilon >0$ such that $B(x,epsilon)subset U$. Let $A_x=B(x,epsilon/2)$. Then the closure of $A_x$ is a compact connected set contained in $U$.
Since ${A_x, xin K}$ is an open conver of $K$, by compactness it has a finite sub-cover, say $A_1,dots,A_n$. Choose $x_0in U$ and connect each $A_i$ to $x_0$ with a path. This is possible because open connectet subset of $mathbb R^n$ are path-connected.
Now set $B_i$ to be the union of $overline{A_i}$ and the path from $A_i$ to $x_0$. It is a connected set because union of (two) connected sets witn nonempty intersection. It is compact because it is a union of two compacts.
Now let $K'$ be the union of the $B_i$'s. It is connected because it is union of connected set with non-empty intersection (they intersect at least at $x_0$) and it is compact because it is a finite union of compact sets.
By construction $K'subset U$.
answered Dec 5 '18 at 11:38
user126154user126154
5,378716
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Since $K$ is compact, there exists $epsilon > 0$ such that $bigcup_{x in K} B(x,epsilon) subset U$.
Now, again since $K$ is compact, there exists a finite set of points $x_{1},ldots,x_{n} in K$ such that $K subset bigcup_{i} B(x_{i},epsilon/2) subset bigcup_{i} overline{B(x_{i},epsilon/2)}$.
$U$ is path connected, hence for each $i < j$ there exist a path $gamma_{i,j} : [0,1] rightarrow U$ such that $gamma(0) = x_{i}$ and $gamma(1) = x_{j}$.
Then
$K' = bigcup_{i} left( overline{B(x_{i},epsilon/2)} cup bigcup_{j > i} gamma_{i,j}([0,1]) right) subset U$
is a finite union of compact sets, hence compact, path connected by construction, hence connected, and contains $K$.
answered Dec 5 '18 at 11:32
David MorrisDavid Morris
661
$endgroup$
add a comment |
$begingroup$
Since $K$ is compact, there exists $epsilon > 0$ such that $bigcup_{x in K} B(x,epsilon) subset U$.
Now, again since $K$ is compact, there exists a finite set of points $x_{1},ldots,x_{n} in K$ such that $K subset bigcup_{i} B(x_{i},epsilon/2) subset bigcup_{i} overline{B(x_{i},epsilon/2)}$.
$U$ is path connected, hence for each $i < j$ there exist a path $gamma_{i,j} : [0,1] rightarrow U$ such that $gamma(0) = x_{i}$ and $gamma(1) = x_{j}$.
Then
$K' = bigcup_{i} left( overline{B(x_{i},epsilon/2)} cup bigcup_{j > i} gamma_{i,j}([0,1]) right) subset U$
is a finite union of compact sets, hence compact, path connected by construction, hence connected, and contains $K$.
answered Dec 5 '18 at 11:32
David MorrisDavid Morris
661
$endgroup$
add a comment |
$begingroup$
Since $K$ is compact, there exists $epsilon > 0$ such that $bigcup_{x in K} B(x,epsilon) subset U$.
Now, again since $K$ is compact, there exists a finite set of points $x_{1},ldots,x_{n} in K$ such that $K subset bigcup_{i} B(x_{i},epsilon/2) subset bigcup_{i} overline{B(x_{i},epsilon/2)}$.
$U$ is path connected, hence for each $i < j$ there exist a path $gamma_{i,j} : [0,1] rightarrow U$ such that $gamma(0) = x_{i}$ and $gamma(1) = x_{j}$.
Then
$K' = bigcup_{i} left( overline{B(x_{i},epsilon/2)} cup bigcup_{j > i} gamma_{i,j}([0,1]) right) subset U$
is a finite union of compact sets, hence compact, path connected by construction, hence connected, and contains $K$.
answered Dec 5 '18 at 11:32
David MorrisDavid Morris
661
$endgroup$
Since $K$ is compact, there exists $epsilon > 0$ such that $bigcup_{x in K} B(x,epsilon) subset U$.
Now, again since $K$ is compact, there exists a finite set of points $x_{1},ldots,x_{n} in K$ such that $K subset bigcup_{i} B(x_{i},epsilon/2) subset bigcup_{i} overline{B(x_{i},epsilon/2)}$.
$U$ is path connected, hence for each $i < j$ there exist a path $gamma_{i,j} : [0,1] rightarrow U$ such that $gamma(0) = x_{i}$ and $gamma(1) = x_{j}$.
Then
$K' = bigcup_{i} left( overline{B(x_{i},epsilon/2)} cup bigcup_{j > i} gamma_{i,j}([0,1]) right) subset U$
is a finite union of compact sets, hence compact, path connected by construction, hence connected, and contains $K$.
answered Dec 5 '18 at 11:32
David MorrisDavid Morris
661
answered Dec 5 '18 at 11:32
David MorrisDavid Morris
661
answered Dec 5 '18 at 11:32
David MorrisDavid Morris
661
answered Dec 5 '18 at 11:32
David MorrisDavid Morris
661
661
add a comment |
add a comment |
$begingroup$
For any $xin K$ there is $epsilon >0$ such that $B(x,epsilon)subset U$. Let $A_x=B(x,epsilon/2)$. Then the closure of $A_x$ is a compact connected set contained in $U$.
Since ${A_x, xin K}$ is an open conver of $K$, by compactness it has a finite sub-cover, say $A_1,dots,A_n$. Choose $x_0in U$ and connect each $A_i$ to $x_0$ with a path. This is possible because open connectet subset of $mathbb R^n$ are path-connected.
Now set $B_i$ to be the union of $overline{A_i}$ and the path from $A_i$ to $x_0$. It is a connected set because union of (two) connected sets witn nonempty intersection. It is compact because it is a union of two compacts.
Now let $K'$ be the union of the $B_i$'s. It is connected because it is union of connected set with non-empty intersection (they intersect at least at $x_0$) and it is compact because it is a finite union of compact sets.
By construction $K'subset U$.
answered Dec 5 '18 at 11:38
user126154user126154
5,378716
$endgroup$
add a comment |
$begingroup$
For any $xin K$ there is $epsilon >0$ such that $B(x,epsilon)subset U$. Let $A_x=B(x,epsilon/2)$. Then the closure of $A_x$ is a compact connected set contained in $U$.
Since ${A_x, xin K}$ is an open conver of $K$, by compactness it has a finite sub-cover, say $A_1,dots,A_n$. Choose $x_0in U$ and connect each $A_i$ to $x_0$ with a path. This is possible because open connectet subset of $mathbb R^n$ are path-connected.
Now set $B_i$ to be the union of $overline{A_i}$ and the path from $A_i$ to $x_0$. It is a connected set because union of (two) connected sets witn nonempty intersection. It is compact because it is a union of two compacts.
Now let $K'$ be the union of the $B_i$'s. It is connected because it is union of connected set with non-empty intersection (they intersect at least at $x_0$) and it is compact because it is a finite union of compact sets.
By construction $K'subset U$.
answered Dec 5 '18 at 11:38
user126154user126154
5,378716
$endgroup$
add a comment |
$begingroup$
For any $xin K$ there is $epsilon >0$ such that $B(x,epsilon)subset U$. Let $A_x=B(x,epsilon/2)$. Then the closure of $A_x$ is a compact connected set contained in $U$.
Since ${A_x, xin K}$ is an open conver of $K$, by compactness it has a finite sub-cover, say $A_1,dots,A_n$. Choose $x_0in U$ and connect each $A_i$ to $x_0$ with a path. This is possible because open connectet subset of $mathbb R^n$ are path-connected.
Now set $B_i$ to be the union of $overline{A_i}$ and the path from $A_i$ to $x_0$. It is a connected set because union of (two) connected sets witn nonempty intersection. It is compact because it is a union of two compacts.
Now let $K'$ be the union of the $B_i$'s. It is connected because it is union of connected set with non-empty intersection (they intersect at least at $x_0$) and it is compact because it is a finite union of compact sets.
By construction $K'subset U$.
answered Dec 5 '18 at 11:38
user126154user126154
5,378716
$endgroup$
For any $xin K$ there is $epsilon >0$ such that $B(x,epsilon)subset U$. Let $A_x=B(x,epsilon/2)$. Then the closure of $A_x$ is a compact connected set contained in $U$.
Since ${A_x, xin K}$ is an open conver of $K$, by compactness it has a finite sub-cover, say $A_1,dots,A_n$. Choose $x_0in U$ and connect each $A_i$ to $x_0$ with a path. This is possible because open connectet subset of $mathbb R^n$ are path-connected.
Now set $B_i$ to be the union of $overline{A_i}$ and the path from $A_i$ to $x_0$. It is a connected set because union of (two) connected sets witn nonempty intersection. It is compact because it is a union of two compacts.
Now let $K'$ be the union of the $B_i$'s. It is connected because it is union of connected set with non-empty intersection (they intersect at least at $x_0$) and it is compact because it is a finite union of compact sets.
By construction $K'subset U$.
answered Dec 5 '18 at 11:38
user126154user126154
5,378716
answered Dec 5 '18 at 11:38
user126154user126154
5,378716
answered Dec 5 '18 at 11:38
user126154user126154
5,378716
answered Dec 5 '18 at 11:38
user126154user126154
5,378716
5,378716
add a comment |
add a comment |
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