Mixing
Prove that any open set of $mathbb{R}^n$ is a countable union of open $n$-rectangles
$begingroup$
I am stuck in proving two parts of this proof.
Let $U$ be the open set in question. Then, for all $x in U$, since $U$ is open in $mathbb{R}^n$, we can build an open ball $B(r,x)$ around $x$ with $r>0$, and construct an $n$-rectangle denoted $R_x = (a_1,b_1)times cdots times (a_n,b_n)$ contained inside $B(r,x)$ where $a_i, b_i in mathbb{Q}$ for all $i in [n]$ since $mathbb{Q}$ is dense in $mathbb{R}$.
One part I have problems with is how can I rigorously prove that
$$bigcup_{xin U} R_x = U$$
and that there is a countable number of rectangles using the fact that all my points $a_i,b_i$ are rational?
Thank you!
real-analysis general-topology analysis
asked Jan 25 at 8:50
The BoscoThe Bosco
613212
$endgroup$
|
show 1 more comment
$begingroup$
I am stuck in proving two parts of this proof.
Let $U$ be the open set in question. Then, for all $x in U$, since $U$ is open in $mathbb{R}^n$, we can build an open ball $B(r,x)$ around $x$ with $r>0$, and construct an $n$-rectangle denoted $R_x = (a_1,b_1)times cdots times (a_n,b_n)$ contained inside $B(r,x)$ where $a_i, b_i in mathbb{Q}$ for all $i in [n]$ since $mathbb{Q}$ is dense in $mathbb{R}$.
One part I have problems with is how can I rigorously prove that
$$bigcup_{xin U} R_x = U$$
and that there is a countable number of rectangles using the fact that all my points $a_i,b_i$ are rational?
Thank you!
real-analysis general-topology analysis
asked Jan 25 at 8:50
The BoscoThe Bosco
613212
$endgroup$
$begingroup$
For all $x$ $R_x subset U$, then $cup R_x subset U$. And there is at most countably many of $R_x$s
$endgroup$
– dEmigOd
Jan 25 at 8:54
$begingroup$
I already stated that in my question. My problem is how to prove it? Or is it evident with what I've got already?
$endgroup$
– The Bosco
Jan 25 at 9:00
$begingroup$
$forall x in U exists R_x : x in R_x Rightarrow forall x in U x in cup R_x Rightarrow U subset cup R_x$ - this gives the equality. Countability comes from knowing the cardinality of $mathbb{Q}^n$
$endgroup$
– dEmigOd
Jan 25 at 9:02
$begingroup$
but for example, in $mathbb{R}$ you could argue this because the subset $U$ is a disjoint countable union of intervals. Therefore, you could argue that each one of them has a unique rational number, but you can't do that in $mathbb{R}^n$ for $n>1$ since now the rectangles have to overlap (although not completely).
$endgroup$
– The Bosco
Jan 25 at 9:06
$begingroup$
nothing in the claim suggests boxes should be disjoint, why bother?
$endgroup$
– dEmigOd
Jan 25 at 9:14
|
show 1 more comment
$begingroup$
I am stuck in proving two parts of this proof.
Let $U$ be the open set in question. Then, for all $x in U$, since $U$ is open in $mathbb{R}^n$, we can build an open ball $B(r,x)$ around $x$ with $r>0$, and construct an $n$-rectangle denoted $R_x = (a_1,b_1)times cdots times (a_n,b_n)$ contained inside $B(r,x)$ where $a_i, b_i in mathbb{Q}$ for all $i in [n]$ since $mathbb{Q}$ is dense in $mathbb{R}$.
One part I have problems with is how can I rigorously prove that
$$bigcup_{xin U} R_x = U$$
and that there is a countable number of rectangles using the fact that all my points $a_i,b_i$ are rational?
Thank you!
real-analysis general-topology analysis
asked Jan 25 at 8:50
The BoscoThe Bosco
613212
$endgroup$
I am stuck in proving two parts of this proof.
Let $U$ be the open set in question. Then, for all $x in U$, since $U$ is open in $mathbb{R}^n$, we can build an open ball $B(r,x)$ around $x$ with $r>0$, and construct an $n$-rectangle denoted $R_x = (a_1,b_1)times cdots times (a_n,b_n)$ contained inside $B(r,x)$ where $a_i, b_i in mathbb{Q}$ for all $i in [n]$ since $mathbb{Q}$ is dense in $mathbb{R}$.
One part I have problems with is how can I rigorously prove that
$$bigcup_{xin U} R_x = U$$
and that there is a countable number of rectangles using the fact that all my points $a_i,b_i$ are rational?
Thank you!
real-analysis general-topology analysis
real-analysis general-topology analysis
asked Jan 25 at 8:50
The BoscoThe Bosco
613212
asked Jan 25 at 8:50
The BoscoThe Bosco
613212
asked Jan 25 at 8:50
The BoscoThe Bosco
613212
asked Jan 25 at 8:50
The BoscoThe Bosco
613212
asked Jan 25 at 8:50
The BoscoThe Bosco
613212
613212
$begingroup$
For all $x$ $R_x subset U$, then $cup R_x subset U$. And there is at most countably many of $R_x$s
$endgroup$
– dEmigOd
Jan 25 at 8:54
$begingroup$
I already stated that in my question. My problem is how to prove it? Or is it evident with what I've got already?
$endgroup$
– The Bosco
Jan 25 at 9:00
$begingroup$
$forall x in U exists R_x : x in R_x Rightarrow forall x in U x in cup R_x Rightarrow U subset cup R_x$ - this gives the equality. Countability comes from knowing the cardinality of $mathbb{Q}^n$
$endgroup$
– dEmigOd
Jan 25 at 9:02
$begingroup$
but for example, in $mathbb{R}$ you could argue this because the subset $U$ is a disjoint countable union of intervals. Therefore, you could argue that each one of them has a unique rational number, but you can't do that in $mathbb{R}^n$ for $n>1$ since now the rectangles have to overlap (although not completely).
$endgroup$
– The Bosco
Jan 25 at 9:06
$begingroup$
nothing in the claim suggests boxes should be disjoint, why bother?
$endgroup$
– dEmigOd
Jan 25 at 9:14
|
show 1 more comment
$begingroup$
For all $x$ $R_x subset U$, then $cup R_x subset U$. And there is at most countably many of $R_x$s
$endgroup$
– dEmigOd
Jan 25 at 8:54
$begingroup$
I already stated that in my question. My problem is how to prove it? Or is it evident with what I've got already?
$endgroup$
– The Bosco
Jan 25 at 9:00
$begingroup$
$forall x in U exists R_x : x in R_x Rightarrow forall x in U x in cup R_x Rightarrow U subset cup R_x$ - this gives the equality. Countability comes from knowing the cardinality of $mathbb{Q}^n$
$endgroup$
– dEmigOd
Jan 25 at 9:02
$begingroup$
but for example, in $mathbb{R}$ you could argue this because the subset $U$ is a disjoint countable union of intervals. Therefore, you could argue that each one of them has a unique rational number, but you can't do that in $mathbb{R}^n$ for $n>1$ since now the rectangles have to overlap (although not completely).
$endgroup$
– The Bosco
Jan 25 at 9:06
$begingroup$
nothing in the claim suggests boxes should be disjoint, why bother?
$endgroup$
– dEmigOd
Jan 25 at 9:14
$begingroup$
For all $x$ $R_x subset U$, then $cup R_x subset U$. And there is at most countably many of $R_x$s
$endgroup$
– dEmigOd
Jan 25 at 8:54
$begingroup$
For all $x$ $R_x subset U$, then $cup R_x subset U$. And there is at most countably many of $R_x$s
$endgroup$
– dEmigOd
Jan 25 at 8:54
$begingroup$
I already stated that in my question. My problem is how to prove it? Or is it evident with what I've got already?
$endgroup$
– The Bosco
Jan 25 at 9:00
$begingroup$
I already stated that in my question. My problem is how to prove it? Or is it evident with what I've got already?
$endgroup$
– The Bosco
Jan 25 at 9:00
$begingroup$
$forall x in U exists R_x : x in R_x Rightarrow forall x in U x in cup R_x Rightarrow U subset cup R_x$ - this gives the equality. Countability comes from knowing the cardinality of $mathbb{Q}^n$
$endgroup$
– dEmigOd
Jan 25 at 9:02
$begingroup$
$forall x in U exists R_x : x in R_x Rightarrow forall x in U x in cup R_x Rightarrow U subset cup R_x$ - this gives the equality. Countability comes from knowing the cardinality of $mathbb{Q}^n$
$endgroup$
– dEmigOd
Jan 25 at 9:02
$begingroup$
but for example, in $mathbb{R}$ you could argue this because the subset $U$ is a disjoint countable union of intervals. Therefore, you could argue that each one of them has a unique rational number, but you can't do that in $mathbb{R}^n$ for $n>1$ since now the rectangles have to overlap (although not completely).
$endgroup$
– The Bosco
Jan 25 at 9:06
$begingroup$
but for example, in $mathbb{R}$ you could argue this because the subset $U$ is a disjoint countable union of intervals. Therefore, you could argue that each one of them has a unique rational number, but you can't do that in $mathbb{R}^n$ for $n>1$ since now the rectangles have to overlap (although not completely).
$endgroup$
– The Bosco
Jan 25 at 9:06
$begingroup$
nothing in the claim suggests boxes should be disjoint, why bother?
$endgroup$
– dEmigOd
Jan 25 at 9:14
$begingroup$
nothing in the claim suggests boxes should be disjoint, why bother?
$endgroup$
– dEmigOd
Jan 25 at 9:14
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
As $R_xsubseteq U$ for all $x in U,$ we have
$bigcup_{xin U} R_x subseteq U.$ Now if $x$ is any point in $U,$ then $xin R_x subseteq bigcup_{xin U} R_x.$ Thus
$$
bigcup_{xin U} R_x = U.
$$
There is only a countable number of rectangles with coordinates in $mathbb{Q}^n$ as $mathbb{Q}^n$ is countable, as is normally shown using Cantor's diagonal argument.
answered Jan 25 at 9:13
Reiner MartinReiner Martin
3,509414
$endgroup$
add a comment |
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1 Answer
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$begingroup$
As $R_xsubseteq U$ for all $x in U,$ we have
$bigcup_{xin U} R_x subseteq U.$ Now if $x$ is any point in $U,$ then $xin R_x subseteq bigcup_{xin U} R_x.$ Thus
$$
bigcup_{xin U} R_x = U.
$$
There is only a countable number of rectangles with coordinates in $mathbb{Q}^n$ as $mathbb{Q}^n$ is countable, as is normally shown using Cantor's diagonal argument.
answered Jan 25 at 9:13
Reiner MartinReiner Martin
3,509414
$endgroup$
add a comment |
$begingroup$
As $R_xsubseteq U$ for all $x in U,$ we have
$bigcup_{xin U} R_x subseteq U.$ Now if $x$ is any point in $U,$ then $xin R_x subseteq bigcup_{xin U} R_x.$ Thus
$$
bigcup_{xin U} R_x = U.
$$
There is only a countable number of rectangles with coordinates in $mathbb{Q}^n$ as $mathbb{Q}^n$ is countable, as is normally shown using Cantor's diagonal argument.
answered Jan 25 at 9:13
Reiner MartinReiner Martin
3,509414
$endgroup$
add a comment |
$begingroup$
As $R_xsubseteq U$ for all $x in U,$ we have
$bigcup_{xin U} R_x subseteq U.$ Now if $x$ is any point in $U,$ then $xin R_x subseteq bigcup_{xin U} R_x.$ Thus
$$
bigcup_{xin U} R_x = U.
$$
There is only a countable number of rectangles with coordinates in $mathbb{Q}^n$ as $mathbb{Q}^n$ is countable, as is normally shown using Cantor's diagonal argument.
answered Jan 25 at 9:13
Reiner MartinReiner Martin
3,509414
$endgroup$
As $R_xsubseteq U$ for all $x in U,$ we have
$bigcup_{xin U} R_x subseteq U.$ Now if $x$ is any point in $U,$ then $xin R_x subseteq bigcup_{xin U} R_x.$ Thus
$$
bigcup_{xin U} R_x = U.
$$
There is only a countable number of rectangles with coordinates in $mathbb{Q}^n$ as $mathbb{Q}^n$ is countable, as is normally shown using Cantor's diagonal argument.
answered Jan 25 at 9:13
Reiner MartinReiner Martin
3,509414
answered Jan 25 at 9:13
Reiner MartinReiner Martin
3,509414
answered Jan 25 at 9:13
Reiner MartinReiner Martin
3,509414
answered Jan 25 at 9:13
Reiner MartinReiner Martin
3,509414
3,509414
add a comment |
add a comment |
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$begingroup$
For all $x$ $R_x subset U$, then $cup R_x subset U$. And there is at most countably many of $R_x$s
$endgroup$
– dEmigOd
Jan 25 at 8:54
$begingroup$
I already stated that in my question. My problem is how to prove it? Or is it evident with what I've got already?
$endgroup$
– The Bosco
Jan 25 at 9:00
$begingroup$
$forall x in U exists R_x : x in R_x Rightarrow forall x in U x in cup R_x Rightarrow U subset cup R_x$ - this gives the equality. Countability comes from knowing the cardinality of $mathbb{Q}^n$
$endgroup$
– dEmigOd
Jan 25 at 9:02
$begingroup$
but for example, in $mathbb{R}$ you could argue this because the subset $U$ is a disjoint countable union of intervals. Therefore, you could argue that each one of them has a unique rational number, but you can't do that in $mathbb{R}^n$ for $n>1$ since now the rectangles have to overlap (although not completely).
$endgroup$
– The Bosco
Jan 25 at 9:06
$begingroup$
nothing in the claim suggests boxes should be disjoint, why bother?
$endgroup$
– dEmigOd
Jan 25 at 9:14