In $mathbb{R}^n$, every open set is a union of closed sets
$begingroup$
I read this statement when reading a proof about "continuous function with compact support is dense in $L^p$". For simplicity, I consider the case $n=1$, but how do I construct a closed covering ${U}_{i=1}^infty$ that make the a open interval $U=cup_{i=1}^infty U_i$. The boundary of the open set can not be covered by closed set and still in $U$, isn't it? For example, the interval $(0,1)$, I can't cover its boundary by closed sets and make the closed sets in it, or should I consider closed sets like $[frac{1}{n},1-frac{1}{n}]$?
Another question is that, I also read another statement:
every open set in $mathbb{R}$ is a countable union of disjoint half-open intervals.
How do I construct this?
real-analysis general-topology analysis
asked Jan 5 at 18:59
ReinherdReinherd
764
$endgroup$
add a comment |
$begingroup$
I read this statement when reading a proof about "continuous function with compact support is dense in $L^p$". For simplicity, I consider the case $n=1$, but how do I construct a closed covering ${U}_{i=1}^infty$ that make the a open interval $U=cup_{i=1}^infty U_i$. The boundary of the open set can not be covered by closed set and still in $U$, isn't it? For example, the interval $(0,1)$, I can't cover its boundary by closed sets and make the closed sets in it, or should I consider closed sets like $[frac{1}{n},1-frac{1}{n}]$?
Another question is that, I also read another statement:
every open set in $mathbb{R}$ is a countable union of disjoint half-open intervals.
How do I construct this?
real-analysis general-topology analysis
asked Jan 5 at 18:59
ReinherdReinherd
764
$endgroup$
2
$begingroup$
The answer to the question in your title is this: every set is a union of singletons, which are closed in $Bbb R^n.$ The answer to your second question is: Given any $a,binBbb R$ with $a<b,$ show that $$(a,b)=bigcup_{n=1}^inftyleft(b-frac{b-a}{2^{n-1}},b-frac{b-a}{2^n}right].$$ Something similar can be done for sets of the form $(-infty, b).$ Readily, $Bbb R$ and sets of the form $(a,infty)$ can be shown to be such a union, as well. (cont'd)
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– Cameron Buie
Jan 5 at 19:18
$begingroup$
Since every non-empty, connected open set is a countably-infinite disjoint union of such half-open intervals, and since any open subset of $Bbb R$ is the union of at most countably-infinitely-many disjoint, connected open sets, then the second claim follows. Unfortunately, I'm not sure I understand the rest of the body of your question.
$endgroup$
– Cameron Buie
Jan 5 at 19:19
add a comment |
$begingroup$
I read this statement when reading a proof about "continuous function with compact support is dense in $L^p$". For simplicity, I consider the case $n=1$, but how do I construct a closed covering ${U}_{i=1}^infty$ that make the a open interval $U=cup_{i=1}^infty U_i$. The boundary of the open set can not be covered by closed set and still in $U$, isn't it? For example, the interval $(0,1)$, I can't cover its boundary by closed sets and make the closed sets in it, or should I consider closed sets like $[frac{1}{n},1-frac{1}{n}]$?
Another question is that, I also read another statement:
every open set in $mathbb{R}$ is a countable union of disjoint half-open intervals.
How do I construct this?
real-analysis general-topology analysis
asked Jan 5 at 18:59
ReinherdReinherd
764
$endgroup$
I read this statement when reading a proof about "continuous function with compact support is dense in $L^p$". For simplicity, I consider the case $n=1$, but how do I construct a closed covering ${U}_{i=1}^infty$ that make the a open interval $U=cup_{i=1}^infty U_i$. The boundary of the open set can not be covered by closed set and still in $U$, isn't it? For example, the interval $(0,1)$, I can't cover its boundary by closed sets and make the closed sets in it, or should I consider closed sets like $[frac{1}{n},1-frac{1}{n}]$?
Another question is that, I also read another statement:
every open set in $mathbb{R}$ is a countable union of disjoint half-open intervals.
How do I construct this?
real-analysis general-topology analysis
real-analysis general-topology analysis
asked Jan 5 at 18:59
ReinherdReinherd
764
asked Jan 5 at 18:59
ReinherdReinherd
764
asked Jan 5 at 18:59
ReinherdReinherd
764
asked Jan 5 at 18:59
ReinherdReinherd
764
asked Jan 5 at 18:59
ReinherdReinherd
764
764
2
$begingroup$
The answer to the question in your title is this: every set is a union of singletons, which are closed in $Bbb R^n.$ The answer to your second question is: Given any $a,binBbb R$ with $a<b,$ show that $$(a,b)=bigcup_{n=1}^inftyleft(b-frac{b-a}{2^{n-1}},b-frac{b-a}{2^n}right].$$ Something similar can be done for sets of the form $(-infty, b).$ Readily, $Bbb R$ and sets of the form $(a,infty)$ can be shown to be such a union, as well. (cont'd)
$endgroup$
– Cameron Buie
Jan 5 at 19:18
$begingroup$
Since every non-empty, connected open set is a countably-infinite disjoint union of such half-open intervals, and since any open subset of $Bbb R$ is the union of at most countably-infinitely-many disjoint, connected open sets, then the second claim follows. Unfortunately, I'm not sure I understand the rest of the body of your question.
$endgroup$
– Cameron Buie
Jan 5 at 19:19
add a comment |
2
$begingroup$
The answer to the question in your title is this: every set is a union of singletons, which are closed in $Bbb R^n.$ The answer to your second question is: Given any $a,binBbb R$ with $a<b,$ show that $$(a,b)=bigcup_{n=1}^inftyleft(b-frac{b-a}{2^{n-1}},b-frac{b-a}{2^n}right].$$ Something similar can be done for sets of the form $(-infty, b).$ Readily, $Bbb R$ and sets of the form $(a,infty)$ can be shown to be such a union, as well. (cont'd)
$endgroup$
– Cameron Buie
Jan 5 at 19:18
$begingroup$
Since every non-empty, connected open set is a countably-infinite disjoint union of such half-open intervals, and since any open subset of $Bbb R$ is the union of at most countably-infinitely-many disjoint, connected open sets, then the second claim follows. Unfortunately, I'm not sure I understand the rest of the body of your question.
$endgroup$
– Cameron Buie
Jan 5 at 19:19
2
2
$begingroup$
The answer to the question in your title is this: every set is a union of singletons, which are closed in $Bbb R^n.$ The answer to your second question is: Given any $a,binBbb R$ with $a<b,$ show that $$(a,b)=bigcup_{n=1}^inftyleft(b-frac{b-a}{2^{n-1}},b-frac{b-a}{2^n}right].$$ Something similar can be done for sets of the form $(-infty, b).$ Readily, $Bbb R$ and sets of the form $(a,infty)$ can be shown to be such a union, as well. (cont'd)
$endgroup$
– Cameron Buie
Jan 5 at 19:18
$begingroup$
The answer to the question in your title is this: every set is a union of singletons, which are closed in $Bbb R^n.$ The answer to your second question is: Given any $a,binBbb R$ with $a<b,$ show that $$(a,b)=bigcup_{n=1}^inftyleft(b-frac{b-a}{2^{n-1}},b-frac{b-a}{2^n}right].$$ Something similar can be done for sets of the form $(-infty, b).$ Readily, $Bbb R$ and sets of the form $(a,infty)$ can be shown to be such a union, as well. (cont'd)
$endgroup$
– Cameron Buie
Jan 5 at 19:18
$begingroup$
Since every non-empty, connected open set is a countably-infinite disjoint union of such half-open intervals, and since any open subset of $Bbb R$ is the union of at most countably-infinitely-many disjoint, connected open sets, then the second claim follows. Unfortunately, I'm not sure I understand the rest of the body of your question.
$endgroup$
– Cameron Buie
Jan 5 at 19:19
$begingroup$
Since every non-empty, connected open set is a countably-infinite disjoint union of such half-open intervals, and since any open subset of $Bbb R$ is the union of at most countably-infinitely-many disjoint, connected open sets, then the second claim follows. Unfortunately, I'm not sure I understand the rest of the body of your question.
$endgroup$
– Cameron Buie
Jan 5 at 19:19
add a comment |
1 Answer
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In $mathbb{R}^n$, if a set $U$ is open then for every $xin U$ you can find an open ball $B_delta(x)$ of appropriately small radius $delta$ centered on $x$ entirely contained inside $U$. Thus, the closure of the ball $B_{delta/2}(x)$ is also contained inside $U$. If you take a union over all of these balls, then you recover $U$.
If you wish to only consider countable unions of closed sets, then we can do something similar but only use balls centered around "completely rational points" with rational radii. The explicit construction is harder to write down, but the idea is the same.
answered Jan 5 at 19:18
ItsJustABallBroItsJustABallBro
112
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$begingroup$
In $mathbb{R}^n$, if a set $U$ is open then for every $xin U$ you can find an open ball $B_delta(x)$ of appropriately small radius $delta$ centered on $x$ entirely contained inside $U$. Thus, the closure of the ball $B_{delta/2}(x)$ is also contained inside $U$. If you take a union over all of these balls, then you recover $U$.
If you wish to only consider countable unions of closed sets, then we can do something similar but only use balls centered around "completely rational points" with rational radii. The explicit construction is harder to write down, but the idea is the same.
answered Jan 5 at 19:18
ItsJustABallBroItsJustABallBro
112
$endgroup$
add a comment |
$begingroup$
In $mathbb{R}^n$, if a set $U$ is open then for every $xin U$ you can find an open ball $B_delta(x)$ of appropriately small radius $delta$ centered on $x$ entirely contained inside $U$. Thus, the closure of the ball $B_{delta/2}(x)$ is also contained inside $U$. If you take a union over all of these balls, then you recover $U$.
If you wish to only consider countable unions of closed sets, then we can do something similar but only use balls centered around "completely rational points" with rational radii. The explicit construction is harder to write down, but the idea is the same.
answered Jan 5 at 19:18
ItsJustABallBroItsJustABallBro
112
$endgroup$
add a comment |
$begingroup$
In $mathbb{R}^n$, if a set $U$ is open then for every $xin U$ you can find an open ball $B_delta(x)$ of appropriately small radius $delta$ centered on $x$ entirely contained inside $U$. Thus, the closure of the ball $B_{delta/2}(x)$ is also contained inside $U$. If you take a union over all of these balls, then you recover $U$.
If you wish to only consider countable unions of closed sets, then we can do something similar but only use balls centered around "completely rational points" with rational radii. The explicit construction is harder to write down, but the idea is the same.
answered Jan 5 at 19:18
ItsJustABallBroItsJustABallBro
112
$endgroup$
In $mathbb{R}^n$, if a set $U$ is open then for every $xin U$ you can find an open ball $B_delta(x)$ of appropriately small radius $delta$ centered on $x$ entirely contained inside $U$. Thus, the closure of the ball $B_{delta/2}(x)$ is also contained inside $U$. If you take a union over all of these balls, then you recover $U$.
If you wish to only consider countable unions of closed sets, then we can do something similar but only use balls centered around "completely rational points" with rational radii. The explicit construction is harder to write down, but the idea is the same.
answered Jan 5 at 19:18
ItsJustABallBroItsJustABallBro
112
edited Jan 5 at 19:25
answered Jan 5 at 19:18
ItsJustABallBroItsJustABallBro
112
answered Jan 5 at 19:18
ItsJustABallBroItsJustABallBro
112
answered Jan 5 at 19:18
ItsJustABallBroItsJustABallBro
112
112
add a comment |
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$begingroup$
The answer to the question in your title is this: every set is a union of singletons, which are closed in $Bbb R^n.$ The answer to your second question is: Given any $a,binBbb R$ with $a<b,$ show that $$(a,b)=bigcup_{n=1}^inftyleft(b-frac{b-a}{2^{n-1}},b-frac{b-a}{2^n}right].$$ Something similar can be done for sets of the form $(-infty, b).$ Readily, $Bbb R$ and sets of the form $(a,infty)$ can be shown to be such a union, as well. (cont'd)
$endgroup$
– Cameron Buie
Jan 5 at 19:18
$begingroup$
Since every non-empty, connected open set is a countably-infinite disjoint union of such half-open intervals, and since any open subset of $Bbb R$ is the union of at most countably-infinitely-many disjoint, connected open sets, then the second claim follows. Unfortunately, I'm not sure I understand the rest of the body of your question.
$endgroup$
– Cameron Buie
Jan 5 at 19:19