In $mathbb{R}^n$, every open set is a union of closed sets












1














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







share|cite|improve this question


















  • 2




    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)
    – Cameron Buie
    2 days ago










  • 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.
    – Cameron Buie
    2 days ago
















1














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







share|cite|improve this question


















  • 2




    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)
    – Cameron Buie
    2 days ago










  • 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.
    – Cameron Buie
    2 days ago














1












1








1







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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 days ago









ReinherdReinherd

714




714








  • 2




    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)
    – Cameron Buie
    2 days ago










  • 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.
    – Cameron Buie
    2 days ago














  • 2




    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)
    – Cameron Buie
    2 days ago










  • 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.
    – Cameron Buie
    2 days ago








2




2




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)
– Cameron Buie
2 days ago




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)
– Cameron Buie
2 days ago












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.
– Cameron Buie
2 days ago




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.
– Cameron Buie
2 days ago










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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.






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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.






    share|cite|improve this answer










    New contributor




    ItsJustABallBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      1














      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.






      share|cite|improve this answer










      New contributor




      ItsJustABallBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





















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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.






        share|cite|improve this answer










        New contributor




        ItsJustABallBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        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.







        share|cite|improve this answer










        New contributor




        ItsJustABallBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        share|cite|improve this answer



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        answered 2 days ago









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