How can I to prove the equality of intervals of open intervals with the equality of the closed interval

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How can I to prove the equality between the following intersection of of open intervals with the following closed interval?



$$
bigcap_n=1^infty left(frac-1n ; 1+frac1nright) = [0,1]
$$



Thank you.






confidence-interval interval-arithmetic







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    Prove that $[0,1]$ is included in the intersection, and no $-epsilon$ nor $1+epsilon$.
    – Yves Daoust
    1 hour ago














up vote
2
down vote

favorite












How can I to prove the equality between the following intersection of of open intervals with the following closed interval?



$$
bigcap_n=1^infty left(frac-1n ; 1+frac1nright) = [0,1]
$$



Thank you.






confidence-interval interval-arithmetic







share|cite|improve this question









New contributor




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















  • 1




    Prove that $[0,1]$ is included in the intersection, and no $-epsilon$ nor $1+epsilon$.
    – Yves Daoust
    1 hour ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite











How can I to prove the equality between the following intersection of of open intervals with the following closed interval?



$$
bigcap_n=1^infty left(frac-1n ; 1+frac1nright) = [0,1]
$$



Thank you.






confidence-interval interval-arithmetic







share|cite|improve this question









New contributor




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











How can I to prove the equality between the following intersection of of open intervals with the following closed interval?



$$
bigcap_n=1^infty left(frac-1n ; 1+frac1nright) = [0,1]
$$



Thank you.






confidence-interval interval-arithmetic




confidence-interval interval-arithmetic






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Theressa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 1 hour ago









amWhy

191k27223436




191k27223436






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asked 2 hours ago









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  • 1




    Prove that $[0,1]$ is included in the intersection, and no $-epsilon$ nor $1+epsilon$.
    – Yves Daoust
    1 hour ago












  • 1




    Prove that $[0,1]$ is included in the intersection, and no $-epsilon$ nor $1+epsilon$.
    – Yves Daoust
    1 hour ago







1




1




Prove that $[0,1]$ is included in the intersection, and no $-epsilon$ nor $1+epsilon$.
– Yves Daoust
1 hour ago




Prove that $[0,1]$ is included in the intersection, and no $-epsilon$ nor $1+epsilon$.
– Yves Daoust
1 hour ago










3 Answers
3






active

oldest

votes

















up vote
2
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  1. If $x in bigcap_n=1^infty left(frac-1n ; 1+frac1nright)$, then

$frac-1n < x <1+frac1n$ for all $n$. With $ n to infty$ we get $0 le x le 1$.



  1. If $0 le x le 1$, then $frac-1n<0 le x le 1 <1+frac1n$ for all $n$, hence $x in bigcap_n=1^infty left(frac-1n ; 1+frac1nright)$ .





share|cite|improve this answer



























    up vote
    2
    down vote













    For all $n$,



    $$left(-frac1n ; 1+frac1nright) cap [0,1]=[0,1]$$



    and for any positive $epsilon$, if $n>dfrac1epsilon$,



    $$0-epsilonnotinleft(-frac1n ; 1+frac1nright)$$



    and



    $$1+epsilonnotinleft(-frac1n ; 1+frac1nright).$$






    share|cite|improve this answer





























      up vote
      1
      down vote













      Use double inclusion, clearly $[0,1]$ is contained in the intersetion since each point $Pin [0,1]$ lies in each open interval $(frac-1n ; 1+frac1n)$. For the other inclusion, pick a point $Q$ outside $[0,1]$ and you need find some $n$ such that $Qnotin (frac-1n ; 1+frac1n)$






      share|cite|improve this answer




















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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

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        active

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        active

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        up vote
        2
        down vote













        1. If $x in bigcap_n=1^infty left(frac-1n ; 1+frac1nright)$, then

        $frac-1n < x <1+frac1n$ for all $n$. With $ n to infty$ we get $0 le x le 1$.



        1. If $0 le x le 1$, then $frac-1n<0 le x le 1 <1+frac1n$ for all $n$, hence $x in bigcap_n=1^infty left(frac-1n ; 1+frac1nright)$ .





        share|cite|improve this answer
























          up vote
          2
          down vote













          1. If $x in bigcap_n=1^infty left(frac-1n ; 1+frac1nright)$, then

          $frac-1n < x <1+frac1n$ for all $n$. With $ n to infty$ we get $0 le x le 1$.



          1. If $0 le x le 1$, then $frac-1n<0 le x le 1 <1+frac1n$ for all $n$, hence $x in bigcap_n=1^infty left(frac-1n ; 1+frac1nright)$ .





          share|cite|improve this answer






















            up vote
            2
            down vote










            up vote
            2
            down vote









            1. If $x in bigcap_n=1^infty left(frac-1n ; 1+frac1nright)$, then

            $frac-1n < x <1+frac1n$ for all $n$. With $ n to infty$ we get $0 le x le 1$.



            1. If $0 le x le 1$, then $frac-1n<0 le x le 1 <1+frac1n$ for all $n$, hence $x in bigcap_n=1^infty left(frac-1n ; 1+frac1nright)$ .





            share|cite|improve this answer












            1. If $x in bigcap_n=1^infty left(frac-1n ; 1+frac1nright)$, then

            $frac-1n < x <1+frac1n$ for all $n$. With $ n to infty$ we get $0 le x le 1$.



            1. If $0 le x le 1$, then $frac-1n<0 le x le 1 <1+frac1n$ for all $n$, hence $x in bigcap_n=1^infty left(frac-1n ; 1+frac1nright)$ .






            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            Fred

            39.9k1338




            39.9k1338




















                up vote
                2
                down vote













                For all $n$,



                $$left(-frac1n ; 1+frac1nright) cap [0,1]=[0,1]$$



                and for any positive $epsilon$, if $n>dfrac1epsilon$,



                $$0-epsilonnotinleft(-frac1n ; 1+frac1nright)$$



                and



                $$1+epsilonnotinleft(-frac1n ; 1+frac1nright).$$






                share|cite|improve this answer


























                  up vote
                  2
                  down vote













                  For all $n$,



                  $$left(-frac1n ; 1+frac1nright) cap [0,1]=[0,1]$$



                  and for any positive $epsilon$, if $n>dfrac1epsilon$,



                  $$0-epsilonnotinleft(-frac1n ; 1+frac1nright)$$



                  and



                  $$1+epsilonnotinleft(-frac1n ; 1+frac1nright).$$






                  share|cite|improve this answer
























                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    For all $n$,



                    $$left(-frac1n ; 1+frac1nright) cap [0,1]=[0,1]$$



                    and for any positive $epsilon$, if $n>dfrac1epsilon$,



                    $$0-epsilonnotinleft(-frac1n ; 1+frac1nright)$$



                    and



                    $$1+epsilonnotinleft(-frac1n ; 1+frac1nright).$$






                    share|cite|improve this answer














                    For all $n$,



                    $$left(-frac1n ; 1+frac1nright) cap [0,1]=[0,1]$$



                    and for any positive $epsilon$, if $n>dfrac1epsilon$,



                    $$0-epsilonnotinleft(-frac1n ; 1+frac1nright)$$



                    and



                    $$1+epsilonnotinleft(-frac1n ; 1+frac1nright).$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 7 mins ago

























                    answered 1 hour ago









                    Yves Daoust

                    118k667214




                    118k667214




















                        up vote
                        1
                        down vote













                        Use double inclusion, clearly $[0,1]$ is contained in the intersetion since each point $Pin [0,1]$ lies in each open interval $(frac-1n ; 1+frac1n)$. For the other inclusion, pick a point $Q$ outside $[0,1]$ and you need find some $n$ such that $Qnotin (frac-1n ; 1+frac1n)$






                        share|cite|improve this answer
























                          up vote
                          1
                          down vote













                          Use double inclusion, clearly $[0,1]$ is contained in the intersetion since each point $Pin [0,1]$ lies in each open interval $(frac-1n ; 1+frac1n)$. For the other inclusion, pick a point $Q$ outside $[0,1]$ and you need find some $n$ such that $Qnotin (frac-1n ; 1+frac1n)$






                          share|cite|improve this answer






















                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            Use double inclusion, clearly $[0,1]$ is contained in the intersetion since each point $Pin [0,1]$ lies in each open interval $(frac-1n ; 1+frac1n)$. For the other inclusion, pick a point $Q$ outside $[0,1]$ and you need find some $n$ such that $Qnotin (frac-1n ; 1+frac1n)$






                            share|cite|improve this answer












                            Use double inclusion, clearly $[0,1]$ is contained in the intersetion since each point $Pin [0,1]$ lies in each open interval $(frac-1n ; 1+frac1n)$. For the other inclusion, pick a point $Q$ outside $[0,1]$ and you need find some $n$ such that $Qnotin (frac-1n ; 1+frac1n)$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 hours ago









                            ALG

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