On a special type of sequence in complete metric space












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$begingroup$


Let ${x_n}$ be a sequence in a complete metric space $X$ such that for every uniformly continuous function $f:X to mathbb R$, the sequence ${f(x_n)}$ is convergent in $mathbb R$. Then is it true that ${x_n}$ is convergent ?



My try: If ${x_n}$ has a convergent subsequence, then ${x_n}$ converges. Indeed, if ${x_{k_n}}$ is a convergent subsequence with limit $yin X$ , then since the function $ f: X to mathbb R $ defined as $f(x)=d(x,y), forall xin X$ is uniformly continuous, so $lim f(x_n)=lim d(x_n,y)$ exists, but since $lim d(x_{k_n},y)=0$, so $lim d(x_n,y)=0$, thus ${x_n}$ converges to $y$.



Unfortunately, I am unable to decide whether ${x_n} $ has a convergent (or for that matter, just Cauchy, since $X$ is complete) subsequence or not.










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$endgroup$

















    6












    $begingroup$


    Let ${x_n}$ be a sequence in a complete metric space $X$ such that for every uniformly continuous function $f:X to mathbb R$, the sequence ${f(x_n)}$ is convergent in $mathbb R$. Then is it true that ${x_n}$ is convergent ?



    My try: If ${x_n}$ has a convergent subsequence, then ${x_n}$ converges. Indeed, if ${x_{k_n}}$ is a convergent subsequence with limit $yin X$ , then since the function $ f: X to mathbb R $ defined as $f(x)=d(x,y), forall xin X$ is uniformly continuous, so $lim f(x_n)=lim d(x_n,y)$ exists, but since $lim d(x_{k_n},y)=0$, so $lim d(x_n,y)=0$, thus ${x_n}$ converges to $y$.



    Unfortunately, I am unable to decide whether ${x_n} $ has a convergent (or for that matter, just Cauchy, since $X$ is complete) subsequence or not.










    share|cite|improve this question











    $endgroup$















      6












      6








      6


      1



      $begingroup$


      Let ${x_n}$ be a sequence in a complete metric space $X$ such that for every uniformly continuous function $f:X to mathbb R$, the sequence ${f(x_n)}$ is convergent in $mathbb R$. Then is it true that ${x_n}$ is convergent ?



      My try: If ${x_n}$ has a convergent subsequence, then ${x_n}$ converges. Indeed, if ${x_{k_n}}$ is a convergent subsequence with limit $yin X$ , then since the function $ f: X to mathbb R $ defined as $f(x)=d(x,y), forall xin X$ is uniformly continuous, so $lim f(x_n)=lim d(x_n,y)$ exists, but since $lim d(x_{k_n},y)=0$, so $lim d(x_n,y)=0$, thus ${x_n}$ converges to $y$.



      Unfortunately, I am unable to decide whether ${x_n} $ has a convergent (or for that matter, just Cauchy, since $X$ is complete) subsequence or not.










      share|cite|improve this question











      $endgroup$




      Let ${x_n}$ be a sequence in a complete metric space $X$ such that for every uniformly continuous function $f:X to mathbb R$, the sequence ${f(x_n)}$ is convergent in $mathbb R$. Then is it true that ${x_n}$ is convergent ?



      My try: If ${x_n}$ has a convergent subsequence, then ${x_n}$ converges. Indeed, if ${x_{k_n}}$ is a convergent subsequence with limit $yin X$ , then since the function $ f: X to mathbb R $ defined as $f(x)=d(x,y), forall xin X$ is uniformly continuous, so $lim f(x_n)=lim d(x_n,y)$ exists, but since $lim d(x_{k_n},y)=0$, so $lim d(x_n,y)=0$, thus ${x_n}$ converges to $y$.



      Unfortunately, I am unable to decide whether ${x_n} $ has a convergent (or for that matter, just Cauchy, since $X$ is complete) subsequence or not.







      metric-spaces uniform-continuity complete-spaces






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      edited Dec 29 '18 at 3:10







      user521337

















      asked Dec 29 '18 at 2:10









      user521337user521337

      1,1981417




      1,1981417






















          2 Answers
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          1












          $begingroup$

          Let's define $a_n = lim_{mto infty} d(x_n, x_m)$. Since distance is uniformly continuous, each of the $a_n$ exists. Eventually all points are contained within a ball of radius $a_1+epsilon$ about $x_1$. This implies that the $a_n$ are eventually bounded above by $2(a_1+epsilon)$, and so by Bolzano-Weierstrass must converge to some nonnegative number $z$. If $z$ is equal to $0$ we can show that the $x_n$ are Cauchy. Otherwise, if $z>0$ then there exists $k_0$ so that $a_k, k geq k_0$ is within $frac{z}{2}$ of $z$.



          Since $a_{k_0} = lim_{mto infty} d(x_{k_0}, x_m) in (frac{z}{2},frac{3z}{2})$, we can now pick $k_1 > k_0$ so that $d(x_{k_0},x_{k_1}) in (frac{z}{2},frac{3z}{2})$. Continue this process inductively, at each step picking a $k_j$ so that $x_{k_j}$ is at least $frac{z}{2}$ from all the previously chosen $x_{k_i}$. This gives us a countable subcollection of the ${x_i}$ who are mutually at least $frac{z}{2}$ distance apart.



          We can construct a uniformly continuous function $f$ equal to $1$ on $x_{k_i}$ for even $i$ and $0$ on $x_{k_i}$ for odd $i$. This is because all the $x_{k_i}$ are suitably spaced out (as a sketch, we can define $f$ equal to $1$ on the ball of radius $frac{z}{4}$ about $x_{k_i}$ for even $i$, and have $f$ decrease to zero in a controlled manner by the time we hit the boundary of the radius $frac{z}{2}$ ball about $x_{k_i}$). This $f$ is equal to $0$ and $1$ infinitely often, contradicting the fact that $z>0$. So $z=0$, and we can pick some convergent subsequence amongst the $x_n$.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Whether the metric $d$ is complete or not, if $Y$ is a non-empty subset of $X$ then $d(x,Y)=inf {d(x,y):yin Y}$ is uniformly continuous. If ${x_n}$ is not a Cauchy sequence then there exists a disjoint pair $A, B$ of infinite subsets of $Bbb N$ such that $inf {d(a,b): ain Aland bin B}>0.$ So let $f(x)=d(x, {x_n:nin A})$. Then $f(x_n)=0$ when $nin A$ but $inf {f(x_n):nin B}>0$ so ${f(x_n)}$ is not convergent.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              how does your $A,B$ depend on ${x_n}$ ? What is $5$ doing there ?
              $endgroup$
              – user521337
              Dec 30 '18 at 1:40










            • $begingroup$
              The 5 is (was) a typo.
              $endgroup$
              – DanielWainfleet
              Dec 30 '18 at 8:32











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            2 Answers
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            2 Answers
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            1












            $begingroup$

            Let's define $a_n = lim_{mto infty} d(x_n, x_m)$. Since distance is uniformly continuous, each of the $a_n$ exists. Eventually all points are contained within a ball of radius $a_1+epsilon$ about $x_1$. This implies that the $a_n$ are eventually bounded above by $2(a_1+epsilon)$, and so by Bolzano-Weierstrass must converge to some nonnegative number $z$. If $z$ is equal to $0$ we can show that the $x_n$ are Cauchy. Otherwise, if $z>0$ then there exists $k_0$ so that $a_k, k geq k_0$ is within $frac{z}{2}$ of $z$.



            Since $a_{k_0} = lim_{mto infty} d(x_{k_0}, x_m) in (frac{z}{2},frac{3z}{2})$, we can now pick $k_1 > k_0$ so that $d(x_{k_0},x_{k_1}) in (frac{z}{2},frac{3z}{2})$. Continue this process inductively, at each step picking a $k_j$ so that $x_{k_j}$ is at least $frac{z}{2}$ from all the previously chosen $x_{k_i}$. This gives us a countable subcollection of the ${x_i}$ who are mutually at least $frac{z}{2}$ distance apart.



            We can construct a uniformly continuous function $f$ equal to $1$ on $x_{k_i}$ for even $i$ and $0$ on $x_{k_i}$ for odd $i$. This is because all the $x_{k_i}$ are suitably spaced out (as a sketch, we can define $f$ equal to $1$ on the ball of radius $frac{z}{4}$ about $x_{k_i}$ for even $i$, and have $f$ decrease to zero in a controlled manner by the time we hit the boundary of the radius $frac{z}{2}$ ball about $x_{k_i}$). This $f$ is equal to $0$ and $1$ infinitely often, contradicting the fact that $z>0$. So $z=0$, and we can pick some convergent subsequence amongst the $x_n$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Let's define $a_n = lim_{mto infty} d(x_n, x_m)$. Since distance is uniformly continuous, each of the $a_n$ exists. Eventually all points are contained within a ball of radius $a_1+epsilon$ about $x_1$. This implies that the $a_n$ are eventually bounded above by $2(a_1+epsilon)$, and so by Bolzano-Weierstrass must converge to some nonnegative number $z$. If $z$ is equal to $0$ we can show that the $x_n$ are Cauchy. Otherwise, if $z>0$ then there exists $k_0$ so that $a_k, k geq k_0$ is within $frac{z}{2}$ of $z$.



              Since $a_{k_0} = lim_{mto infty} d(x_{k_0}, x_m) in (frac{z}{2},frac{3z}{2})$, we can now pick $k_1 > k_0$ so that $d(x_{k_0},x_{k_1}) in (frac{z}{2},frac{3z}{2})$. Continue this process inductively, at each step picking a $k_j$ so that $x_{k_j}$ is at least $frac{z}{2}$ from all the previously chosen $x_{k_i}$. This gives us a countable subcollection of the ${x_i}$ who are mutually at least $frac{z}{2}$ distance apart.



              We can construct a uniformly continuous function $f$ equal to $1$ on $x_{k_i}$ for even $i$ and $0$ on $x_{k_i}$ for odd $i$. This is because all the $x_{k_i}$ are suitably spaced out (as a sketch, we can define $f$ equal to $1$ on the ball of radius $frac{z}{4}$ about $x_{k_i}$ for even $i$, and have $f$ decrease to zero in a controlled manner by the time we hit the boundary of the radius $frac{z}{2}$ ball about $x_{k_i}$). This $f$ is equal to $0$ and $1$ infinitely often, contradicting the fact that $z>0$. So $z=0$, and we can pick some convergent subsequence amongst the $x_n$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Let's define $a_n = lim_{mto infty} d(x_n, x_m)$. Since distance is uniformly continuous, each of the $a_n$ exists. Eventually all points are contained within a ball of radius $a_1+epsilon$ about $x_1$. This implies that the $a_n$ are eventually bounded above by $2(a_1+epsilon)$, and so by Bolzano-Weierstrass must converge to some nonnegative number $z$. If $z$ is equal to $0$ we can show that the $x_n$ are Cauchy. Otherwise, if $z>0$ then there exists $k_0$ so that $a_k, k geq k_0$ is within $frac{z}{2}$ of $z$.



                Since $a_{k_0} = lim_{mto infty} d(x_{k_0}, x_m) in (frac{z}{2},frac{3z}{2})$, we can now pick $k_1 > k_0$ so that $d(x_{k_0},x_{k_1}) in (frac{z}{2},frac{3z}{2})$. Continue this process inductively, at each step picking a $k_j$ so that $x_{k_j}$ is at least $frac{z}{2}$ from all the previously chosen $x_{k_i}$. This gives us a countable subcollection of the ${x_i}$ who are mutually at least $frac{z}{2}$ distance apart.



                We can construct a uniformly continuous function $f$ equal to $1$ on $x_{k_i}$ for even $i$ and $0$ on $x_{k_i}$ for odd $i$. This is because all the $x_{k_i}$ are suitably spaced out (as a sketch, we can define $f$ equal to $1$ on the ball of radius $frac{z}{4}$ about $x_{k_i}$ for even $i$, and have $f$ decrease to zero in a controlled manner by the time we hit the boundary of the radius $frac{z}{2}$ ball about $x_{k_i}$). This $f$ is equal to $0$ and $1$ infinitely often, contradicting the fact that $z>0$. So $z=0$, and we can pick some convergent subsequence amongst the $x_n$.






                share|cite|improve this answer









                $endgroup$



                Let's define $a_n = lim_{mto infty} d(x_n, x_m)$. Since distance is uniformly continuous, each of the $a_n$ exists. Eventually all points are contained within a ball of radius $a_1+epsilon$ about $x_1$. This implies that the $a_n$ are eventually bounded above by $2(a_1+epsilon)$, and so by Bolzano-Weierstrass must converge to some nonnegative number $z$. If $z$ is equal to $0$ we can show that the $x_n$ are Cauchy. Otherwise, if $z>0$ then there exists $k_0$ so that $a_k, k geq k_0$ is within $frac{z}{2}$ of $z$.



                Since $a_{k_0} = lim_{mto infty} d(x_{k_0}, x_m) in (frac{z}{2},frac{3z}{2})$, we can now pick $k_1 > k_0$ so that $d(x_{k_0},x_{k_1}) in (frac{z}{2},frac{3z}{2})$. Continue this process inductively, at each step picking a $k_j$ so that $x_{k_j}$ is at least $frac{z}{2}$ from all the previously chosen $x_{k_i}$. This gives us a countable subcollection of the ${x_i}$ who are mutually at least $frac{z}{2}$ distance apart.



                We can construct a uniformly continuous function $f$ equal to $1$ on $x_{k_i}$ for even $i$ and $0$ on $x_{k_i}$ for odd $i$. This is because all the $x_{k_i}$ are suitably spaced out (as a sketch, we can define $f$ equal to $1$ on the ball of radius $frac{z}{4}$ about $x_{k_i}$ for even $i$, and have $f$ decrease to zero in a controlled manner by the time we hit the boundary of the radius $frac{z}{2}$ ball about $x_{k_i}$). This $f$ is equal to $0$ and $1$ infinitely often, contradicting the fact that $z>0$. So $z=0$, and we can pick some convergent subsequence amongst the $x_n$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 29 '18 at 7:18









                wfawwerwfawwer

                414




                414























                    0












                    $begingroup$

                    Whether the metric $d$ is complete or not, if $Y$ is a non-empty subset of $X$ then $d(x,Y)=inf {d(x,y):yin Y}$ is uniformly continuous. If ${x_n}$ is not a Cauchy sequence then there exists a disjoint pair $A, B$ of infinite subsets of $Bbb N$ such that $inf {d(a,b): ain Aland bin B}>0.$ So let $f(x)=d(x, {x_n:nin A})$. Then $f(x_n)=0$ when $nin A$ but $inf {f(x_n):nin B}>0$ so ${f(x_n)}$ is not convergent.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      how does your $A,B$ depend on ${x_n}$ ? What is $5$ doing there ?
                      $endgroup$
                      – user521337
                      Dec 30 '18 at 1:40










                    • $begingroup$
                      The 5 is (was) a typo.
                      $endgroup$
                      – DanielWainfleet
                      Dec 30 '18 at 8:32
















                    0












                    $begingroup$

                    Whether the metric $d$ is complete or not, if $Y$ is a non-empty subset of $X$ then $d(x,Y)=inf {d(x,y):yin Y}$ is uniformly continuous. If ${x_n}$ is not a Cauchy sequence then there exists a disjoint pair $A, B$ of infinite subsets of $Bbb N$ such that $inf {d(a,b): ain Aland bin B}>0.$ So let $f(x)=d(x, {x_n:nin A})$. Then $f(x_n)=0$ when $nin A$ but $inf {f(x_n):nin B}>0$ so ${f(x_n)}$ is not convergent.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      how does your $A,B$ depend on ${x_n}$ ? What is $5$ doing there ?
                      $endgroup$
                      – user521337
                      Dec 30 '18 at 1:40










                    • $begingroup$
                      The 5 is (was) a typo.
                      $endgroup$
                      – DanielWainfleet
                      Dec 30 '18 at 8:32














                    0












                    0








                    0





                    $begingroup$

                    Whether the metric $d$ is complete or not, if $Y$ is a non-empty subset of $X$ then $d(x,Y)=inf {d(x,y):yin Y}$ is uniformly continuous. If ${x_n}$ is not a Cauchy sequence then there exists a disjoint pair $A, B$ of infinite subsets of $Bbb N$ such that $inf {d(a,b): ain Aland bin B}>0.$ So let $f(x)=d(x, {x_n:nin A})$. Then $f(x_n)=0$ when $nin A$ but $inf {f(x_n):nin B}>0$ so ${f(x_n)}$ is not convergent.






                    share|cite|improve this answer











                    $endgroup$



                    Whether the metric $d$ is complete or not, if $Y$ is a non-empty subset of $X$ then $d(x,Y)=inf {d(x,y):yin Y}$ is uniformly continuous. If ${x_n}$ is not a Cauchy sequence then there exists a disjoint pair $A, B$ of infinite subsets of $Bbb N$ such that $inf {d(a,b): ain Aland bin B}>0.$ So let $f(x)=d(x, {x_n:nin A})$. Then $f(x_n)=0$ when $nin A$ but $inf {f(x_n):nin B}>0$ so ${f(x_n)}$ is not convergent.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 30 '18 at 8:32

























                    answered Dec 29 '18 at 15:38









                    DanielWainfleetDanielWainfleet

                    35.6k31648




                    35.6k31648












                    • $begingroup$
                      how does your $A,B$ depend on ${x_n}$ ? What is $5$ doing there ?
                      $endgroup$
                      – user521337
                      Dec 30 '18 at 1:40










                    • $begingroup$
                      The 5 is (was) a typo.
                      $endgroup$
                      – DanielWainfleet
                      Dec 30 '18 at 8:32


















                    • $begingroup$
                      how does your $A,B$ depend on ${x_n}$ ? What is $5$ doing there ?
                      $endgroup$
                      – user521337
                      Dec 30 '18 at 1:40










                    • $begingroup$
                      The 5 is (was) a typo.
                      $endgroup$
                      – DanielWainfleet
                      Dec 30 '18 at 8:32
















                    $begingroup$
                    how does your $A,B$ depend on ${x_n}$ ? What is $5$ doing there ?
                    $endgroup$
                    – user521337
                    Dec 30 '18 at 1:40




                    $begingroup$
                    how does your $A,B$ depend on ${x_n}$ ? What is $5$ doing there ?
                    $endgroup$
                    – user521337
                    Dec 30 '18 at 1:40












                    $begingroup$
                    The 5 is (was) a typo.
                    $endgroup$
                    – DanielWainfleet
                    Dec 30 '18 at 8:32




                    $begingroup$
                    The 5 is (was) a typo.
                    $endgroup$
                    – DanielWainfleet
                    Dec 30 '18 at 8:32


















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