why a subspace is closed?Why This Map is Closed?Exercise involving topological vector spaces, linear maps,...

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why a subspace is closed?



why a subspace is closed?


Why This Map is Closed?Exercise involving topological vector spaces, linear maps, and the quotient mapSubspaces of a Topological Vector SpacesQuestion about the basis for the topology on an inductive limit of Frechet spaces$C_c(X)$ is dense in $C_0(X)$What happens to a Banach space of functions when the domain is not compact?Convergence in quotient topological vector spaceWeakly compact subset of a normed space is sequentially compact, norm bounded, and proximalIs the following application of the Banach-Alaoglu theorem sound?Rudin's functional analysis theorem 3.21.













1












$begingroup$


Let $E$ be $K-$vector space with a norm $|cdot|$, and $F$ a subspace with dimension $n$. Show that $F$ is a closed set .



I am trying to show that any convergent sequence of elements $(x_n)_{n in N}$ of $F$ converge to an element $x in F$.



This means I need to show that $||x_n-x|| rightarrow 0$



I see that like a distance which is attended by the fuction $y rightarrow ||x-y||$ .



How can I show that $ x in F$?



I have been reading a solution which I don't really understand the intuition behind it :



let $x$ be in $E$ and let $B={ y in F ||y-x|| leq ||x||}$
after showing that its a compact, they tried to show that $inf_{y in F}(||x-y||)= lambda $ exists.



by definition of $lambda$ it is the distance between $x$ and $F$










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I am voting to close, because it is not at all clear to me what you are asking. Are you asking for the intuition behind closed sets, or are you asking how to show that a finite dimensional subspace is closed?
    $endgroup$
    – K.Power
    4 hours ago












  • $begingroup$
    @K.Power actually I am looking for both
    $endgroup$
    – user515918
    4 hours ago










  • $begingroup$
    If it is the latter then you will find such a question has been asked before on this site. I should also point out that you need to show that any convergent sequence in $F$ has its limit in $F$.
    $endgroup$
    – K.Power
    4 hours ago










  • $begingroup$
    @K.Power I've been searching but I dont have any idea about the methode they using
    $endgroup$
    – user515918
    4 hours ago










  • $begingroup$
    As these are very distinct questions you should definitely ask these in two separate posts.
    $endgroup$
    – K.Power
    4 hours ago
















1












$begingroup$


Let $E$ be $K-$vector space with a norm $|cdot|$, and $F$ a subspace with dimension $n$. Show that $F$ is a closed set .



I am trying to show that any convergent sequence of elements $(x_n)_{n in N}$ of $F$ converge to an element $x in F$.



This means I need to show that $||x_n-x|| rightarrow 0$



I see that like a distance which is attended by the fuction $y rightarrow ||x-y||$ .



How can I show that $ x in F$?



I have been reading a solution which I don't really understand the intuition behind it :



let $x$ be in $E$ and let $B={ y in F ||y-x|| leq ||x||}$
after showing that its a compact, they tried to show that $inf_{y in F}(||x-y||)= lambda $ exists.



by definition of $lambda$ it is the distance between $x$ and $F$










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I am voting to close, because it is not at all clear to me what you are asking. Are you asking for the intuition behind closed sets, or are you asking how to show that a finite dimensional subspace is closed?
    $endgroup$
    – K.Power
    4 hours ago












  • $begingroup$
    @K.Power actually I am looking for both
    $endgroup$
    – user515918
    4 hours ago










  • $begingroup$
    If it is the latter then you will find such a question has been asked before on this site. I should also point out that you need to show that any convergent sequence in $F$ has its limit in $F$.
    $endgroup$
    – K.Power
    4 hours ago










  • $begingroup$
    @K.Power I've been searching but I dont have any idea about the methode they using
    $endgroup$
    – user515918
    4 hours ago










  • $begingroup$
    As these are very distinct questions you should definitely ask these in two separate posts.
    $endgroup$
    – K.Power
    4 hours ago














1












1








1





$begingroup$


Let $E$ be $K-$vector space with a norm $|cdot|$, and $F$ a subspace with dimension $n$. Show that $F$ is a closed set .



I am trying to show that any convergent sequence of elements $(x_n)_{n in N}$ of $F$ converge to an element $x in F$.



This means I need to show that $||x_n-x|| rightarrow 0$



I see that like a distance which is attended by the fuction $y rightarrow ||x-y||$ .



How can I show that $ x in F$?



I have been reading a solution which I don't really understand the intuition behind it :



let $x$ be in $E$ and let $B={ y in F ||y-x|| leq ||x||}$
after showing that its a compact, they tried to show that $inf_{y in F}(||x-y||)= lambda $ exists.



by definition of $lambda$ it is the distance between $x$ and $F$










share|cite|improve this question











$endgroup$




Let $E$ be $K-$vector space with a norm $|cdot|$, and $F$ a subspace with dimension $n$. Show that $F$ is a closed set .



I am trying to show that any convergent sequence of elements $(x_n)_{n in N}$ of $F$ converge to an element $x in F$.



This means I need to show that $||x_n-x|| rightarrow 0$



I see that like a distance which is attended by the fuction $y rightarrow ||x-y||$ .



How can I show that $ x in F$?



I have been reading a solution which I don't really understand the intuition behind it :



let $x$ be in $E$ and let $B={ y in F ||y-x|| leq ||x||}$
after showing that its a compact, they tried to show that $inf_{y in F}(||x-y||)= lambda $ exists.



by definition of $lambda$ it is the distance between $x$ and $F$







topological-vector-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 hours ago







user515918

















asked 4 hours ago









user515918user515918

515




515








  • 1




    $begingroup$
    I am voting to close, because it is not at all clear to me what you are asking. Are you asking for the intuition behind closed sets, or are you asking how to show that a finite dimensional subspace is closed?
    $endgroup$
    – K.Power
    4 hours ago












  • $begingroup$
    @K.Power actually I am looking for both
    $endgroup$
    – user515918
    4 hours ago










  • $begingroup$
    If it is the latter then you will find such a question has been asked before on this site. I should also point out that you need to show that any convergent sequence in $F$ has its limit in $F$.
    $endgroup$
    – K.Power
    4 hours ago










  • $begingroup$
    @K.Power I've been searching but I dont have any idea about the methode they using
    $endgroup$
    – user515918
    4 hours ago










  • $begingroup$
    As these are very distinct questions you should definitely ask these in two separate posts.
    $endgroup$
    – K.Power
    4 hours ago














  • 1




    $begingroup$
    I am voting to close, because it is not at all clear to me what you are asking. Are you asking for the intuition behind closed sets, or are you asking how to show that a finite dimensional subspace is closed?
    $endgroup$
    – K.Power
    4 hours ago












  • $begingroup$
    @K.Power actually I am looking for both
    $endgroup$
    – user515918
    4 hours ago










  • $begingroup$
    If it is the latter then you will find such a question has been asked before on this site. I should also point out that you need to show that any convergent sequence in $F$ has its limit in $F$.
    $endgroup$
    – K.Power
    4 hours ago










  • $begingroup$
    @K.Power I've been searching but I dont have any idea about the methode they using
    $endgroup$
    – user515918
    4 hours ago










  • $begingroup$
    As these are very distinct questions you should definitely ask these in two separate posts.
    $endgroup$
    – K.Power
    4 hours ago








1




1




$begingroup$
I am voting to close, because it is not at all clear to me what you are asking. Are you asking for the intuition behind closed sets, or are you asking how to show that a finite dimensional subspace is closed?
$endgroup$
– K.Power
4 hours ago






$begingroup$
I am voting to close, because it is not at all clear to me what you are asking. Are you asking for the intuition behind closed sets, or are you asking how to show that a finite dimensional subspace is closed?
$endgroup$
– K.Power
4 hours ago














$begingroup$
@K.Power actually I am looking for both
$endgroup$
– user515918
4 hours ago




$begingroup$
@K.Power actually I am looking for both
$endgroup$
– user515918
4 hours ago












$begingroup$
If it is the latter then you will find such a question has been asked before on this site. I should also point out that you need to show that any convergent sequence in $F$ has its limit in $F$.
$endgroup$
– K.Power
4 hours ago




$begingroup$
If it is the latter then you will find such a question has been asked before on this site. I should also point out that you need to show that any convergent sequence in $F$ has its limit in $F$.
$endgroup$
– K.Power
4 hours ago












$begingroup$
@K.Power I've been searching but I dont have any idea about the methode they using
$endgroup$
– user515918
4 hours ago




$begingroup$
@K.Power I've been searching but I dont have any idea about the methode they using
$endgroup$
– user515918
4 hours ago












$begingroup$
As these are very distinct questions you should definitely ask these in two separate posts.
$endgroup$
– K.Power
4 hours ago




$begingroup$
As these are very distinct questions you should definitely ask these in two separate posts.
$endgroup$
– K.Power
4 hours ago










5 Answers
5






active

oldest

votes


















2












$begingroup$

Hint:




  1. Use that all norms on $F$ are equivalent to show that all bounded subsets of $F$ which are closed in $F$ are compact.


  2. Show that your sequence is contained in a bounded subset of $F$ which is closed in $F$.


  3. Find a subsequence which has a limit in $F$. What is the limit?







share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Another approach will be to show that a finite-dimensional normed space (over $mathbb R$ or $mathbb C$ or indeed over any complete normed field) is complete. Then conclude that it is closed in a larger normed space.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      To elaborate a bit on GEdgar's suggestion: We know that $F$ has a basis ${e_1,dots,e_n}$. Using Bolzano-Weierstrass you can show that we have the existence of some $r>0$ such that for any choice of scalars $c_1,dots,c_nin K$ we have
      $$|sum_{i=1}^nc_ie_i|geq rsum_{i=1}^n|c_i|.$$
      Using this you can show that $F$ is complete. Consider a cauchy sequence $(x_m)subset F$. We know that each $x_m=sum_{i=1}^nc_i^{(m)}e_i$. Using the inequality above and the cauchyness of $(x_m)$ try and show that each sequence of scalars $(c_i^{(m)})$ is also Cauchy. As $K$ is complete we know that these sequences each have a limit in $K$. Show that $(x_m)$ converges to the element of $F$ defined by these scalar limits.






      share|cite|improve this answer











      $endgroup$





















        0












        $begingroup$

        You haven't correctly stated what you need to prove -- you need to show that any sequence of elements of $F$ that converges in $E$ in fact has its limit in $F$.



        But I think it may be easier to show that the complement of F is open. Consider an element $x in E setminus F$. It has a distance $gamma gt 0$ from $F$, given by $||z||$, where $x = y+z$ with $y in F$ and $z perp F$. Then $B(x, gamma /2) subseteq E setminus F$ so $x$ is in the interior of $E setminus F$. We chose $x$ arbitrarily, so this proves $E setminus F$ is open so that $F$ is closed.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          how can we write x=y+z ? E is not an euclidien space to talk about the orthogonal of a subspace !
          $endgroup$
          – user515918
          4 hours ago










        • $begingroup$
          Aren't you assuming that the OP knows that finite dimensional subspaces are complemented? This implies they already know that they are closed.
          $endgroup$
          – K.Power
          4 hours ago



















        0












        $begingroup$

        I think K.Power's answer is the best elementary proof but if you don't know the inequality used in his/her answer, you can argue successfully without it:



        Let $F=text{span}{e^1,cdots, e^n},$ where the $|e^i|=1; 1le ile n.$



        Now, suppose $Fsupset (x_n)to xin E$. Then, $x_i=a^i_1e^1+cdots a^i_ne^n; 1le ile n.$



        Of course, $(x_n)$ is a Cauchy sequence in $E$, so since



        $|x_m-x_k|=|(a^m_1-a^k_1)e^1+cdots (a^m_n-a^k_n)e^n|$



        the triangle inequality shows that each $(a^m_i); 1le ile n$ is a Cauchy sequence in $K$.



        Assuming now that $K$ is complete, we have $a^m_ito a_iin K; 1le ile n$.



        To finish, consider the vector $x=a_1e^1+cdots +a_ne^n$.



        Clearly, $xin F$ and now, another application of the triangle inequality shows that



        $|x-x_n|to 0$, from which it follows that $x_nto xin F$






        share|cite|improve this answer











        $endgroup$













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          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Hint:




          1. Use that all norms on $F$ are equivalent to show that all bounded subsets of $F$ which are closed in $F$ are compact.


          2. Show that your sequence is contained in a bounded subset of $F$ which is closed in $F$.


          3. Find a subsequence which has a limit in $F$. What is the limit?







          share|cite|improve this answer











          $endgroup$


















            2












            $begingroup$

            Hint:




            1. Use that all norms on $F$ are equivalent to show that all bounded subsets of $F$ which are closed in $F$ are compact.


            2. Show that your sequence is contained in a bounded subset of $F$ which is closed in $F$.


            3. Find a subsequence which has a limit in $F$. What is the limit?







            share|cite|improve this answer











            $endgroup$
















              2












              2








              2





              $begingroup$

              Hint:




              1. Use that all norms on $F$ are equivalent to show that all bounded subsets of $F$ which are closed in $F$ are compact.


              2. Show that your sequence is contained in a bounded subset of $F$ which is closed in $F$.


              3. Find a subsequence which has a limit in $F$. What is the limit?







              share|cite|improve this answer











              $endgroup$



              Hint:




              1. Use that all norms on $F$ are equivalent to show that all bounded subsets of $F$ which are closed in $F$ are compact.


              2. Show that your sequence is contained in a bounded subset of $F$ which is closed in $F$.


              3. Find a subsequence which has a limit in $F$. What is the limit?








              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 3 hours ago

























              answered 4 hours ago









              triitrii

              1515




              1515























                  1












                  $begingroup$

                  Another approach will be to show that a finite-dimensional normed space (over $mathbb R$ or $mathbb C$ or indeed over any complete normed field) is complete. Then conclude that it is closed in a larger normed space.






                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    Another approach will be to show that a finite-dimensional normed space (over $mathbb R$ or $mathbb C$ or indeed over any complete normed field) is complete. Then conclude that it is closed in a larger normed space.






                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      Another approach will be to show that a finite-dimensional normed space (over $mathbb R$ or $mathbb C$ or indeed over any complete normed field) is complete. Then conclude that it is closed in a larger normed space.






                      share|cite|improve this answer









                      $endgroup$



                      Another approach will be to show that a finite-dimensional normed space (over $mathbb R$ or $mathbb C$ or indeed over any complete normed field) is complete. Then conclude that it is closed in a larger normed space.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 4 hours ago









                      GEdgarGEdgar

                      62.7k267171




                      62.7k267171























                          1












                          $begingroup$

                          To elaborate a bit on GEdgar's suggestion: We know that $F$ has a basis ${e_1,dots,e_n}$. Using Bolzano-Weierstrass you can show that we have the existence of some $r>0$ such that for any choice of scalars $c_1,dots,c_nin K$ we have
                          $$|sum_{i=1}^nc_ie_i|geq rsum_{i=1}^n|c_i|.$$
                          Using this you can show that $F$ is complete. Consider a cauchy sequence $(x_m)subset F$. We know that each $x_m=sum_{i=1}^nc_i^{(m)}e_i$. Using the inequality above and the cauchyness of $(x_m)$ try and show that each sequence of scalars $(c_i^{(m)})$ is also Cauchy. As $K$ is complete we know that these sequences each have a limit in $K$. Show that $(x_m)$ converges to the element of $F$ defined by these scalar limits.






                          share|cite|improve this answer











                          $endgroup$


















                            1












                            $begingroup$

                            To elaborate a bit on GEdgar's suggestion: We know that $F$ has a basis ${e_1,dots,e_n}$. Using Bolzano-Weierstrass you can show that we have the existence of some $r>0$ such that for any choice of scalars $c_1,dots,c_nin K$ we have
                            $$|sum_{i=1}^nc_ie_i|geq rsum_{i=1}^n|c_i|.$$
                            Using this you can show that $F$ is complete. Consider a cauchy sequence $(x_m)subset F$. We know that each $x_m=sum_{i=1}^nc_i^{(m)}e_i$. Using the inequality above and the cauchyness of $(x_m)$ try and show that each sequence of scalars $(c_i^{(m)})$ is also Cauchy. As $K$ is complete we know that these sequences each have a limit in $K$. Show that $(x_m)$ converges to the element of $F$ defined by these scalar limits.






                            share|cite|improve this answer











                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              To elaborate a bit on GEdgar's suggestion: We know that $F$ has a basis ${e_1,dots,e_n}$. Using Bolzano-Weierstrass you can show that we have the existence of some $r>0$ such that for any choice of scalars $c_1,dots,c_nin K$ we have
                              $$|sum_{i=1}^nc_ie_i|geq rsum_{i=1}^n|c_i|.$$
                              Using this you can show that $F$ is complete. Consider a cauchy sequence $(x_m)subset F$. We know that each $x_m=sum_{i=1}^nc_i^{(m)}e_i$. Using the inequality above and the cauchyness of $(x_m)$ try and show that each sequence of scalars $(c_i^{(m)})$ is also Cauchy. As $K$ is complete we know that these sequences each have a limit in $K$. Show that $(x_m)$ converges to the element of $F$ defined by these scalar limits.






                              share|cite|improve this answer











                              $endgroup$



                              To elaborate a bit on GEdgar's suggestion: We know that $F$ has a basis ${e_1,dots,e_n}$. Using Bolzano-Weierstrass you can show that we have the existence of some $r>0$ such that for any choice of scalars $c_1,dots,c_nin K$ we have
                              $$|sum_{i=1}^nc_ie_i|geq rsum_{i=1}^n|c_i|.$$
                              Using this you can show that $F$ is complete. Consider a cauchy sequence $(x_m)subset F$. We know that each $x_m=sum_{i=1}^nc_i^{(m)}e_i$. Using the inequality above and the cauchyness of $(x_m)$ try and show that each sequence of scalars $(c_i^{(m)})$ is also Cauchy. As $K$ is complete we know that these sequences each have a limit in $K$. Show that $(x_m)$ converges to the element of $F$ defined by these scalar limits.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited 3 hours ago

























                              answered 3 hours ago









                              K.PowerK.Power

                              3,055926




                              3,055926























                                  0












                                  $begingroup$

                                  You haven't correctly stated what you need to prove -- you need to show that any sequence of elements of $F$ that converges in $E$ in fact has its limit in $F$.



                                  But I think it may be easier to show that the complement of F is open. Consider an element $x in E setminus F$. It has a distance $gamma gt 0$ from $F$, given by $||z||$, where $x = y+z$ with $y in F$ and $z perp F$. Then $B(x, gamma /2) subseteq E setminus F$ so $x$ is in the interior of $E setminus F$. We chose $x$ arbitrarily, so this proves $E setminus F$ is open so that $F$ is closed.






                                  share|cite|improve this answer









                                  $endgroup$













                                  • $begingroup$
                                    how can we write x=y+z ? E is not an euclidien space to talk about the orthogonal of a subspace !
                                    $endgroup$
                                    – user515918
                                    4 hours ago










                                  • $begingroup$
                                    Aren't you assuming that the OP knows that finite dimensional subspaces are complemented? This implies they already know that they are closed.
                                    $endgroup$
                                    – K.Power
                                    4 hours ago
















                                  0












                                  $begingroup$

                                  You haven't correctly stated what you need to prove -- you need to show that any sequence of elements of $F$ that converges in $E$ in fact has its limit in $F$.



                                  But I think it may be easier to show that the complement of F is open. Consider an element $x in E setminus F$. It has a distance $gamma gt 0$ from $F$, given by $||z||$, where $x = y+z$ with $y in F$ and $z perp F$. Then $B(x, gamma /2) subseteq E setminus F$ so $x$ is in the interior of $E setminus F$. We chose $x$ arbitrarily, so this proves $E setminus F$ is open so that $F$ is closed.






                                  share|cite|improve this answer









                                  $endgroup$













                                  • $begingroup$
                                    how can we write x=y+z ? E is not an euclidien space to talk about the orthogonal of a subspace !
                                    $endgroup$
                                    – user515918
                                    4 hours ago










                                  • $begingroup$
                                    Aren't you assuming that the OP knows that finite dimensional subspaces are complemented? This implies they already know that they are closed.
                                    $endgroup$
                                    – K.Power
                                    4 hours ago














                                  0












                                  0








                                  0





                                  $begingroup$

                                  You haven't correctly stated what you need to prove -- you need to show that any sequence of elements of $F$ that converges in $E$ in fact has its limit in $F$.



                                  But I think it may be easier to show that the complement of F is open. Consider an element $x in E setminus F$. It has a distance $gamma gt 0$ from $F$, given by $||z||$, where $x = y+z$ with $y in F$ and $z perp F$. Then $B(x, gamma /2) subseteq E setminus F$ so $x$ is in the interior of $E setminus F$. We chose $x$ arbitrarily, so this proves $E setminus F$ is open so that $F$ is closed.






                                  share|cite|improve this answer









                                  $endgroup$



                                  You haven't correctly stated what you need to prove -- you need to show that any sequence of elements of $F$ that converges in $E$ in fact has its limit in $F$.



                                  But I think it may be easier to show that the complement of F is open. Consider an element $x in E setminus F$. It has a distance $gamma gt 0$ from $F$, given by $||z||$, where $x = y+z$ with $y in F$ and $z perp F$. Then $B(x, gamma /2) subseteq E setminus F$ so $x$ is in the interior of $E setminus F$. We chose $x$ arbitrarily, so this proves $E setminus F$ is open so that $F$ is closed.







                                  share|cite|improve this answer












                                  share|cite|improve this answer



                                  share|cite|improve this answer










                                  answered 4 hours ago









                                  Robert ShoreRobert Shore

                                  1,719115




                                  1,719115












                                  • $begingroup$
                                    how can we write x=y+z ? E is not an euclidien space to talk about the orthogonal of a subspace !
                                    $endgroup$
                                    – user515918
                                    4 hours ago










                                  • $begingroup$
                                    Aren't you assuming that the OP knows that finite dimensional subspaces are complemented? This implies they already know that they are closed.
                                    $endgroup$
                                    – K.Power
                                    4 hours ago


















                                  • $begingroup$
                                    how can we write x=y+z ? E is not an euclidien space to talk about the orthogonal of a subspace !
                                    $endgroup$
                                    – user515918
                                    4 hours ago










                                  • $begingroup$
                                    Aren't you assuming that the OP knows that finite dimensional subspaces are complemented? This implies they already know that they are closed.
                                    $endgroup$
                                    – K.Power
                                    4 hours ago
















                                  $begingroup$
                                  how can we write x=y+z ? E is not an euclidien space to talk about the orthogonal of a subspace !
                                  $endgroup$
                                  – user515918
                                  4 hours ago




                                  $begingroup$
                                  how can we write x=y+z ? E is not an euclidien space to talk about the orthogonal of a subspace !
                                  $endgroup$
                                  – user515918
                                  4 hours ago












                                  $begingroup$
                                  Aren't you assuming that the OP knows that finite dimensional subspaces are complemented? This implies they already know that they are closed.
                                  $endgroup$
                                  – K.Power
                                  4 hours ago




                                  $begingroup$
                                  Aren't you assuming that the OP knows that finite dimensional subspaces are complemented? This implies they already know that they are closed.
                                  $endgroup$
                                  – K.Power
                                  4 hours ago











                                  0












                                  $begingroup$

                                  I think K.Power's answer is the best elementary proof but if you don't know the inequality used in his/her answer, you can argue successfully without it:



                                  Let $F=text{span}{e^1,cdots, e^n},$ where the $|e^i|=1; 1le ile n.$



                                  Now, suppose $Fsupset (x_n)to xin E$. Then, $x_i=a^i_1e^1+cdots a^i_ne^n; 1le ile n.$



                                  Of course, $(x_n)$ is a Cauchy sequence in $E$, so since



                                  $|x_m-x_k|=|(a^m_1-a^k_1)e^1+cdots (a^m_n-a^k_n)e^n|$



                                  the triangle inequality shows that each $(a^m_i); 1le ile n$ is a Cauchy sequence in $K$.



                                  Assuming now that $K$ is complete, we have $a^m_ito a_iin K; 1le ile n$.



                                  To finish, consider the vector $x=a_1e^1+cdots +a_ne^n$.



                                  Clearly, $xin F$ and now, another application of the triangle inequality shows that



                                  $|x-x_n|to 0$, from which it follows that $x_nto xin F$






                                  share|cite|improve this answer











                                  $endgroup$


















                                    0












                                    $begingroup$

                                    I think K.Power's answer is the best elementary proof but if you don't know the inequality used in his/her answer, you can argue successfully without it:



                                    Let $F=text{span}{e^1,cdots, e^n},$ where the $|e^i|=1; 1le ile n.$



                                    Now, suppose $Fsupset (x_n)to xin E$. Then, $x_i=a^i_1e^1+cdots a^i_ne^n; 1le ile n.$



                                    Of course, $(x_n)$ is a Cauchy sequence in $E$, so since



                                    $|x_m-x_k|=|(a^m_1-a^k_1)e^1+cdots (a^m_n-a^k_n)e^n|$



                                    the triangle inequality shows that each $(a^m_i); 1le ile n$ is a Cauchy sequence in $K$.



                                    Assuming now that $K$ is complete, we have $a^m_ito a_iin K; 1le ile n$.



                                    To finish, consider the vector $x=a_1e^1+cdots +a_ne^n$.



                                    Clearly, $xin F$ and now, another application of the triangle inequality shows that



                                    $|x-x_n|to 0$, from which it follows that $x_nto xin F$






                                    share|cite|improve this answer











                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      I think K.Power's answer is the best elementary proof but if you don't know the inequality used in his/her answer, you can argue successfully without it:



                                      Let $F=text{span}{e^1,cdots, e^n},$ where the $|e^i|=1; 1le ile n.$



                                      Now, suppose $Fsupset (x_n)to xin E$. Then, $x_i=a^i_1e^1+cdots a^i_ne^n; 1le ile n.$



                                      Of course, $(x_n)$ is a Cauchy sequence in $E$, so since



                                      $|x_m-x_k|=|(a^m_1-a^k_1)e^1+cdots (a^m_n-a^k_n)e^n|$



                                      the triangle inequality shows that each $(a^m_i); 1le ile n$ is a Cauchy sequence in $K$.



                                      Assuming now that $K$ is complete, we have $a^m_ito a_iin K; 1le ile n$.



                                      To finish, consider the vector $x=a_1e^1+cdots +a_ne^n$.



                                      Clearly, $xin F$ and now, another application of the triangle inequality shows that



                                      $|x-x_n|to 0$, from which it follows that $x_nto xin F$






                                      share|cite|improve this answer











                                      $endgroup$



                                      I think K.Power's answer is the best elementary proof but if you don't know the inequality used in his/her answer, you can argue successfully without it:



                                      Let $F=text{span}{e^1,cdots, e^n},$ where the $|e^i|=1; 1le ile n.$



                                      Now, suppose $Fsupset (x_n)to xin E$. Then, $x_i=a^i_1e^1+cdots a^i_ne^n; 1le ile n.$



                                      Of course, $(x_n)$ is a Cauchy sequence in $E$, so since



                                      $|x_m-x_k|=|(a^m_1-a^k_1)e^1+cdots (a^m_n-a^k_n)e^n|$



                                      the triangle inequality shows that each $(a^m_i); 1le ile n$ is a Cauchy sequence in $K$.



                                      Assuming now that $K$ is complete, we have $a^m_ito a_iin K; 1le ile n$.



                                      To finish, consider the vector $x=a_1e^1+cdots +a_ne^n$.



                                      Clearly, $xin F$ and now, another application of the triangle inequality shows that



                                      $|x-x_n|to 0$, from which it follows that $x_nto xin F$







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited 3 hours ago

























                                      answered 3 hours ago









                                      MatematletaMatematleta

                                      11.4k2920




                                      11.4k2920






























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