$bigcup_{alphain C}alpha A$ is compact in the topological vector space $X$












0












$begingroup$


Let $X$ be a topological vector space, with $A subset X$ and $C subset mathbb{K}$ both compact subsets.



Why is $bigcup_{alphain C}alpha A$ compact in $X$?






general-topology topological-vector-spaces







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$endgroup$

















    0












    $begingroup$


    Let $X$ be a topological vector space, with $A subset X$ and $C subset mathbb{K}$ both compact subsets.



    Why is $bigcup_{alphain C}alpha A$ compact in $X$?






    general-topology topological-vector-spaces







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $X$ be a topological vector space, with $A subset X$ and $C subset mathbb{K}$ both compact subsets.



      Why is $bigcup_{alphain C}alpha A$ compact in $X$?






      general-topology topological-vector-spaces







      share|cite|improve this question











      $endgroup$




      Let $X$ be a topological vector space, with $A subset X$ and $C subset mathbb{K}$ both compact subsets.



      Why is $bigcup_{alphain C}alpha A$ compact in $X$?






      general-topology topological-vector-spaces




      general-topology topological-vector-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 9 at 10:20









      egreg

      186k1486208




      186k1486208










      asked Jan 9 at 10:17









      Anas BOUALIIAnas BOUALII

      1408




      1408






















          2 Answers
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          $begingroup$

          $cup_{alpha in C} alpha A$ is the image of $Ctimes A$ under the continuous map $(c,y) to cy$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Hint: $Ctimes A$ is compact in $mathbb{K}times X$ and multiplication by scalars is continuous.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Yes, we always have $C. A= cup_{alphain C}alpha A$. But what if $A$ is just closed ?
              $endgroup$
              – Anas BOUALII
              Jan 9 at 10:24












            • $begingroup$
              @AnasBOUALII $A=X$ is closed and $C={1}$ is compact.
              $endgroup$
              – egreg
              Jan 9 at 10:26










            • $begingroup$
              Right, probably in case $A$ is closed, then $cup_{alphain C}alpha A$ is closed too.
              $endgroup$
              – Anas BOUALII
              Jan 9 at 10:33












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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            $cup_{alpha in C} alpha A$ is the image of $Ctimes A$ under the continuous map $(c,y) to cy$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              $cup_{alpha in C} alpha A$ is the image of $Ctimes A$ under the continuous map $(c,y) to cy$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                $cup_{alpha in C} alpha A$ is the image of $Ctimes A$ under the continuous map $(c,y) to cy$.






                share|cite|improve this answer









                $endgroup$



                $cup_{alpha in C} alpha A$ is the image of $Ctimes A$ under the continuous map $(c,y) to cy$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 9 at 10:27









                Kavi Rama MurthyKavi Rama Murthy

                75.2k53270




                75.2k53270























                    1












                    $begingroup$

                    Hint: $Ctimes A$ is compact in $mathbb{K}times X$ and multiplication by scalars is continuous.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Yes, we always have $C. A= cup_{alphain C}alpha A$. But what if $A$ is just closed ?
                      $endgroup$
                      – Anas BOUALII
                      Jan 9 at 10:24












                    • $begingroup$
                      @AnasBOUALII $A=X$ is closed and $C={1}$ is compact.
                      $endgroup$
                      – egreg
                      Jan 9 at 10:26










                    • $begingroup$
                      Right, probably in case $A$ is closed, then $cup_{alphain C}alpha A$ is closed too.
                      $endgroup$
                      – Anas BOUALII
                      Jan 9 at 10:33
















                    1












                    $begingroup$

                    Hint: $Ctimes A$ is compact in $mathbb{K}times X$ and multiplication by scalars is continuous.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Yes, we always have $C. A= cup_{alphain C}alpha A$. But what if $A$ is just closed ?
                      $endgroup$
                      – Anas BOUALII
                      Jan 9 at 10:24












                    • $begingroup$
                      @AnasBOUALII $A=X$ is closed and $C={1}$ is compact.
                      $endgroup$
                      – egreg
                      Jan 9 at 10:26










                    • $begingroup$
                      Right, probably in case $A$ is closed, then $cup_{alphain C}alpha A$ is closed too.
                      $endgroup$
                      – Anas BOUALII
                      Jan 9 at 10:33














                    1












                    1








                    1





                    $begingroup$

                    Hint: $Ctimes A$ is compact in $mathbb{K}times X$ and multiplication by scalars is continuous.






                    share|cite|improve this answer









                    $endgroup$



                    Hint: $Ctimes A$ is compact in $mathbb{K}times X$ and multiplication by scalars is continuous.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 9 at 10:22









                    egregegreg

                    186k1486208




                    186k1486208












                    • $begingroup$
                      Yes, we always have $C. A= cup_{alphain C}alpha A$. But what if $A$ is just closed ?
                      $endgroup$
                      – Anas BOUALII
                      Jan 9 at 10:24












                    • $begingroup$
                      @AnasBOUALII $A=X$ is closed and $C={1}$ is compact.
                      $endgroup$
                      – egreg
                      Jan 9 at 10:26










                    • $begingroup$
                      Right, probably in case $A$ is closed, then $cup_{alphain C}alpha A$ is closed too.
                      $endgroup$
                      – Anas BOUALII
                      Jan 9 at 10:33


















                    • $begingroup$
                      Yes, we always have $C. A= cup_{alphain C}alpha A$. But what if $A$ is just closed ?
                      $endgroup$
                      – Anas BOUALII
                      Jan 9 at 10:24












                    • $begingroup$
                      @AnasBOUALII $A=X$ is closed and $C={1}$ is compact.
                      $endgroup$
                      – egreg
                      Jan 9 at 10:26










                    • $begingroup$
                      Right, probably in case $A$ is closed, then $cup_{alphain C}alpha A$ is closed too.
                      $endgroup$
                      – Anas BOUALII
                      Jan 9 at 10:33
















                    $begingroup$
                    Yes, we always have $C. A= cup_{alphain C}alpha A$. But what if $A$ is just closed ?
                    $endgroup$
                    – Anas BOUALII
                    Jan 9 at 10:24






                    $begingroup$
                    Yes, we always have $C. A= cup_{alphain C}alpha A$. But what if $A$ is just closed ?
                    $endgroup$
                    – Anas BOUALII
                    Jan 9 at 10:24














                    $begingroup$
                    @AnasBOUALII $A=X$ is closed and $C={1}$ is compact.
                    $endgroup$
                    – egreg
                    Jan 9 at 10:26




                    $begingroup$
                    @AnasBOUALII $A=X$ is closed and $C={1}$ is compact.
                    $endgroup$
                    – egreg
                    Jan 9 at 10:26












                    $begingroup$
                    Right, probably in case $A$ is closed, then $cup_{alphain C}alpha A$ is closed too.
                    $endgroup$
                    – Anas BOUALII
                    Jan 9 at 10:33




                    $begingroup$
                    Right, probably in case $A$ is closed, then $cup_{alphain C}alpha A$ is closed too.
                    $endgroup$
                    – Anas BOUALII
                    Jan 9 at 10:33


















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