${x} times B subset mathbb{R}^{n+m}$ is compact if $B subset mathbb{R}^m$ is compact and $x in mathbb{R}^n$












1












$begingroup$


This is from Spivak Calculus on manifolds page 14.
I cannot use the fact that the cartesian product of two compact sets which are subsets of Euclidean space is compact because he uses what is written in the title to prove that fact.



It seems that I am only allowed to use the definition of compactness here, which is that a set is compact if every open cover of it has a finite subvoer.
(The book only deals with Euclidean space here, and Spivak defines an open set $U$ to be such that for every element $x in U$, there is an open rectangle $A$ such that $x in A subset U$.)



As he says in the book, this must be very easy to see, but I just cannot figure out.



Thanks in advance.






real-analysis general-topology analysis







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    This is from Spivak Calculus on manifolds page 14.
    I cannot use the fact that the cartesian product of two compact sets which are subsets of Euclidean space is compact because he uses what is written in the title to prove that fact.



    It seems that I am only allowed to use the definition of compactness here, which is that a set is compact if every open cover of it has a finite subvoer.
    (The book only deals with Euclidean space here, and Spivak defines an open set $U$ to be such that for every element $x in U$, there is an open rectangle $A$ such that $x in A subset U$.)



    As he says in the book, this must be very easy to see, but I just cannot figure out.



    Thanks in advance.






    real-analysis general-topology analysis







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      This is from Spivak Calculus on manifolds page 14.
      I cannot use the fact that the cartesian product of two compact sets which are subsets of Euclidean space is compact because he uses what is written in the title to prove that fact.



      It seems that I am only allowed to use the definition of compactness here, which is that a set is compact if every open cover of it has a finite subvoer.
      (The book only deals with Euclidean space here, and Spivak defines an open set $U$ to be such that for every element $x in U$, there is an open rectangle $A$ such that $x in A subset U$.)



      As he says in the book, this must be very easy to see, but I just cannot figure out.



      Thanks in advance.






      real-analysis general-topology analysis







      share|cite|improve this question









      $endgroup$




      This is from Spivak Calculus on manifolds page 14.
      I cannot use the fact that the cartesian product of two compact sets which are subsets of Euclidean space is compact because he uses what is written in the title to prove that fact.



      It seems that I am only allowed to use the definition of compactness here, which is that a set is compact if every open cover of it has a finite subvoer.
      (The book only deals with Euclidean space here, and Spivak defines an open set $U$ to be such that for every element $x in U$, there is an open rectangle $A$ such that $x in A subset U$.)



      As he says in the book, this must be very easy to see, but I just cannot figure out.



      Thanks in advance.






      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 Jan 9 at 5:11









      Hunnam Hunnam

      735




      735






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Well, I think this comes directly from the definition. For any set of cover ${U_k}_{kinmathcal{K}}$, then we can project it to the later $mathbb{R}^m$ sub-space, set: $$
          V_k:=mathbb{P}(U_k),quadmathtt{where}quadmathbb{P}:mathbb{R}^{n+m}mapsto mathbb{R}^m
          $$
          which maintains a set of open set in $mathbb{R}^m$ and it forms a cover of B.



          Due to B is compact, there exists a finite sub-cover $V_{k_i}$. Then by the axiom of Choice, pick up the corresponding $U_{k_i}$ and it forms a finite sub-cover of ${x}times B$.



          I think this should work. :)






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks so much for your help! But could you explain more about why the projection still preserves the openness of $V_k$?
            $endgroup$
            – Hunnam
            Jan 9 at 5:30










          • $begingroup$
            I think by the definition of open set will do. For any point $xin V_k$, then it comes from some point $(x,y)in U_k$. Due to $U_k$ is open, hence there exists a small ball containing $(x,y)$. After the projection, the ball becomes a ball or a line (whatsoever), this makes x still an inside point of $V_k$. Hence $V_k$ is open.
            $endgroup$
            – Zixiao_Liu
            Jan 9 at 5:43












          • $begingroup$
            I would appreciate it If you could answer just one more question. I think I understand everything but the last part. After figuring out that $V_k$ forms a finite sub cover of $B$, how do we know the corresponding $U_k$ forms a finite sub cover of ${x} times B$? This really looks obvious, but I just don't understand
            $endgroup$
            – Hunnam
            Jan 11 at 5:42










          • $begingroup$
            By the definition of "cover", we have that $Bsubsetbigcup_{k=1}^NV_k$. Note that for the original cover of ${x}times B$, we may assume WLOG that each $U_kcap {x}times Bnot=varnothing$. Hence this makes the subcover corresponding $V_k$ satisfies ${x}times V_ksubset U_k$. Now we combine these stuffs together and obtaine $${x}times Bsubset {x}times (bigcup_{k=1}^NV_k)subset bigcup_{k=1}^NU_k$$. I hope this is enough for you. :)
            $endgroup$
            – Zixiao_Liu
            Jan 12 at 13:29





















          1












          $begingroup$

          The function $f:Bto mathbb R^{m+n}$ defined by $ymapsto xtimes y$ is continuous. Therefore, as $B$ is compact so is $f(B)={x}times B$.



          If you want to use covers, but avoid projections, then note that a cover of ${x}times B$, endowed with the subspace topology, is a collection $mathscr A={V_icap ({x}times B):iin I}$ where $V_i$ are open sets in $mathbb R^{n+m}$. Each $V_i$ is a union of basis elements $(U_1timescdotstimes U_n)times U_{n+1}timescdotstimes U_{n+m}$ so we may assume without loss of generality that our cover is a collection of these in the first place. i.e. $mathscr A=(U_{1i}timescdotstimes U_{ni})times U_{(n+1)i}timescdotstimes U_{(n+m)i})_i:iin I}.$ Now, then $Bsubseteq bigcup_i (U_{(n+1)i}timescdotstimes U_{(n+m)i})_i$ and as $B$ is compact, it is covered by finitely many elements ${U_{(n+1)i}timescdotstimes U_{(n+m)i}}^N_{i=1}$. Then, ${x}times B$ is covered by ${U_{1i}timescdotstimes U_{ni})times U_{(n+1)i}timescdotstimes U_{(n+m)i}}^N_{i=1}$.






          share|cite|improve this answer











          $endgroup$













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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Well, I think this comes directly from the definition. For any set of cover ${U_k}_{kinmathcal{K}}$, then we can project it to the later $mathbb{R}^m$ sub-space, set: $$
            V_k:=mathbb{P}(U_k),quadmathtt{where}quadmathbb{P}:mathbb{R}^{n+m}mapsto mathbb{R}^m
            $$
            which maintains a set of open set in $mathbb{R}^m$ and it forms a cover of B.



            Due to B is compact, there exists a finite sub-cover $V_{k_i}$. Then by the axiom of Choice, pick up the corresponding $U_{k_i}$ and it forms a finite sub-cover of ${x}times B$.



            I think this should work. :)






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks so much for your help! But could you explain more about why the projection still preserves the openness of $V_k$?
              $endgroup$
              – Hunnam
              Jan 9 at 5:30










            • $begingroup$
              I think by the definition of open set will do. For any point $xin V_k$, then it comes from some point $(x,y)in U_k$. Due to $U_k$ is open, hence there exists a small ball containing $(x,y)$. After the projection, the ball becomes a ball or a line (whatsoever), this makes x still an inside point of $V_k$. Hence $V_k$ is open.
              $endgroup$
              – Zixiao_Liu
              Jan 9 at 5:43












            • $begingroup$
              I would appreciate it If you could answer just one more question. I think I understand everything but the last part. After figuring out that $V_k$ forms a finite sub cover of $B$, how do we know the corresponding $U_k$ forms a finite sub cover of ${x} times B$? This really looks obvious, but I just don't understand
              $endgroup$
              – Hunnam
              Jan 11 at 5:42










            • $begingroup$
              By the definition of "cover", we have that $Bsubsetbigcup_{k=1}^NV_k$. Note that for the original cover of ${x}times B$, we may assume WLOG that each $U_kcap {x}times Bnot=varnothing$. Hence this makes the subcover corresponding $V_k$ satisfies ${x}times V_ksubset U_k$. Now we combine these stuffs together and obtaine $${x}times Bsubset {x}times (bigcup_{k=1}^NV_k)subset bigcup_{k=1}^NU_k$$. I hope this is enough for you. :)
              $endgroup$
              – Zixiao_Liu
              Jan 12 at 13:29


















            1












            $begingroup$

            Well, I think this comes directly from the definition. For any set of cover ${U_k}_{kinmathcal{K}}$, then we can project it to the later $mathbb{R}^m$ sub-space, set: $$
            V_k:=mathbb{P}(U_k),quadmathtt{where}quadmathbb{P}:mathbb{R}^{n+m}mapsto mathbb{R}^m
            $$
            which maintains a set of open set in $mathbb{R}^m$ and it forms a cover of B.



            Due to B is compact, there exists a finite sub-cover $V_{k_i}$. Then by the axiom of Choice, pick up the corresponding $U_{k_i}$ and it forms a finite sub-cover of ${x}times B$.



            I think this should work. :)






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks so much for your help! But could you explain more about why the projection still preserves the openness of $V_k$?
              $endgroup$
              – Hunnam
              Jan 9 at 5:30










            • $begingroup$
              I think by the definition of open set will do. For any point $xin V_k$, then it comes from some point $(x,y)in U_k$. Due to $U_k$ is open, hence there exists a small ball containing $(x,y)$. After the projection, the ball becomes a ball or a line (whatsoever), this makes x still an inside point of $V_k$. Hence $V_k$ is open.
              $endgroup$
              – Zixiao_Liu
              Jan 9 at 5:43












            • $begingroup$
              I would appreciate it If you could answer just one more question. I think I understand everything but the last part. After figuring out that $V_k$ forms a finite sub cover of $B$, how do we know the corresponding $U_k$ forms a finite sub cover of ${x} times B$? This really looks obvious, but I just don't understand
              $endgroup$
              – Hunnam
              Jan 11 at 5:42










            • $begingroup$
              By the definition of "cover", we have that $Bsubsetbigcup_{k=1}^NV_k$. Note that for the original cover of ${x}times B$, we may assume WLOG that each $U_kcap {x}times Bnot=varnothing$. Hence this makes the subcover corresponding $V_k$ satisfies ${x}times V_ksubset U_k$. Now we combine these stuffs together and obtaine $${x}times Bsubset {x}times (bigcup_{k=1}^NV_k)subset bigcup_{k=1}^NU_k$$. I hope this is enough for you. :)
              $endgroup$
              – Zixiao_Liu
              Jan 12 at 13:29
















            1












            1








            1





            $begingroup$

            Well, I think this comes directly from the definition. For any set of cover ${U_k}_{kinmathcal{K}}$, then we can project it to the later $mathbb{R}^m$ sub-space, set: $$
            V_k:=mathbb{P}(U_k),quadmathtt{where}quadmathbb{P}:mathbb{R}^{n+m}mapsto mathbb{R}^m
            $$
            which maintains a set of open set in $mathbb{R}^m$ and it forms a cover of B.



            Due to B is compact, there exists a finite sub-cover $V_{k_i}$. Then by the axiom of Choice, pick up the corresponding $U_{k_i}$ and it forms a finite sub-cover of ${x}times B$.



            I think this should work. :)






            share|cite|improve this answer









            $endgroup$



            Well, I think this comes directly from the definition. For any set of cover ${U_k}_{kinmathcal{K}}$, then we can project it to the later $mathbb{R}^m$ sub-space, set: $$
            V_k:=mathbb{P}(U_k),quadmathtt{where}quadmathbb{P}:mathbb{R}^{n+m}mapsto mathbb{R}^m
            $$
            which maintains a set of open set in $mathbb{R}^m$ and it forms a cover of B.



            Due to B is compact, there exists a finite sub-cover $V_{k_i}$. Then by the axiom of Choice, pick up the corresponding $U_{k_i}$ and it forms a finite sub-cover of ${x}times B$.



            I think this should work. :)







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 9 at 5:18









            Zixiao_LiuZixiao_Liu

            566




            566












            • $begingroup$
              Thanks so much for your help! But could you explain more about why the projection still preserves the openness of $V_k$?
              $endgroup$
              – Hunnam
              Jan 9 at 5:30










            • $begingroup$
              I think by the definition of open set will do. For any point $xin V_k$, then it comes from some point $(x,y)in U_k$. Due to $U_k$ is open, hence there exists a small ball containing $(x,y)$. After the projection, the ball becomes a ball or a line (whatsoever), this makes x still an inside point of $V_k$. Hence $V_k$ is open.
              $endgroup$
              – Zixiao_Liu
              Jan 9 at 5:43












            • $begingroup$
              I would appreciate it If you could answer just one more question. I think I understand everything but the last part. After figuring out that $V_k$ forms a finite sub cover of $B$, how do we know the corresponding $U_k$ forms a finite sub cover of ${x} times B$? This really looks obvious, but I just don't understand
              $endgroup$
              – Hunnam
              Jan 11 at 5:42










            • $begingroup$
              By the definition of "cover", we have that $Bsubsetbigcup_{k=1}^NV_k$. Note that for the original cover of ${x}times B$, we may assume WLOG that each $U_kcap {x}times Bnot=varnothing$. Hence this makes the subcover corresponding $V_k$ satisfies ${x}times V_ksubset U_k$. Now we combine these stuffs together and obtaine $${x}times Bsubset {x}times (bigcup_{k=1}^NV_k)subset bigcup_{k=1}^NU_k$$. I hope this is enough for you. :)
              $endgroup$
              – Zixiao_Liu
              Jan 12 at 13:29




















            • $begingroup$
              Thanks so much for your help! But could you explain more about why the projection still preserves the openness of $V_k$?
              $endgroup$
              – Hunnam
              Jan 9 at 5:30










            • $begingroup$
              I think by the definition of open set will do. For any point $xin V_k$, then it comes from some point $(x,y)in U_k$. Due to $U_k$ is open, hence there exists a small ball containing $(x,y)$. After the projection, the ball becomes a ball or a line (whatsoever), this makes x still an inside point of $V_k$. Hence $V_k$ is open.
              $endgroup$
              – Zixiao_Liu
              Jan 9 at 5:43












            • $begingroup$
              I would appreciate it If you could answer just one more question. I think I understand everything but the last part. After figuring out that $V_k$ forms a finite sub cover of $B$, how do we know the corresponding $U_k$ forms a finite sub cover of ${x} times B$? This really looks obvious, but I just don't understand
              $endgroup$
              – Hunnam
              Jan 11 at 5:42










            • $begingroup$
              By the definition of "cover", we have that $Bsubsetbigcup_{k=1}^NV_k$. Note that for the original cover of ${x}times B$, we may assume WLOG that each $U_kcap {x}times Bnot=varnothing$. Hence this makes the subcover corresponding $V_k$ satisfies ${x}times V_ksubset U_k$. Now we combine these stuffs together and obtaine $${x}times Bsubset {x}times (bigcup_{k=1}^NV_k)subset bigcup_{k=1}^NU_k$$. I hope this is enough for you. :)
              $endgroup$
              – Zixiao_Liu
              Jan 12 at 13:29


















            $begingroup$
            Thanks so much for your help! But could you explain more about why the projection still preserves the openness of $V_k$?
            $endgroup$
            – Hunnam
            Jan 9 at 5:30




            $begingroup$
            Thanks so much for your help! But could you explain more about why the projection still preserves the openness of $V_k$?
            $endgroup$
            – Hunnam
            Jan 9 at 5:30












            $begingroup$
            I think by the definition of open set will do. For any point $xin V_k$, then it comes from some point $(x,y)in U_k$. Due to $U_k$ is open, hence there exists a small ball containing $(x,y)$. After the projection, the ball becomes a ball or a line (whatsoever), this makes x still an inside point of $V_k$. Hence $V_k$ is open.
            $endgroup$
            – Zixiao_Liu
            Jan 9 at 5:43






            $begingroup$
            I think by the definition of open set will do. For any point $xin V_k$, then it comes from some point $(x,y)in U_k$. Due to $U_k$ is open, hence there exists a small ball containing $(x,y)$. After the projection, the ball becomes a ball or a line (whatsoever), this makes x still an inside point of $V_k$. Hence $V_k$ is open.
            $endgroup$
            – Zixiao_Liu
            Jan 9 at 5:43














            $begingroup$
            I would appreciate it If you could answer just one more question. I think I understand everything but the last part. After figuring out that $V_k$ forms a finite sub cover of $B$, how do we know the corresponding $U_k$ forms a finite sub cover of ${x} times B$? This really looks obvious, but I just don't understand
            $endgroup$
            – Hunnam
            Jan 11 at 5:42




            $begingroup$
            I would appreciate it If you could answer just one more question. I think I understand everything but the last part. After figuring out that $V_k$ forms a finite sub cover of $B$, how do we know the corresponding $U_k$ forms a finite sub cover of ${x} times B$? This really looks obvious, but I just don't understand
            $endgroup$
            – Hunnam
            Jan 11 at 5:42












            $begingroup$
            By the definition of "cover", we have that $Bsubsetbigcup_{k=1}^NV_k$. Note that for the original cover of ${x}times B$, we may assume WLOG that each $U_kcap {x}times Bnot=varnothing$. Hence this makes the subcover corresponding $V_k$ satisfies ${x}times V_ksubset U_k$. Now we combine these stuffs together and obtaine $${x}times Bsubset {x}times (bigcup_{k=1}^NV_k)subset bigcup_{k=1}^NU_k$$. I hope this is enough for you. :)
            $endgroup$
            – Zixiao_Liu
            Jan 12 at 13:29






            $begingroup$
            By the definition of "cover", we have that $Bsubsetbigcup_{k=1}^NV_k$. Note that for the original cover of ${x}times B$, we may assume WLOG that each $U_kcap {x}times Bnot=varnothing$. Hence this makes the subcover corresponding $V_k$ satisfies ${x}times V_ksubset U_k$. Now we combine these stuffs together and obtaine $${x}times Bsubset {x}times (bigcup_{k=1}^NV_k)subset bigcup_{k=1}^NU_k$$. I hope this is enough for you. :)
            $endgroup$
            – Zixiao_Liu
            Jan 12 at 13:29













            1












            $begingroup$

            The function $f:Bto mathbb R^{m+n}$ defined by $ymapsto xtimes y$ is continuous. Therefore, as $B$ is compact so is $f(B)={x}times B$.



            If you want to use covers, but avoid projections, then note that a cover of ${x}times B$, endowed with the subspace topology, is a collection $mathscr A={V_icap ({x}times B):iin I}$ where $V_i$ are open sets in $mathbb R^{n+m}$. Each $V_i$ is a union of basis elements $(U_1timescdotstimes U_n)times U_{n+1}timescdotstimes U_{n+m}$ so we may assume without loss of generality that our cover is a collection of these in the first place. i.e. $mathscr A=(U_{1i}timescdotstimes U_{ni})times U_{(n+1)i}timescdotstimes U_{(n+m)i})_i:iin I}.$ Now, then $Bsubseteq bigcup_i (U_{(n+1)i}timescdotstimes U_{(n+m)i})_i$ and as $B$ is compact, it is covered by finitely many elements ${U_{(n+1)i}timescdotstimes U_{(n+m)i}}^N_{i=1}$. Then, ${x}times B$ is covered by ${U_{1i}timescdotstimes U_{ni})times U_{(n+1)i}timescdotstimes U_{(n+m)i}}^N_{i=1}$.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              The function $f:Bto mathbb R^{m+n}$ defined by $ymapsto xtimes y$ is continuous. Therefore, as $B$ is compact so is $f(B)={x}times B$.



              If you want to use covers, but avoid projections, then note that a cover of ${x}times B$, endowed with the subspace topology, is a collection $mathscr A={V_icap ({x}times B):iin I}$ where $V_i$ are open sets in $mathbb R^{n+m}$. Each $V_i$ is a union of basis elements $(U_1timescdotstimes U_n)times U_{n+1}timescdotstimes U_{n+m}$ so we may assume without loss of generality that our cover is a collection of these in the first place. i.e. $mathscr A=(U_{1i}timescdotstimes U_{ni})times U_{(n+1)i}timescdotstimes U_{(n+m)i})_i:iin I}.$ Now, then $Bsubseteq bigcup_i (U_{(n+1)i}timescdotstimes U_{(n+m)i})_i$ and as $B$ is compact, it is covered by finitely many elements ${U_{(n+1)i}timescdotstimes U_{(n+m)i}}^N_{i=1}$. Then, ${x}times B$ is covered by ${U_{1i}timescdotstimes U_{ni})times U_{(n+1)i}timescdotstimes U_{(n+m)i}}^N_{i=1}$.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                The function $f:Bto mathbb R^{m+n}$ defined by $ymapsto xtimes y$ is continuous. Therefore, as $B$ is compact so is $f(B)={x}times B$.



                If you want to use covers, but avoid projections, then note that a cover of ${x}times B$, endowed with the subspace topology, is a collection $mathscr A={V_icap ({x}times B):iin I}$ where $V_i$ are open sets in $mathbb R^{n+m}$. Each $V_i$ is a union of basis elements $(U_1timescdotstimes U_n)times U_{n+1}timescdotstimes U_{n+m}$ so we may assume without loss of generality that our cover is a collection of these in the first place. i.e. $mathscr A=(U_{1i}timescdotstimes U_{ni})times U_{(n+1)i}timescdotstimes U_{(n+m)i})_i:iin I}.$ Now, then $Bsubseteq bigcup_i (U_{(n+1)i}timescdotstimes U_{(n+m)i})_i$ and as $B$ is compact, it is covered by finitely many elements ${U_{(n+1)i}timescdotstimes U_{(n+m)i}}^N_{i=1}$. Then, ${x}times B$ is covered by ${U_{1i}timescdotstimes U_{ni})times U_{(n+1)i}timescdotstimes U_{(n+m)i}}^N_{i=1}$.






                share|cite|improve this answer











                $endgroup$



                The function $f:Bto mathbb R^{m+n}$ defined by $ymapsto xtimes y$ is continuous. Therefore, as $B$ is compact so is $f(B)={x}times B$.



                If you want to use covers, but avoid projections, then note that a cover of ${x}times B$, endowed with the subspace topology, is a collection $mathscr A={V_icap ({x}times B):iin I}$ where $V_i$ are open sets in $mathbb R^{n+m}$. Each $V_i$ is a union of basis elements $(U_1timescdotstimes U_n)times U_{n+1}timescdotstimes U_{n+m}$ so we may assume without loss of generality that our cover is a collection of these in the first place. i.e. $mathscr A=(U_{1i}timescdotstimes U_{ni})times U_{(n+1)i}timescdotstimes U_{(n+m)i})_i:iin I}.$ Now, then $Bsubseteq bigcup_i (U_{(n+1)i}timescdotstimes U_{(n+m)i})_i$ and as $B$ is compact, it is covered by finitely many elements ${U_{(n+1)i}timescdotstimes U_{(n+m)i}}^N_{i=1}$. Then, ${x}times B$ is covered by ${U_{1i}timescdotstimes U_{ni})times U_{(n+1)i}timescdotstimes U_{(n+m)i}}^N_{i=1}$.







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                edited Jan 9 at 6:23

























                answered Jan 9 at 5:26









                MatematletaMatematleta

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                10.2k2918






























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