A diifficulty in understanding a sentence in a paragraph in Guillemin and Pollack p.77











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The paragraph is given below:



enter image description here



But I have a difficulty in understanding the sentence starting in the forth line by "If we furthur ...." until its end, could anyone explain it for me please?



thanks!






general-topology differential-topology compactness







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    up vote
    1
    down vote

    favorite












    The paragraph is given below:



    enter image description here



    But I have a difficulty in understanding the sentence starting in the forth line by "If we furthur ...." until its end, could anyone explain it for me please?



    thanks!






    general-topology differential-topology compactness







    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      The paragraph is given below:



      enter image description here



      But I have a difficulty in understanding the sentence starting in the forth line by "If we furthur ...." until its end, could anyone explain it for me please?



      thanks!






      general-topology differential-topology compactness







      share|cite|improve this question













      The paragraph is given below:



      enter image description here



      But I have a difficulty in understanding the sentence starting in the forth line by "If we furthur ...." until its end, could anyone explain it for me please?



      thanks!






      general-topology differential-topology compactness




      general-topology differential-topology compactness






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked 1 hour ago









      hopefully

      125112




      125112






















          2 Answers
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          This is all basic point-set stuff. $X cap Z$ is a $0$-dimensional manifold, so it is discrete. If both $X$ and $Z$ are closed then so is $X cap Z$. If, say, $X$ is compact, then $X cap Z subseteq X$. Hence $X cap Z$ is a closed subset of a compact space, so it is compact. Since it is compact and discrete, it is finite.






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            If $X$ and $Z$ are closed and at least one of them is compact, then $Xcap Z$ is closed and compact (the intersection of closed sets is closed, and a closed subset of a compact set is compact). Then, the statement is that compact zero-dimensional submanifolds must be finite, which is clear. Zero-dimensional manifolds are discrete, and discrete compact sets are finite.






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              2 Answers
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              2 Answers
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              This is all basic point-set stuff. $X cap Z$ is a $0$-dimensional manifold, so it is discrete. If both $X$ and $Z$ are closed then so is $X cap Z$. If, say, $X$ is compact, then $X cap Z subseteq X$. Hence $X cap Z$ is a closed subset of a compact space, so it is compact. Since it is compact and discrete, it is finite.






              share|cite|improve this answer

























                up vote
                4
                down vote













                This is all basic point-set stuff. $X cap Z$ is a $0$-dimensional manifold, so it is discrete. If both $X$ and $Z$ are closed then so is $X cap Z$. If, say, $X$ is compact, then $X cap Z subseteq X$. Hence $X cap Z$ is a closed subset of a compact space, so it is compact. Since it is compact and discrete, it is finite.






                share|cite|improve this answer























                  up vote
                  4
                  down vote










                  up vote
                  4
                  down vote









                  This is all basic point-set stuff. $X cap Z$ is a $0$-dimensional manifold, so it is discrete. If both $X$ and $Z$ are closed then so is $X cap Z$. If, say, $X$ is compact, then $X cap Z subseteq X$. Hence $X cap Z$ is a closed subset of a compact space, so it is compact. Since it is compact and discrete, it is finite.






                  share|cite|improve this answer












                  This is all basic point-set stuff. $X cap Z$ is a $0$-dimensional manifold, so it is discrete. If both $X$ and $Z$ are closed then so is $X cap Z$. If, say, $X$ is compact, then $X cap Z subseteq X$. Hence $X cap Z$ is a closed subset of a compact space, so it is compact. Since it is compact and discrete, it is finite.







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                  answered 1 hour ago









                  Randall

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                  8,77811128






















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                      If $X$ and $Z$ are closed and at least one of them is compact, then $Xcap Z$ is closed and compact (the intersection of closed sets is closed, and a closed subset of a compact set is compact). Then, the statement is that compact zero-dimensional submanifolds must be finite, which is clear. Zero-dimensional manifolds are discrete, and discrete compact sets are finite.






                      share|cite|improve this answer

























                        up vote
                        2
                        down vote













                        If $X$ and $Z$ are closed and at least one of them is compact, then $Xcap Z$ is closed and compact (the intersection of closed sets is closed, and a closed subset of a compact set is compact). Then, the statement is that compact zero-dimensional submanifolds must be finite, which is clear. Zero-dimensional manifolds are discrete, and discrete compact sets are finite.






                        share|cite|improve this answer























                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          If $X$ and $Z$ are closed and at least one of them is compact, then $Xcap Z$ is closed and compact (the intersection of closed sets is closed, and a closed subset of a compact set is compact). Then, the statement is that compact zero-dimensional submanifolds must be finite, which is clear. Zero-dimensional manifolds are discrete, and discrete compact sets are finite.






                          share|cite|improve this answer












                          If $X$ and $Z$ are closed and at least one of them is compact, then $Xcap Z$ is closed and compact (the intersection of closed sets is closed, and a closed subset of a compact set is compact). Then, the statement is that compact zero-dimensional submanifolds must be finite, which is clear. Zero-dimensional manifolds are discrete, and discrete compact sets are finite.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 1 hour ago









                          Rolf Hoyer

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                          10.9k31629






























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