A diifficulty in understanding a sentence in a paragraph in Guillemin and Pollack p.77
$begingroup$
The paragraph is given below:
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
asked Dec 8 '18 at 1:49
hopefullyhopefully
311113
$endgroup$
add a comment |
$begingroup$
The paragraph is given below:
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
asked Dec 8 '18 at 1:49
hopefullyhopefully
311113
$endgroup$
add a comment |
$begingroup$
The paragraph is given below:
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
asked Dec 8 '18 at 1:49
hopefullyhopefully
311113
$endgroup$
The paragraph is given below:
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
asked Dec 8 '18 at 1:49
hopefullyhopefully
311113
asked Dec 8 '18 at 1:49
hopefullyhopefully
311113
asked Dec 8 '18 at 1:49
hopefullyhopefully
311113
asked Dec 8 '18 at 1:49
hopefullyhopefully
311113
asked Dec 8 '18 at 1:49
hopefullyhopefully
311113
311113
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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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.
answered Dec 8 '18 at 1:58
RandallRandall
9,65611230
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add a comment |
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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.
answered Dec 8 '18 at 1:57
Rolf HoyerRolf Hoyer
11.2k31629
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Dec 8 '18 at 1:58
RandallRandall
9,65611230
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 8 '18 at 1:58
RandallRandall
9,65611230
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 8 '18 at 1:58
RandallRandall
9,65611230
$endgroup$
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.
answered Dec 8 '18 at 1:58
RandallRandall
9,65611230
answered Dec 8 '18 at 1:58
RandallRandall
9,65611230
answered Dec 8 '18 at 1:58
RandallRandall
9,65611230
answered Dec 8 '18 at 1:58
RandallRandall
9,65611230
9,65611230
add a comment |
add a comment |
$begingroup$
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.
answered Dec 8 '18 at 1:57
Rolf HoyerRolf Hoyer
11.2k31629
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 8 '18 at 1:57
Rolf HoyerRolf Hoyer
11.2k31629
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 8 '18 at 1:57
Rolf HoyerRolf Hoyer
11.2k31629
$endgroup$
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.
answered Dec 8 '18 at 1:57
Rolf HoyerRolf Hoyer
11.2k31629
answered Dec 8 '18 at 1:57
Rolf HoyerRolf Hoyer
11.2k31629
answered Dec 8 '18 at 1:57
Rolf HoyerRolf Hoyer
11.2k31629
answered Dec 8 '18 at 1:57
Rolf HoyerRolf Hoyer
11.2k31629
11.2k31629
add a comment |
add a comment |
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