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I feel blocked with this claim - it sounds intuitively true, just thinking as a jellyfish entering a real line, the intersection of her legs with the real line is certainly finite since the jellyfish is compact - but I stuck with why.

$X$ and $Z$ are closed submanifolds inside $Y$ with complementary dimension. If at lease one of them, say $X$, is compact, and $X \pitchfork Z$, then $X \cap Z$ must be a finite set of points.

I understand that $X \cap Z$ is a zero-dimensional manifold. So it must be a series of disjoint points.

Then I start to guess: this conclusion is perhaps related to each sequence in a compact set has finite subsequence? So let $X \cap Z$ be the sequence, and hence it needs to be finite?

The statement is from Guillemin and Pollack's Differential Topology.

1LiterTears
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2 Answers2

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AAs you say, $X\cap Z$ is zero dimensional. If $X$ is compact and $Z$ is closed, then $X\cap Z$ is a closed set in $X$. If it is not finite, then there is a point $x\in X\cap Z$ and a sequence $(x_n)_{n\geq1}$ with values in $X\cap Z$ and all distinct from $x$ such that $x_n\to x$ as $n\to\infty$.

Can you reach a contradition from this?

Notice that $x$ is a point of transverse intersection, so you know how the whole thing is near $x$.

Mariano Suárez-Álvarez
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  • Oh finally got it.. So since $n$ can not really reach $\infty$, $x_n$ can not really reach $x$. That is saying $X \cap Z$ does not contain the limit point $x$, which contradicts with the definition of closed set. (or contradicts with the fact that you assumed $x \in X \cap Z$.) – 1LiterTears Jul 13 '13 at 21:14
  • Huh? You should probably pick a textbook on calculus and review the chapter on sequences and their limits before continuing with Guillemin and Pollack. – Mariano Suárez-Álvarez Jul 13 '13 at 22:11
  • I tried to check the book on calculus and still checking, haven't found content inspiring to this yet. Just to rephrase my comment: given $x_n \to n$ as $n \to \infty$, $x$ will never be assumed. Hence contradicts with the assumption that $x \in X \cap Z$. Is this still wildly wrong...? – 1LiterTears Jul 13 '13 at 23:37
  • You need to think about what a $0$-dimensional manifold must look like topologically. – Ted Shifrin Jul 13 '13 at 23:40
  • I know it is a series of disjoint points. Should I look for more topological properties...? – 1LiterTears Jul 13 '13 at 23:43
  • oh.... so if there are infinite disjoint points in the 0-dimensional manifold, their covering would be infinite, which contradicts with the fact that compact set have finite subcovering!! – 1LiterTears Jul 13 '13 at 23:44
  • Indeed, the subset must be discrete — each point must be open in the subspace topology. That's where you get what you just said. – Ted Shifrin Jul 14 '13 at 00:48
  • At some moment transversality must be used. Without it, the result is false. Why is trhe intersection discrete? – Mariano Suárez-Álvarez Jul 14 '13 at 01:01
  • @MarianoSuárez-Alvarez: Transversality is telling us that it's a $0$-dimensional submanifold of $X$. – Ted Shifrin Jul 14 '13 at 02:05
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Let me clarify this statement. In fact, the derivation based on the definition of zero-manifold is not necessary and non-intrinsic. Just like what Mariano Suárez-Alvarez suggested, we should trace back to the proof of transversality (page 28 of GP) to fill the gap.

Recall that $X$ is compact, $Z$ is closed, and $\dim X+\dim Z=\dim Y$. Consider the inclusion $i:X\rightarrow Y(\supseteq Z)$ just like what is done on page 29 of GP. $X\pitchfork Z$ means that $\forall x\in X\cap Z$, $T_x(X)+T_x(Z)=T_x(Y)$, which is equivalent to that (see page 28) $g\circ i: X\rightarrow \mathbb{R}^{l}$ is a submerssion at the point $x\in i^{-1}(Z)=X\cap Z$, where $l=\dim Y-\dim Z$. Also consider the surjective map $\mathrm{d}(g\circ i)_x:T_x(X)\rightarrow \mathbb{R}^l$, since $\dim T_x(X)=\dim X=\dim Y-\dim Z=l$, we conclude that $\mathrm{d}(g\circ i)_x$ is a isomorphism. Now comes the inverse function theorem, which asserts the existence of a diffeomorphism between a neighborhood of $x$ and a neighborhood of $g\circ i(x)$, i.e., the local diffeomorphism.

If the set $X\cap Z$ is infinite, the compactness establishes a limit point, say $x$. Performing the same derivation for this $x$ as above, we get a neighborhood $U$ of $x$ diffeomorphic to the neighborhood of its image $g\circ i(x)$. But $U$ contains infinite points mapped to $g\circ i(x)$, which is a contradiction.

zenos
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