TITLE: Finite set on compact manifolds
QUESTION [0 upvotes]: 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.

REPLY [0 votes]: 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.
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