Doubt Regarding Problem $6.14$ in Lee's Smooth Manifolds

Question

I am trying to solve Problem 6.14 in Lee's smooth manifolds, which states that:

Let $S=\mathbb R^n\times\{0\}\subset \mathbb R^{n+1}$, and suppose $A\subset S$ is any arbitrarily closed subset, then there exists a properly embedded hypersurface $S'$ such that $S\cap S'=A$.

Before solving this, I tried to find some examples, but I literally can think of none that make sense to me.

Unless I am drastically misremembering my point set topology, the graph of $|x|$ should be a closed subset of $\mathbb R^2$, so let $A$ denote this subset. Then it follows by the problem of above that there should exist a hypersurface $S'\subset \mathbb R^3$ such that $\mathbb R^2\times\{0\}\cap S'=A$. But wouldn't this imply that $S'$ has some sort of cusp? The only hypersurface I can think of of $\mathbb R^3$ that would satisfy this is some form of cone, but that is only immersed.

Similarly, if we replaces $A$ with the figure $8$ curve, wouldn't this imply that the hypersurface self intersects itself so it couldn't be embedded?

You're picturing a transverse intersection. Remember (the very nontrivial result) that an arbitrary closed subset of $\Bbb R$ is the zero set of a smooth function — of course, $0$ will not be a regular value. — Ted Shifrin, May 10 '24 at 21:43
@TedShifrin Oh I see, so something like $f(x,y)=x^2-y^2$ satisfies satisfies $A=f^{-1}(0)$, and the graph $(x,y,f(x,y))$ is a smooth hypersurface? — Chris, May 10 '24 at 21:53
Would you mind posting a proof sketch after you solve it? I’m curious how to do this one. Thanks! — random0620, May 10 '24 at 22:20
@SamKirkiles I posted an answer to the question, the hard part is really Theorem 2.29. — Chris, May 10 '24 at 22:51

score 2 · Accepted Answer · edited May 11 '24 at 00:12

I think the example part is clear from the comments, and to show that any closed subset is $f^{-1}(0)$ for some smooth nonnegative function $f:M\rightarrow \mathbb R$ is Theorem 2.29 from Lee's smooth manifolds.

To prove the statement, let $A\subset \mathbb R^n\times\{0\}$ be a closed subset, and suppose it is the level set of a smooth function $f:\mathbb R^n\rightarrow \mathbb R$. Then the graph $\Gamma_f$ is an embedded submanifold of $\mathbb R^{n+1}$, and the intersection $\Gamma_f\cap \mathbb R^n\times\{0\}=A$ as the intersection are the set of all point that map to zero, i.e. $f^{-1}(0)$ which as stated earlier was equal to $A$.

Doubt Regarding Problem $6.14$ in Lee's Smooth Manifolds

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