If $f(x)\le g(x)\le h(x)$ for all $x$ and $\lim f(x)=\lim h(x)=L$ then $\lim g(x)=L$, right?
Don't think there's anything else we need
Well, if we have $\lim_{x\to a}$, then $a$ needs to be a limit point of the domains of everything, but that's just needed for the limits to be defined in the first place
@AkivaWeinberger then when take the limit tends to 0 for this inequality and you will see that squeeze theorem doesn't work for this inequality but why?
@MikeMiller Sorry, I left. No, not particularly. Just letting you know that I got work done today (I also worked through some exercises from chapter 1 in no particular order).
I skipped the chapter on Morse theory. Should I read it?
Given a set $A$ of real numbers, let the set $B$ be the set of all real numbers that are each less than every element in $A$. The greatest element in $B$ is the $\inf{A}$.
@BalarkaSen The only extension of the embedding result you got that you should know and isn't in G&P is a function space result. If $\dim N \geq 2\dim M + 1$ (and $M$ is compact), then $\text{Emb}(M,N)$ is dense in $C^\infty(M,N)$. Same with immersions, and you can drop the RHS to $2 \dim M$.
u19: I need to get on an airplane, so I may have to leave suddenly. Let me say something about the second step: a very standard-sounding proof starts with "Suppose, for the sake of contradiction, that $-y'$ is an upper bound of $A$ with $-y'<y$. … "
@AndyMiles I took a look and I think I would really need to dig deep in the notation to understand what's going on (like, as in looking at the book). I did upvote your question earlier, so hopefully someone else will take a look at it.
