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The question is given below:

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And here is exercise (16):

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And here is the solution to exercise(17)

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But I have difficulties in understanding the following parts of the solution:

1-Why the codomain of the defined $h$ in the second line is $\mathbb{R}$?

2-Why $h$ is clearly smooth as stated in the third line?

3-Why we are using U x {0}, what is the importance of using the singleton $0$?

4-Why K x {0} is compact? and why this leads to that $h > 2\delta$ for some $\delta > 0$?

5-And by which property of continuity of h, there exists an open set $U'$ such that $h > \delta$ on $U'$?

6-When usually Tube lemma is applied, when we need what, we apply it?

Thank you!

Intuition
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1 Answers1

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  1. Because $\det(Hf_t|_x)^2 + \sum_{i=1}^k \frac{\partial f_t}{\partial x_i}$ is a real number.
  2. Presumably, there is a blanket assumption of smoothness for "homotopic families of functions" in this text. In general, this wording is not sufficient to suppose that $f_t$ is differentiable at all, much less smooth. However, since the solution immediately takes derivatives without any mention of conditions, I'm guessing that the authors have already placed a blanket restriction on such families to be be smooth. Differentiating, summing and multiplying smooth functions results in smooth functions, so $h$ would also be smooth.
  3. Because it is given in the question that $f_0$ is Morse. That is the starting point of their logic: when $t = 0$.
  4. Because $K$ is compact. And because $h$ is continuous on a compact set, and therefore must have a minimum.
  5. $h^{-1}((\delta, \infty))$ is open.
  6. I have no idea what you are asking. Please clarify.
Paul Sinclair
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  • For 6 I do not understand why the author used tube lemma? what is he doing with it? the main idea – Intuition Nov 11 '18 at 23:15
  • They have just shown when $t = 0, h > \delta$. They are using it to show that there is some $\epsilon > 0$ such that $h > \delta$ for all $t \in [0, \epsilon]$. – Paul Sinclair Nov 11 '18 at 23:20
  • why are the partial derivatives real numbers? is there a hidden assumption that the function is linear in its coordinates? – Intuition Nov 12 '18 at 00:57
  • any singleton set is closed and bounded and hence compact ..... correct? – Intuition Nov 12 '18 at 01:40
  • @hopefully it's certainly a real number at each point, which is what a real-valued function is. –  Nov 12 '18 at 02:31
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    There is no way to take a derivative of real function with respect to a real variable and get anything other than a real number as the result (assuming the derivative exists, which is part of what "smooth" means). In a metric space, singleton points are closed and bounded and thus are compact. But more generally, in any topology, any finite set is compact. This follows immediately from the general definition of "compact". – Paul Sinclair Nov 12 '18 at 04:04
  • In this link Ted Shfirin has referred to the use of the definition of continuity in topological spaces , how it is used in the above answer ? – Intuition Nov 12 '18 at 06:08
  • https://math.stackexchange.com/questions/402479/a-basic-proof-on-morse-function – Intuition Nov 12 '18 at 06:22
  • Also how in the above solution this is an application of the tube lemma? the tube lemma that I know is here https://en.wikipedia.org/wiki/Tube_lemma – Intuition Nov 12 '18 at 11:11
  • why we know that the open set U does not contain 0? or it may contain it? – Intuition Nov 12 '18 at 11:24
  • why the solution said $h > 2 \delta $ and not $h > \delta$ only? – Intuition Nov 12 '18 at 11:26
  • In the third line from below in the answer I do not understand what property of continuity say this "By continuity of h, there exists an open set U' containing K x {0} such that $h > \delta$ on U' " – Intuition Nov 12 '18 at 11:32
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    (1) No, Ted does not. And it makes no difference what definition of continuity you use, they are all the same in this context. (2)That is the lemma that provides the result I mentioned in my first comment on this post above (3) $U$ is a subset of $\Bbb R^k$. $0$ itself is not in $\Bbb R^k$. $(0,0,...,0) \in \Bbb R^k$, but there is no significance whatsoever to $(0,0,...,0)$ in this proof, so it is immaterial whether $U$ contains it. (4) Because we have $h \ge 2\delta$, not $h > 2\delta$. They need a smaller number to have $>$, which they need to make $h^{-1}((\delta, \infty))$ open. – Paul Sinclair Nov 12 '18 at 13:42
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    (5) I already answered that in 5. above. Since $h$ is continuous, $h^{-1}$ of any open set is also open. – Paul Sinclair Nov 12 '18 at 13:42
  • I think the @hopefully are speaking about the comments under the answer of Ted Shfirin – Idonotknow Nov 12 '18 at 14:20