# 0708-1300/Class notes for Tuesday, October 16

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## Dror's Computer Program for C+${\displaystyle \alpha }$ C

Our handout today is a printout of a Mathematica notebook that computes the measure of the projection of ${\displaystyle C\times C}$ in a direction ${\displaystyle t\in [0,\pi /2]}$ (where ${\displaystyle C}$ is the standard Cantor set). Here's the notebook, and here's a PDF version. Also, here's the main picture on that notebook:

## Typed Class Notes

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

### First Hour

Today's Agenda:

Proof of Sard's Theorem. That is, for ${\displaystyle f:M^{m}\rightarrow N^{n}}$ being smooth, the measure ${\displaystyle \mu }$(critical values of f) = 0.

1) In our counterexample to Sard's Theorem for the case of ${\displaystyle C^{1}}$ functions it was emphasized that there are functions f from a Cantor set C' to the Cantor set C. We then let ${\displaystyle g(x,y)=f(x)+f(y)}$ and hence the critical values are ${\displaystyle C+C=[0,2]}$ as was shown last time. The sketch of such an f was the same as last class.

Furthermore, in general we can find a ${\displaystyle C^{n}}$ such function where we make the "bumps" in f smoother as needed and so ${\displaystyle f(C')=C''}$ where ${\displaystyle C''}$ is a "very thin" Cantor set. But now let ${\displaystyle g(x,y,z,\ldots )=f(x)+f(y)+f(z)+\ldots }$ which will have an image of ${\displaystyle C''+C''+C''+\ldots =}$ an interval.

2) The code, and what the program does, for Dror's program (above) was described. It is impractical to describe it here in detail and so I will only comment that it computes the measure of ${\displaystyle C+\alpha C}$ for various alpha and that the methodology relied on the 2nd method of proof regarding C+C done last class.

Proof of Sard's Theorem

Firstly, it is enough to argue locally, since a manifold is second countable (that is, a manifold has a countable basis) and a countable union of sets of measure zero has measure zero.

Further, the technical assumption about manifolds that until now has been largely ignored is that our M must be second countable. Recall that this means that there is a countable basis for the topology on M.

As a counterexample to Sard's Theorem when M is NOT second countable consider the real line with the discrete topology, a zero dimensional manifold, mapping via the identity onto the real line with the normal topology. Every point in the real line is thus a critical point and the real line has non zero measure.

We can restrict our neighborhoods so that we can assume ${\displaystyle M=\mathbb {R} ^{m}}$ and ${\displaystyle N=\mathbb {R} ^{n}}$

The general idea here is that if we consider a function g=f' that is nonzero at p but that f is zero at p, the inverse image is (in some chart) a straight line (a manifold). As such, we will inductively reduce the dimension from m down to zero. For m=0 there is nothing to prove. Hence we assume true for m-1.

Now, set ${\displaystyle D_{k}=\{p|}$ all partial derivities of f of order ${\displaystyle \leq k}$ vanish} for ${\displaystyle k\geq 1}$

Set ${\displaystyle D_{0}=\{p\ |\ df_{p}}$ is not onto }. This is just the critical points.

Clearly ${\displaystyle D_{0}\supset D_{1}\supset \ldots \supset D_{i}\supset \ldots \supset D_{m}}$

We will show by backwards induction that ${\displaystyle \mu (F(D_{k}))=0}$

Comment:

We have not actually defined the measure ${\displaystyle \mu }$. We use it merely to denote that ${\displaystyle F(D_{k}))}$ has measure zero, a concept that we DID define.

### Second Hour

Claim 1

${\displaystyle F(D_{m})}$ has measure 0.

Proof

W.L.O.G (without loss of generality) we can assume n=1. Intuitively this is reasonable as in lower dimensions the theorem is harder to prove; indeed, a set of size ${\displaystyle \epsilon }$ in 1D becomes a smaller set of size ${\displaystyle \epsilon ^{2}}$ in 2D etc. More precisely, for ${\displaystyle f=(f_{1},\ldots ,f_{n})}$, ${\displaystyle f(D_{m})\subset f_{1}(D_{m})\times {R}^{n-1}}$. Applying the proposition that if A is of measure zero in ${\displaystyle \mathbb {R} }$ then ${\displaystyle A\times \mathbb {R} ^{n-1}}$ is measure zero in ${\displaystyle \mathbb {R} ^{n}}$ we now see that assuming n=1 is justified.

Reminder

Taylor's Theorem: for smooth enough ${\displaystyle g:\mathbb {R} \rightarrow \mathbb {R} }$ and some ${\displaystyle x_{0}}$ then ${\displaystyle g(x)=\sum _{j=0}^{m}{\frac {g^{(j)}(x_{0})}{j!}}(x-x_{0})^{j}+{\frac {g^{(m+1)}(t)}{(m+1)!}}(x-x_{0})^{m+1}}$ for some t between x and ${\displaystyle x_{0}}$.

For ${\displaystyle x_{0}\in D_{m}}$ all but the last term vanishes and so we can conclude that f(x) is bounded by a constant times ${\displaystyle (x-x_{0})^{m+1}}$.

Now let us consider a box B in ${\displaystyle \mathbb {R} ^{m}}$ containing a section of ${\displaystyle D_{m}}$. We divide B into ${\displaystyle C_{1}\lambda ^{m}}$ boxes of side ${\displaystyle 1/\lambda }$.

By Taylor's Theorem, ${\displaystyle f(B_{i})\subset }$ of an interval of length ${\displaystyle C_{2}{\frac {1}{\lambda ^{m+1}}}}$ where the constant ${\displaystyle C_{2}}$ is determined by Taylor's Theorem. Call this interval ${\displaystyle I_{i}}$

Hence,

${\displaystyle f(D_{m})\subset \bigcup _{i:B_{i}\cap D_{m}\neq 0}f(B_{i})\subset \bigcup _{i:B_{i}\cap D_{m}\neq 0}I_{i}}$

But ${\displaystyle \sum _{i}length(I_{i})\leq C_{1}\lambda ^{m}C_{2}{\frac {1}{\lambda ^{m+1}}}}$ which tends to zero as ${\displaystyle \lambda }$ tends to infinity.

Q.E.D for Claim 1

Claim 2

${\displaystyle f(D_{k})}$ has measure zero for ${\displaystyle k\geq 1}$. We just proved this for k=m.

Now, W.L.O.G. ${\displaystyle D_{k+1}}$ is the empty set. If not, just consider ${\displaystyle M^{m}-D_{k+1}}$ which is still a manifold as ${\displaystyle D_{k+1}}$ is closed (as it is determined by the "closed" condition that a determinant equals zero)

So, there is some kth derivative g of f such that ${\displaystyle dg\neq 0}$.

So ${\displaystyle D_{k}\subset g^{-1}(0)}$ but ${\displaystyle g^{-1}(0)}$ is at least locally a manifold of dimension 1 less. So, ${\displaystyle f(D_{k})\subset f(D_{k}\cap g^{-1}(0))}$ which has measure zero due to our induction hypothesis.

Q.E.D for Claim 2

Claim 3

${\displaystyle f(D_{0})}$ is of measure zero.

Recall ${\displaystyle D_{0}}$ is defined differently from the ${\displaystyle D_{k}}$ and so requires a different technique to prove.

W.L.O.G. assume that ${\displaystyle D_{1}}$ is the empty set. So, some derivative of f is not zero. W.L.O.G. ${\displaystyle {\frac {\partial f_{1}}{\partial x_{1}}}}$ is non zero near some point p. We can simply move to a coordinate system where this is true.

The idea here is to prove that the intersection with any "slice" has measure zero where we will then invoke a theorem that will claim everything has measure zero.

So, let U be an open neighborhood of a point ${\displaystyle p\in M}$. Consider ${\displaystyle f_{1}:U\rightarrow \mathbb {R} }$ and let ${\displaystyle df_{1}}$ be onto. Using our previous theorem for the local structure of such a submersion W.L.O.G. let us assume ${\displaystyle f_{1}:\mathbb {R} ^{m}\rightarrow \mathbb {R} }$ via ${\displaystyle (x_{1},\ldots ,x_{m})\mapsto x_{1}}$. That is, ${\displaystyle f_{1}=x_{1}}$.

Our differential df then is just the matrix whose first row consists of ${\displaystyle (1,0,\ldots ,0)}$. Then df is onto if the submatrix consisting of all but the first row and first column is invertible.

For notational convenience let us say ${\displaystyle f=(f_{1},f_{rest})}$.

Now define ${\displaystyle U_{t}=\{t\}\times \mathbb {R} ^{m-1}}$. Also, let us denote "critical points of f" by CP(f) and "critical values of f" by CV(f).

Claim 4

The ${\displaystyle CP(f)=\bigcup _{t\in \mathbb {R} }\{t\}\times CP(f_{rest}|_{U_{t}})}$

${\displaystyle \Rightarrow }$

${\displaystyle CV(f)=\bigcup _{t\in \mathbb {R} }\{t\}\times CV(f_{rest}|_{U_{t}})}$.

But ${\displaystyle CV(f_{rest}|_{U_{t}})}$ has measure zero by our induction.

Lemma 1

If ${\displaystyle A\subset I^{2}}$ is closed and has ${\displaystyle \mu (A\cap (\{t\}\times I))=0\ \forall t}$ then ${\displaystyle \mu (A)=0}$.

Proof

Note: We prove this significantly differently then in Bredon

Sublemma

If ${\displaystyle \{t\}\times U}$ for an open U cover ${\displaystyle \{t\}\times I\cap A}$ for a closed A then ${\displaystyle \exists \epsilon >0}$ such that ${\displaystyle (t-\epsilon ,t+\epsilon )\times U\supset (t-\epsilon ,t+\epsilon )\times I\cap A}$

Indeed, let ${\displaystyle d:A-(I\times U)\rightarrow \mathbb {R} }$ be ${\displaystyle d(x)=|x_{1}-t|}$ then d is a continuous function of a compact set and so obtains a minimum and since d>0 then ${\displaystyle min(d)>0\rightarrow d>\epsilon >0}$. But this ${\displaystyle \epsilon }$ works for the claim. Q.E.D

The rest of the proof of Lemma 1, and of Sard's Theorem will be left until next class