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Lemma: Let $E$ be a closed bounded subset of $\Omega$ and let $T$ be an injective transformation of class $C'$ on $\mathbb{R}^3$. Also the Jacobian does not vanish. Define $v(K)$ to be the volume of $K.$ Then $$\lim_{C\downarrow p} \frac{v(T(C))}{v(C)} = J(p)$$ where $C$ ranges over the family of cubes lying in $\Omega$ and having center $p$, and the limit is uniform for all $p\in E.$

In the proof of the above lemma, it says

If $p$ lies in the sphere $S$ with center $p_0$ and radius $r<\epsilon$, then $T(p)$ lies in the sphere whose center is $T(p_0)$ and whose radius is $(1+\epsilon)r.$ Moreover, when $p$ lies on the boundary of $S$, $T(p)$ lies outside the sphere with center $T(p_0)$ and with radius $(1-\epsilon)r.$

That stuff is fine and I understand the derivations which are in the book.

However, it then says

Since $T$ is injective and takes open sets into open sets, we see that $T(S)$ must contain the smaller sphere of radius $(1-\epsilon)r$.

I do not understand why injectivity and taking open sets to open sets implies that $T(S)$ contains the smaller sphere. That is where I am stuck.


And why did this get downvoted??

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"Boundary of $S$"? The boundary of a sphere is itself... Spheres are usually assumed to be hollow. What definition of "sphere" are you using? Do you mean "ball"? – Jesse Madnick Jun 27 '11 at 5:53
Also, what kind of a function is $T$? What is its domain? Is it continuous? Linear? – Jesse Madnick Jun 27 '11 at 5:55
Glassjawed: Echoing Jesse Madnick somewhat, will you please provide more context? The proof of what? – Jonas Meyer Jun 27 '11 at 6:47
Okay I guess I meant ball here. $T$ is a $C^1$ function, not necessarily linear. And I only want to know why the last line holds. This is part of a proof taken from Buck's Advanced Calculus. – Clark Kent Jun 27 '11 at 13:55
@Glassjawed: I cannot speak for the downvoter, but it is tempting to downvote for lack of clarity. You can still edit your question to fix this. Buck's Advanced Calculus has many proofs, so saying that it is from that book doesn't help much. Will you please give a precise reference, and/or say precisely what it is a proof of? – Jonas Meyer Jun 28 '11 at 0:19
up vote 1 down vote accepted

Not familiar with the book, but I'll answer just to stop this question from being rolled over by the hardworking Community hamster.

  1. Compose $T$ with a linear transformation (the inverse of $T'(p_0)$) so that the composition, denoted $F$, satisfies $F'(p_0)=I$. Assuming we understand what Jacobians of linear maps have to do with volume, it remains to prove that $v(F(C))/v(C)\to 1$ as $C$ shrinks to $p_0$.

  2. Since $F$ is $C^1$ smooth, for every $\epsilon>0$ there exists a neighborhood $N$ of $p_0$ in which $\|F'\|\le 1+\epsilon$. In this neighborhood $F$ is $(1+\epsilon)$-Lipschitz: that is, $|F(p)-F(q)|\le (1+\epsilon)|p-q|$ for any pair of points $p,q$.

  3. The Lipschitz condition implies $v(F(C))\le (1+\epsilon)^n v(C)$ for any (measurable) set $C\subset N$. This follows from the definition of measure in terms of coverings. Or, for nice geometric shapes such as balls, this follows by a geometric argument based on containment.

  4. Everything said in 2 and 3 applies to the inverse $F^{-1}$ as well, by the inverse function theorem. Hence $v(F(C))\ge (1+\epsilon)^{-n} v(C)$.

  5. Taken together, 3 and 4 imply that $v(F(C))/v(C)\to 1$. $\Box$

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Warning: I'm using norm notation for distances (because it's easier to write) but the existence of a norm is not a prerequisite for this to work, just replace it with your distance metric if you'd like (which has to exist for "sphere" to be meaningful):

Suppose that $\exists p : \|p-p_0\| \leq (1-\epsilon)r \land p \not\in T(S)$ (suppose there's a point $p$ in the smaller sphere that's not in $T(S)$).

If you can somehow prove that $T^{-1}$ is defined at $p$, the rest gets you to a contradiction...

Then $q=T^{-1}(p) \not \in S$ (the inverse exists because $T$ is injective). If $\exists q : T(q) = p$, pick $r_2 = \|q-p_0\| \geq r$ (because $q \not\in S$). $q$ lies on the boundary of a sphere of radius $r_2$. By the above, $\|p-p_0\| = \|T(q)-p_0\| \geq (1-\epsilon)r_2 \geq (1-\epsilon)r$ which contradicts the choice of $p$.

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Yeah it looks like we still need some help. – Clark Kent Jun 28 '11 at 13:48

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