continuous mappings of open sets question 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.
Thanks.
And why did this get downvoted??
 A: Not familiar with the book, but I'll answer just to stop this question from being rolled over by the hardworking Community hamster. 


*

*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 $p_0$. 

*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 $p,q$. 

*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. 

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

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