Give an example of a set $\mathcal{X}$ of $N$ points for which no scaled projection onto a subspace of dimension $m \ll \log{N}$ is an approximate isometry.
Context and hint: this is exercise 5.3.4 from Vershynin's book (HDP). Link to text (https://www.math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.pdf).
Approximate isometry refers to a linear map $f$ such that $(1-\epsilon)|x-y|<|f(x)-f(y)|<(1+\epsilon)|x-y|$. A hint given by the author reads: Set $\mathcal{X}$ be an orthogonal basis and show that the projected set defines a packing.
To understand "scaled projection", see Theorem 5.3.1 in Vershynin's book where projection is scaled by $\sqrt{n/m}$.
Attempt: I'm not quite sure what I'm supposed to do, especially about dealing with $\epsilon$. Maybe I want to first fix an $\epsilon$ and show that we have this set of N points (as orthogonal basis) such that for sufficiently large N there is at least two points such that their projections depart from each other by more than $\epsilon$ no matter which subspace we choose?
In other words, fix $\epsilon>0$ and fix projection $P$ of size $N\times m$ with columns being orthonormal basis for the new subspace. Each row of $P$ is image of a point in $\mathcal{X}$ in the new subspace. How do I show that there are two rows $r_1, r_2$ of $P$ such that $\|r_1-r_2\|_2>\epsilon$ when $N-m$ is large?
I'm sorry but this question seems to be hard to answer without first reading Vershynin's book. I have some further thoughts regarding the hint about "packing" and seem to be able to give a solution.
First, a lemma in the other section of the textbook (Proposition 4.2.12) may be of use. For a ball of radius $r$ in $\mathbb{R}^k$, at most we can place $(2r/t+1)^k$ points so that each pair of points distance from each other by more than $2 t$. This is called "packing (number)".
So let's consider $N=n+1$ points in $\mathbb{R}^n$: $\{0, e_1,...,e_n\}$ (standard basis). Suppose to the contrary the subspace $m\ll\log N$ exists so that a scaled projection to this subspace is indeed an approximate isometry. Then the image of all points must be contained in a ball of radius $(1+\epsilon)$ certered at $0$. And each pair of points distance from each other by more than $2 \frac{1-\epsilon}{2}.$
So I tried to $t:=(1-\epsilon)/2$, $r=1+\epsilon$ and $k:=m$. By the lemma, we can place at most $(\frac{4(1+\epsilon)}{1-\epsilon}+1)^m$ points. So we have $N\le(\frac{4(1+\epsilon)}{1-\epsilon}+1)^m$. Re-manage this expression yields $m\ge \log(\frac{4(1+\epsilon)}{1-\epsilon}+1)^{-1} \log N$.
This seems to solve the problem in the intended way.