Let $\lambda$ be Lebesgue measure on $[0,1]$. For $\mathbf{x}=(x\_1,x\_2,..,x\_k)\in[0,1]^k$, define $$A(\mathbf{x}):=\{(y\_1,\dots,y\_k)\in [0,1]^k: \text{there exist intervals }I\_1,\dots,I\_k \text{ in }[0,1]$$ 
$$\text{ such that } x\_i,y\_i\in I\_i \text{ and } \lambda(\cup\_iI\_i)\leq\frac12\}$$ 


My question: Is it true that the maximum number $n$ such that there exists $\mathbf{x}\_1,..,\mathbf{x}\_n\in[0,1]^k$ that satisfy $A(\mathbf{x}\_1),..,A(\mathbf{x}\_n)$ are mutually exclusive has at most exponentially growth speed in $k$?


That is if we denote $$n^\*=\max\{n\in\mathbb{N}:\exists \text{ $\mathbf{x}\_1,..,\mathbf{x}\_n\in[0,1]^k$ that $A(\mathbf{x}\_1),..,A(\mathbf{x}\_n)$ are mutually exclusive }\} ,$$ can we prove $$n^\*\leq C^k$$for some constant $C>1$?




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First I tried a simple volume argument. I guessed for any $\mathbf{x}\in[0,1]^k$ we have $\text{vol}(A(\mathbf{x}))\geq c^k$ for some constant $c<1$. But it turns out not true. See the answer here [lower bound volume of a set](https://mathoverflow.net/questions/280823/lower-bound-volume-of-a-set).


Any thoughts and comments are welcome!


Thanks for your help! 



packing with special sets in high dimensional Euclidean space
**[EDIT]** This is an **affirmative** answer to the original question, with $C=2$. The negative answer to the version where $1/2$ is replaced by a smaller constant is kept below.


For every $\mathbf x\in[0,1]^k$, we define $\mathbf a(\mathbf x)\in\{0,1\}^k$ by $a(x)\_i=0\iff x\_i\leq 1/2$. Now, assume that $A(\mathbf x^1),\dots,A(\mathbf x^n)$ are disjoint and $N>2^k$; then there are $\mathbf x=\mathbf x^i$ and $\mathbf z=\mathbf x^j$ with $i\neq j$ but $\mathbf a(\mathbf x)=\mathbf a(\mathbf z)=\mathbf a$.


Now define $\mathbf y\in[0,1]^k$ as follows: If $a\_i=0$ then $y\_i=x\_i(\leq 1/2)$, otherwise $y\_i=z\_i(>1/2)$. Then $\mathbf y\in A(\mathbf x)$ due to the following system of segments: $I\_i=[x\_i,x\_i]$ if $a\_i=0$, and $I\_i=[1/2,1]$ otherwise. Similarly, $\mathbf y\in A(\mathbf z)$ due to the following system of segments: $I\_i=[0,1/2]$ if $a\_i=0$, and $I\_i=[z\_i,z\_i]$ otherwise. Thus $A(\mathbf x)\cap A(\mathbf z)\neq\varnothing$ --- a contradiction.


Thus $C=2$ fits.


  



**[ORIGINAL] The rest of the text shows what happens if we replace $1/2$ by a smaller constant.**


For $\mathbf x,\mathbf y\in[0,1]^k$, denote by $d(\mathbf x,\mathbf y)$ the minimal total length of a system of (disjoint) segments such that for every $i$, the numbers $x\_i$ and $y\_i$ lie in one of the segments. Clearly, $d$ is a metric. If we replace $1/2$ by some $\alpha<1/2$, then we are interested in finding many mutually disjoint balls $B\_\alpha(\mathbf x)$ of radius $\alpha$ one can find in $[0.1]^k$. 


For that purpose, it suffices to find many points $\mathbf x^i$ with pairwise distance $>2\alpha$. We will seek for such points of the form 
$$
 \mathbf x^\sigma=\left(\frac{\sigma(0)}{k-1},\dots,\frac{\sigma(k-1)}{k-1}\right),
$$
where $\sigma$ is a permutation of $\{0,1,\dots,k-1\}$.


Let $V$ be the number of points having the form $\mathbf x^\sigma$ and lying in a ball $B\_{2\alpha}(\mathbf x^\tau)$ for some fixed $\tau$; the number $V$ does not depend on $\tau$, since all our points are equivalent via permutation of coordinates. Then we may choose the ``distant'' set of at least $k!/V$ such points, choosing one by one (since each chosen point prohibits at most $V$ points, including itself). So we are to estimate $V$ now, asssuming that $\tau=\mathrm{id}$ and denoting $\mathbf x=\mathbf x^\tau$.


Set $\ell=[2\alpha(k-1)]$. Then $d(\mathbf x,\mathbf x^\sigma)\leq 2\alpha$ iff there exists a set of (disjoint) segments of the form $[a/(k-1),b/(k-1)]$ with total length exactly $\ell/(k-1)$ such that $i/(k-1)$ and $x^\sigma\_i$ are always in the same segment (we allow $a=b$, in which case the length of the segment is $0$). 


Such a set of segments can be chosen in ${k-1\choose \ell}$ ways (choosing which elementary segments are covered). Consider any such set with lengths $m\_1,\dots,m\_s$; then there are exactly $(m\_1+1)!(m\_2+1)!\dots(m\_s+1)!$ points $\mathbf x^\sigma$ ``compatible'' with this set.


Finally, notice that for any $m,n\geq 1$ we have $m!n!=m!\cdot 2\cdots n\leq (m+n-1)!$. Thus
$$
 (m\_1+1)!\dots(m\_s+1)!\leq (m\_1+\dots+m\_s+1)!=(\ell+1)!.
$$
This means that $V\leq{k-1\choose \ell}(\ell+1)!=\frac{(k-1)!}{(k-\ell-1)!}(\ell+1)$.


Therefore, we can choose at least
$$
 \frac{k!}V\geq \frac{k(k-\ell-1)!}{\ell+1}\geq (k-\ell-1)!
$$
points with pairwise distances $>2\alpha$; 
this number is superexponential in $k$ (since $k-\ell-1\approx[k(1-2\alpha)]$).


