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Let $\lambda$ be Lebesgue measure on [0,1]. For any $x_{1},\dots,x_{k}$ in $[0,1]$, define $$A(x_1,..,x_k):=\{(y_1,\dots,y_k)\in [0,1]^k: \text{there exist intervals }I_1,\dots,I_k \text{ in }[0,1]$$ $$\text{ such that } x_i,y_i\in I_i \text{ and } \lambda(\cup_iI_i)\leq\frac12\}$$

My question is: Is it true that there exists a constant $c\in(0,1)$ such that for any $x_{1},\dots,x_{k}$ in $[0,1]$ $$\lambda^{\otimes k}(A(x_1,..,x_k))\geq c^k$$

(Note that when $x_1,..,x_k$ can be covered by a interval of length not larger than 1/2, then $\lambda(A(x_1,..,x_k))\geq (\frac12)^k$. Thus the question is when $x_1,..,x_k$ is quite spread out on $[0,1]$, like for $x_i=\frac{i}{k+1}$ , is the volume still decay at most exponentially? For $x_i=\frac{i}{k+1},i=1,2,..,k$, we can construct interval $I_i=[x_i,x_i+\frac{1}{2k}]$, thus $\lambda(A(x_1,..,x_k))\geq (\frac{1}{2k})^k$. But of course we can construct other intervals and union all situations. )

Let $\lambda$ be Lebesgue measure on [0,1]. For any $x_{1},\dots,x_{k}$ in $[0,1]$, define $$A(x_1,..,x_k):=\{(y_1,\dots,y_k)\in [0,1]^k: \text{there exist intervals }I_1,\dots,I_k \text{ in }[0,1]$$ $$\text{ such that } x_i,y_i\in I_i \text{ and } \lambda(\cup_iI_i)\leq\frac12\}$$

My question is: Is it true that there exists a constant $c\in(0,1)$ such that for any $x_{1},\dots,x_{k}$ in $[0,1]$ $$\lambda^{\otimes k}(A(x_1,..,x_k))\geq c^k$$

(Note that when $x_1,..,x_k$ can be covered by a interval of length not larger than 1/2, then $\lambda^{\otimes k}(A(x_1,..,x_k))\geq (\frac12)^k$. Thus the question is when $x_1,..,x_k$ is quite spread out on $[0,1]$, like for $x_i=\frac{i}{k+1}$ , is the volume still decay at most exponentially? For $x_i=\frac{i}{k+1},i=1,2,..,k$, we can construct interval $I_i=[x_i,x_i+\frac{1}{2k}]$, thus $\lambda^{\otimes k}(A(x_1,..,x_k))\geq (\frac{1}{2k})^k$. But of course we can construct other intervals and union all situations. )

Let $\lambda$ be Lebesgue measure on [0,1]. For any $x_{1},..,x_{k}\in [0,1]$, define $$A(x_1,..,x_k):=\{y_1,.,y_k\in [0,1]: \text{there exist intervals }I_1,..,I_k \text{ on }[0,1]\text{ such that }x_i,y_i\in I_i \text{ and } \lambda(\cup_iI_i)\leq\frac12\}$$

My question is: Is it true that there exists a constant $c\in(0,1)$ such that for any $x_{1},..,x_{k}\in [0,1]$ $$\lambda(A(x_1,..,x_k))\geq c^k$$

(Note that when $x_1,..,x_k$ can be covered by a interval of length not larger than 1/2, then $\lambda(A(x_1,..,x_k))\geq (\frac12)^k$. Thus the question is when $x_1,..,x_k$ is quite spread out on $[0,1]$, like for $x_i=\frac{i}{k+1}$ , is the volume still decay at most exponentially? For $x_i=\frac{i}{k+1},i=1,2,..,k$, we can construct interval $I_i=[x_i,x_i+\frac{1}{2k}]$, thus $\lambda(A(x_1,..,x_k))\geq (\frac{1}{2k})^k$. But of course we can construct other intervals and union all situations. )

Let $\lambda$ be Lebesgue measure on [0,1]. For any $x_{1},\dots,x_{k}$ in $[0,1]$, define $$A(x_1,..,x_k):=\{(y_1,\dots,y_k)\in [0,1]^k: \text{there exist intervals }I_1,\dots,I_k \text{ in }[0,1]$$ $$\text{ such that } x_i,y_i\in I_i \text{ and } \lambda(\cup_iI_i)\leq\frac12\}$$

My question is: Is it true that there exists a constant $c\in(0,1)$ such that for any $x_{1},\dots,x_{k}$ in $[0,1]$ $$\lambda^{\otimes k}(A(x_1,..,x_k))\geq c^k$$

(Note that when $x_1,..,x_k$ can be covered by a interval of length not larger than 1/2, then $\lambda(A(x_1,..,x_k))\geq (\frac12)^k$. Thus the question is when $x_1,..,x_k$ is quite spread out on $[0,1]$, like for $x_i=\frac{i}{k+1}$ , is the volume still decay at most exponentially? For $x_i=\frac{i}{k+1},i=1,2,..,k$, we can construct interval $I_i=[x_i,x_i+\frac{1}{2k}]$, thus $\lambda(A(x_1,..,x_k))\geq (\frac{1}{2k})^k$. But of course we can construct other intervals and union all situations. )