Suppose $$E_1,E_2 \subseteq \Bbb R^m$$ are closed sets and at least one of them is a bounded set.

Prove that there exist $$x_0\in E_1,y_0\in E_2$$,such that $$\rho (x_0,y_0)=\rho(E_1,E_2)$$

Attempt :I think $$\rho(E_1,E_2)$$ is $$\inf\{\rho(x,y)|x\in E_1,y \in E_2\}$$

Then for $$\epsilon_n =\frac{1}{n}> 0$$,there exist $$x_n,y_n$$,s.t $$\rho(E_1,E_2)\ge \rho(x_n,y_n)\ge \rho(E_1,E_2)-\epsilon_n$$

I want to use Weierstrass theorem ,but this set doesn’t satisfy the condition of the theorem.

can someone see my prove is right or not

prove

Now for $$\epsilon_n =\frac{1}{n}> 0$$,there exists $$\{y_n\}\in E_2,\{x_n\}\in E_1$$,such that $$\rho(E_1,E_2)\le \rho(x_n,y_n)\le \rho(E_1,E_2)+\epsilon_n.$$ Since $$x_n$$ is bounded,$$\rho(x_n,x_m)\le M$$ Also note that \begin{align} \rho(y_n,y_m)&\leq \rho(y_n,x_n)+\rho(x_m,y_m)+\rho(x_m,x_n)\\ &\leq 2\rho(E_1,E_2)+2+M,\forall m,n \in \Bbb N. \end{align} So $$y_n$$ has convergence sub consequence $$y_{n_k}$$ converse to $$y_0 \in E_2$$ So $$\rho(E_1,E_2)\le \rho(x_{n_k},y_{n_k})\le \rho(E_1,E_2)+\epsilon_{n_k}.$$ Also $$x_{n_k}$$ has convergence sub consequence $$x_{n_{k_l}}$$ converse to $$x_0 \in E_1$$ so $$\rho (x_0,y_0)=\rho (E_1,E_2)$$

• $\rho(E_1,E_2)=inf\{\rho(x,y)|x \in E_1,y\in E_2\}$ – mm-crj Jan 1 '19 at 11:33

WLOG,let $$E_1$$ be a bounded set. By Heine-Borel Theorem, $$E_1$$ is compact. Define $$f:E_1 \to \Bbb R$$ such that $$x\mapsto \rho (x,E_2)$$. Clearly, $$\rho (E_1,E_2)\leq f(x), \forall x \in E_1$$. Since a continuous function from a compact set attains its bounds, there exists $$x_0 \in E_1$$ such that $$f(x_0)=\rho (x_0,E_2)=\rho (E_1,E_2)$$.

Now for $$\epsilon_n =\frac{1}{n}> 0$$,there exists $$\{y_n\}\in E_2$$,such that $$\rho(E_1,E_2)\le \rho(x_0,y_n)\le \rho(E_1,E_2)+\epsilon_n.$$

Also note that \begin{align} \rho(y_n,y_m)&\leq \rho(y_n,x_0)+\rho(x_0,y_m)\\ &\leq 2\rho(E_1,E_2)+2,\forall m,n \in \Bbb N. \end{align} So $$\{y_n\}$$ is a bounded sequence and hence has a convergent sub-sequence. Can you complete the proof?

• Can you use the weierstrass theorem – jackson Jan 1 '19 at 11:41
• I don’t understand the last sentence in your proof – jackson Jan 1 '19 at 11:47
• math.stackexchange.com/questions/109548/… – Thomas Shelby Jan 1 '19 at 11:50
• but you can’t say 0 is f’s bounds,it maybe 1 or other number be it’s bounds – jackson Jan 2 '19 at 4:37
• like $E_1:x^2+y^2=1$ $E_2:2\le x\le 3$ and I fix $y_0=(0,3)$,obviously $\rho(E_1,E_2)=1$ but $1\le f$ – jackson Jan 2 '19 at 4:44

Now for $$\epsilon_n =\frac{1}{n}> 0$$,there exists $$\{y_n\}\in E_2,\{x_n\}\in E_1$$,such that $$\rho(E_1,E_2)\le \rho(x_n,y_n)\le \rho(E_1,E_2)+\epsilon_n.$$ for $$E_1$$ is bounded ,and $$x_n\in E_1$$,so $$x_n$$ is bounded Since $$x_n$$ is bounded,$$\rho(x_n,x_m)\le M$$ Also note that \begin{align} \rho(y_n,y_m)&\leq \rho(y_n,x_n)+\rho(x_m,y_m)+\rho(x_m,x_n)\\ &\leq 2\rho(E_1,E_2)+2+M,\forall m,n \in \Bbb N. \end{align} So $$y_n$$ has a convergent subsequence $$y_{n_k}$$ converging to $$y_0 \in E_2$$. So $$\rho(E_1,E_2)\le \rho(x_{n_k},y_{n_k})\le \rho(E_1,E_2)+\epsilon_{n_k}.$$ Also $$x_{n_k}$$ has a convergent subsequence $$x_{n_{k_l}}$$ converging to $$x_0 \in E_1$$. So $$\rho (x_0,y_0)=\rho (E_1,E_2)$$

• @ThomasShelby for $E_1$ is bound and $x_n\in E_1$ so $x_n$ is bounded right? – jackson Jan 2 '19 at 13:46
• @ThomasShelby thanks a lot – jackson Jan 2 '19 at 13:52