2
$\begingroup$

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 $\rho (x_0,y_0)=\rho (E_1,E_2)$

$\endgroup$
  • 1
    $\begingroup$ $\rho(E_1,E_2)=inf\{\rho(x,y)|x \in E_1,y\in E_2\}$ $\endgroup$ – mm-crj Jan 1 '19 at 11:33
4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Can you use the weierstrass theorem $\endgroup$ – jackson Jan 1 '19 at 11:41
  • $\begingroup$ I don’t understand the last sentence in your proof $\endgroup$ – jackson Jan 1 '19 at 11:47
  • $\begingroup$ math.stackexchange.com/questions/109548/… $\endgroup$ – Thomas Shelby Jan 1 '19 at 11:50
  • $\begingroup$ but you can’t say 0 is f’s bounds,it maybe 1 or other number be it’s bounds $\endgroup$ – jackson Jan 2 '19 at 4:37
  • $\begingroup$ 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 $1\le f$ $\endgroup$ – jackson Jan 2 '19 at 4:44
1
$\begingroup$

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 $\rho (x_0,y_0)=\rho (E_1,E_2)$

$\endgroup$
  • $\begingroup$ @ThomasShelby for $E_1$ is bound and $x_n\in E_1$ so $x_n $ is bounded right? $\endgroup$ – jackson Jan 2 '19 at 13:46
  • 1
    $\begingroup$ @ThomasShelby thanks a lot $\endgroup$ – jackson Jan 2 '19 at 13:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.