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Question

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

$\rho(E_1,E_2)=inf{\rho(x,y)|x \in E_1,y\in E_2}$ – mmcrjx Jan 01 '19 at 11:33 — mmcrjx, Jan 01 '19 at 11:33

cqfd · Accepted Answer · 2019-01-02T06:42:06.690

4

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?

edited Jan 02 '19 at 06:42

answered Jan 01 '19 at 11:32

cqfd

12,974

Can you use the weierstrass theorem – jackson Jan 01 '19 at 11:41
I don’t understand the last sentence in your proof – jackson Jan 01 '19 at 11:47
https://math.stackexchange.com/questions/109548/x-compact-metric-space-fx-rightarrow-mathbbr-continuous-attains-max-min – cqfd Jan 01 '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 02 '19 at 04:37
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 02 '19 at 04:44
@jackson Thank you. Edited the answer. If there is any issue, please let me know. – cqfd Jan 02 '19 at 06:43
Can you see my prove ,is it right? – jackson Jan 02 '19 at 07:27

jackson · Answer 2 · 2019-01-02T13:43:59.383

1

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

edited Jan 02 '19 at 13:43

answered Jan 02 '19 at 07:25

jackson

453

@ThomasShelby for $E_1$ is bound and $x_n\in E_1$ so $x_n $ is bounded right? – jackson Jan 02 '19 at 13:46
1

@ThomasShelby thanks a lot – jackson Jan 02 '19 at 13:52

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