about Euclidean Space












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 $x_0in E_1,y_0in E_2$,such that $rho (x_0,y_0)=rho(E_1,E_2)$



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



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 2rho(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 $rho (x_0,y_0)=rho (E_1,E_2)$






calculus linear-algebra sequences-and-series functional-analysis analysis







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$endgroup$








  • 1




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


















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 $x_0in E_1,y_0in E_2$,such that $rho (x_0,y_0)=rho(E_1,E_2)$



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



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 2rho(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 $rho (x_0,y_0)=rho (E_1,E_2)$






calculus linear-algebra sequences-and-series functional-analysis analysis







share|cite|improve this question











$endgroup$








  • 1




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
















2












2








2


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 $x_0in E_1,y_0in E_2$,such that $rho (x_0,y_0)=rho(E_1,E_2)$



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



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 2rho(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 $rho (x_0,y_0)=rho (E_1,E_2)$






calculus linear-algebra sequences-and-series functional-analysis analysis







share|cite|improve this question











$endgroup$




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 $x_0in E_1,y_0in E_2$,such that $rho (x_0,y_0)=rho(E_1,E_2)$



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



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 2rho(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 $rho (x_0,y_0)=rho (E_1,E_2)$






calculus linear-algebra sequences-and-series functional-analysis analysis




calculus linear-algebra sequences-and-series functional-analysis analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 13:05







jackson

















asked Jan 1 at 11:08









jacksonjackson

1379




1379








  • 1




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
















  • 1




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










1




1




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






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












2 Answers
2






active

oldest

votes


















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 $xmapsto rho (x,E_2)$. 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 2rho(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?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can you use the weierstrass theorem
    $endgroup$
    – jackson
    Jan 1 at 11:41










  • $begingroup$
    I don’t understand the last sentence in your proof
    $endgroup$
    – jackson
    Jan 1 at 11:47










  • $begingroup$
    math.stackexchange.com/questions/109548/…
    $endgroup$
    – Thomas Shelby
    Jan 1 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 at 4:37










  • $begingroup$
    like $E_1:x^2+y^2=1$ $E_2:2le xle 3$ and I fix $y_0=(0,3)$,obviously $rho(E_1,E_2)=1$ but $1le f$
    $endgroup$
    – jackson
    Jan 2 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 $x_nin E_1$,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 2rho(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 $rho (x_0,y_0)=rho (E_1,E_2)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @ThomasShelby for $E_1$ is bound and $x_nin E_1$ so $x_n $ is bounded right?
    $endgroup$
    – jackson
    Jan 2 at 13:46












  • $begingroup$
    Your proof looks OK to me.
    $endgroup$
    – Thomas Shelby
    Jan 2 at 13:51






  • 1




    $begingroup$
    @ThomasShelby thanks a lot
    $endgroup$
    – jackson
    Jan 2 at 13:52











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









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 $xmapsto rho (x,E_2)$. 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 2rho(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?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can you use the weierstrass theorem
    $endgroup$
    – jackson
    Jan 1 at 11:41










  • $begingroup$
    I don’t understand the last sentence in your proof
    $endgroup$
    – jackson
    Jan 1 at 11:47










  • $begingroup$
    math.stackexchange.com/questions/109548/…
    $endgroup$
    – Thomas Shelby
    Jan 1 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 at 4:37










  • $begingroup$
    like $E_1:x^2+y^2=1$ $E_2:2le xle 3$ and I fix $y_0=(0,3)$,obviously $rho(E_1,E_2)=1$ but $1le f$
    $endgroup$
    – jackson
    Jan 2 at 4:44


















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 $xmapsto rho (x,E_2)$. 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 2rho(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?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can you use the weierstrass theorem
    $endgroup$
    – jackson
    Jan 1 at 11:41










  • $begingroup$
    I don’t understand the last sentence in your proof
    $endgroup$
    – jackson
    Jan 1 at 11:47










  • $begingroup$
    math.stackexchange.com/questions/109548/…
    $endgroup$
    – Thomas Shelby
    Jan 1 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 at 4:37










  • $begingroup$
    like $E_1:x^2+y^2=1$ $E_2:2le xle 3$ and I fix $y_0=(0,3)$,obviously $rho(E_1,E_2)=1$ but $1le f$
    $endgroup$
    – jackson
    Jan 2 at 4:44
















4












4








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 $xmapsto rho (x,E_2)$. 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 2rho(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?






share|cite|improve this answer











$endgroup$



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 $xmapsto rho (x,E_2)$. 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 2rho(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?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 2 at 6:42

























answered Jan 1 at 11:32









Thomas ShelbyThomas Shelby

4,4592726




4,4592726












  • $begingroup$
    Can you use the weierstrass theorem
    $endgroup$
    – jackson
    Jan 1 at 11:41










  • $begingroup$
    I don’t understand the last sentence in your proof
    $endgroup$
    – jackson
    Jan 1 at 11:47










  • $begingroup$
    math.stackexchange.com/questions/109548/…
    $endgroup$
    – Thomas Shelby
    Jan 1 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 at 4:37










  • $begingroup$
    like $E_1:x^2+y^2=1$ $E_2:2le xle 3$ and I fix $y_0=(0,3)$,obviously $rho(E_1,E_2)=1$ but $1le f$
    $endgroup$
    – jackson
    Jan 2 at 4:44




















  • $begingroup$
    Can you use the weierstrass theorem
    $endgroup$
    – jackson
    Jan 1 at 11:41










  • $begingroup$
    I don’t understand the last sentence in your proof
    $endgroup$
    – jackson
    Jan 1 at 11:47










  • $begingroup$
    math.stackexchange.com/questions/109548/…
    $endgroup$
    – Thomas Shelby
    Jan 1 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 at 4:37










  • $begingroup$
    like $E_1:x^2+y^2=1$ $E_2:2le xle 3$ and I fix $y_0=(0,3)$,obviously $rho(E_1,E_2)=1$ but $1le f$
    $endgroup$
    – jackson
    Jan 2 at 4:44


















$begingroup$
Can you use the weierstrass theorem
$endgroup$
– jackson
Jan 1 at 11:41




$begingroup$
Can you use the weierstrass theorem
$endgroup$
– jackson
Jan 1 at 11:41












$begingroup$
I don’t understand the last sentence in your proof
$endgroup$
– jackson
Jan 1 at 11:47




$begingroup$
I don’t understand the last sentence in your proof
$endgroup$
– jackson
Jan 1 at 11:47












$begingroup$
math.stackexchange.com/questions/109548/…
$endgroup$
– Thomas Shelby
Jan 1 at 11:50




$begingroup$
math.stackexchange.com/questions/109548/…
$endgroup$
– Thomas Shelby
Jan 1 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 at 4:37




$begingroup$
but you can’t say 0 is f’s bounds,it maybe 1 or other number be it’s bounds
$endgroup$
– jackson
Jan 2 at 4:37












$begingroup$
like $E_1:x^2+y^2=1$ $E_2:2le xle 3$ and I fix $y_0=(0,3)$,obviously $rho(E_1,E_2)=1$ but $1le f$
$endgroup$
– jackson
Jan 2 at 4:44






$begingroup$
like $E_1:x^2+y^2=1$ $E_2:2le xle 3$ and I fix $y_0=(0,3)$,obviously $rho(E_1,E_2)=1$ but $1le f$
$endgroup$
– jackson
Jan 2 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 $x_nin E_1$,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 2rho(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 $rho (x_0,y_0)=rho (E_1,E_2)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @ThomasShelby for $E_1$ is bound and $x_nin E_1$ so $x_n $ is bounded right?
    $endgroup$
    – jackson
    Jan 2 at 13:46












  • $begingroup$
    Your proof looks OK to me.
    $endgroup$
    – Thomas Shelby
    Jan 2 at 13:51






  • 1




    $begingroup$
    @ThomasShelby thanks a lot
    $endgroup$
    – jackson
    Jan 2 at 13:52
















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 $x_nin E_1$,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 2rho(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 $rho (x_0,y_0)=rho (E_1,E_2)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @ThomasShelby for $E_1$ is bound and $x_nin E_1$ so $x_n $ is bounded right?
    $endgroup$
    – jackson
    Jan 2 at 13:46












  • $begingroup$
    Your proof looks OK to me.
    $endgroup$
    – Thomas Shelby
    Jan 2 at 13:51






  • 1




    $begingroup$
    @ThomasShelby thanks a lot
    $endgroup$
    – jackson
    Jan 2 at 13:52














1












1








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 $x_nin E_1$,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 2rho(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 $rho (x_0,y_0)=rho (E_1,E_2)$






share|cite|improve this answer











$endgroup$



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 $x_nin E_1$,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 2rho(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 $rho (x_0,y_0)=rho (E_1,E_2)$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 2 at 13:43

























answered Jan 2 at 7:25









jacksonjackson

1379




1379












  • $begingroup$
    @ThomasShelby for $E_1$ is bound and $x_nin E_1$ so $x_n $ is bounded right?
    $endgroup$
    – jackson
    Jan 2 at 13:46












  • $begingroup$
    Your proof looks OK to me.
    $endgroup$
    – Thomas Shelby
    Jan 2 at 13:51






  • 1




    $begingroup$
    @ThomasShelby thanks a lot
    $endgroup$
    – jackson
    Jan 2 at 13:52


















  • $begingroup$
    @ThomasShelby for $E_1$ is bound and $x_nin E_1$ so $x_n $ is bounded right?
    $endgroup$
    – jackson
    Jan 2 at 13:46












  • $begingroup$
    Your proof looks OK to me.
    $endgroup$
    – Thomas Shelby
    Jan 2 at 13:51






  • 1




    $begingroup$
    @ThomasShelby thanks a lot
    $endgroup$
    – jackson
    Jan 2 at 13:52
















$begingroup$
@ThomasShelby for $E_1$ is bound and $x_nin E_1$ so $x_n $ is bounded right?
$endgroup$
– jackson
Jan 2 at 13:46






$begingroup$
@ThomasShelby for $E_1$ is bound and $x_nin E_1$ so $x_n $ is bounded right?
$endgroup$
– jackson
Jan 2 at 13:46














$begingroup$
Your proof looks OK to me.
$endgroup$
– Thomas Shelby
Jan 2 at 13:51




$begingroup$
Your proof looks OK to me.
$endgroup$
– Thomas Shelby
Jan 2 at 13:51




1




1




$begingroup$
@ThomasShelby thanks a lot
$endgroup$
– jackson
Jan 2 at 13:52




$begingroup$
@ThomasShelby thanks a lot
$endgroup$
– jackson
Jan 2 at 13:52


















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