Let (a_n,b_n) be a sequence of nested open intervals. Assume that intersection of all (a_n, b_n) is empty. Let A=Sup a_n=Inf b_n Show that there exists an n_0 such that for all n > n_0 either a_n=A or b_n=A.
Proof: If r,s >= n_0 the x_r, x_s both belong to (a_n0, b_n0) . therefore |x_r - x_s| <= b_n0 - a_n0 since lim(b_n - a_n)=0 . Furthermore, since b_n - a_n =0 it follows that x_r is Cauchy and thus converges so x_r -> x. We know that A=Sup a_n=Inf b_n thus the subsequence x_r in (a_n0, b_n0) converges to the single value A for all n > n_0 and thus so do a_n and b_n . That is, either a_n=A or b_n=A. Q.E.D
On Oct 6, 12:59 pm, TheGist <theg...@nospam.net> wrote:
> Let (a_n,b_n) be a sequence of nested open intervals. Assume > that intersection of all (a_n, b_n) is empty. > Let A=Sup a_n=Inf b_n > Show that there exists an n_0 such that for all n > n_0 either a_n=A or > b_n=A.
> Proof: > If r,s >= n_0 the x_r, x_s both belong to > (a_n0, b_n0) . therefore |x_r - x_s| <= b_n0 - a_n0
It looks to me like you are assuming the existence of n_0 with the properties that you want and then proving it exists. It looks this way because you start out, "If r,s >= n_0 the ...". What is n_0 in this context? You are supposed to prove the existence of a certain n_0 with certain properties. Yet you are already using n_0 and yet you haven't even defined it!
> since lim(b_n - a_n)=0 . Furthermore, since b_n - a_n =0 > it follows that x_r is Cauchy and thus converges so x_r -> x. > We know that A=Sup a_n=Inf b_n thus the subsequence x_r in (a_n0, b_n0) > converges to the single value A for all n > n_0 and thus so do a_n and > b_n . That is, either a_n=A or b_n=A. Q.E.D
> Problems? Am I close or is it all wrong?
I am afraid it is all wrong, unless I am very much mistaken which is far from improbable. But you should be able go get this problem. It might even involve some theorem or other about nested closed intervals, although I don't know as I haven't tried to do the problem.
> Let (a_n,b_n) be a sequence of nested open intervals. Assume > that intersection of all (a_n, b_n) is empty. > Let A=Sup a_n=Inf b_n > Show that there exists an n_0 such that for all n > n_0 either a_n=A or > b_n=A.
> Proof: > If r,s >= n_0
What is n_0? Aren't you supposed to prove that _there is_ some n_0 such that... ?
> the x_r, x_s
What are x_r and x_s?
> both belong to > (a_n0, b_n0) . therefore |x_r - x_s| <= b_n0 - a_n0 > since lim(b_n - a_n)=0 .
If _x_ and _y_ both belong to (a,b), then |x - y| < b - a. Why do you think you need the fact that lim_n (b_n - a_n) = 0? Besides, why do you believe that this is true?
> Furthermore, since b_n - a_n =0
Why? What is _n_?
> it follows that x_r is Cauchy and thus converges so x_r -> x.
What is _x_?
> We know that A=Sup a_n=Inf b_n thus the subsequence x_r in (a_n0, b_n0)
x_r is a subsequence of which sequence?
> converges to the single value A for all n > n_0
Just for those? Not for the others values of _n_?
> and thus so do a_n and b_n .That is, either a_n=A or b_n=A. Q.E.D
TheGist wrote: > Let (a_n,b_n) be a sequence of nested open intervals. Assume > that intersection of all (a_n, b_n) is empty. > Let A=Sup a_n=Inf b_n > Show that there exists an n_0 such that for all n > n_0 either a_n=A or > b_n=A.
> Proof: > If r,s >= n_0 the x_r, x_s both belong to > (a_n0, b_n0) . therefore |x_r - x_s| <= b_n0 - a_n0 > since lim(b_n - a_n)=0 . Furthermore, since b_n - a_n =0 > it follows that x_r is Cauchy and thus converges so x_r -> x. > We know that A=Sup a_n=Inf b_n thus the subsequence x_r in (a_n0, b_n0) > converges to the single value A for all n > n_0 and thus so do a_n and > b_n . That is, either a_n=A or b_n=A. Q.E.D
> Problems? Am I close or is it all wrong?
I am afraid I am very stuck on this one. I need to start at square one again. Can anyone provide me with a hint to get started in the right direction?
> TheGist wrote: > > Let (a_n,b_n) be a sequence of nested open intervals. Assume > > that intersection of all (a_n, b_n) is empty. > > Let A=Sup a_n=Inf b_n > > Show that there exists an n_0 such that for all n > n_0 either a_n=A or > > b_n=A.
> > Proof: > > If r,s >= n_0 the x_r, x_s both belong to > > (a_n0, b_n0) . therefore |x_r - x_s| <= b_n0 - a_n0 > > since lim(b_n - a_n)=0 . Furthermore, since b_n - a_n =0 > > it follows that x_r is Cauchy and thus converges so x_r -> x. > > We know that A=Sup a_n=Inf b_n thus the subsequence x_r in (a_n0, b_n0) > > converges to the single value A for all n > n_0 and thus so do a_n and > > b_n . That is, either a_n=A or b_n=A. Q.E.D
> > Problems? Am I close or is it all wrong?
> I am afraid I am very stuck on this one. > I need to start at square one again. > Can anyone provide me with a hint to get started in the > right direction?- Hide quoted text -
> - Show quoted text -
No problem. You know that a_1 <= a_2 <= ... and b_1 >- b_2 >= ... . You also know (previous post) that sup a_n = sup b_n. Now if one either of these sequences becomes stationary, say a_56 is the final a, then there is nothing in the interesection because the b's march all the way down to a_56, sweeping past every number in their path on the way to their goal of a_56. On the other hand, if neither the a's nor the b's becomes stationary, can this happen, or must there be an intersection?
> On Oct 6, 5:58 pm, TheGist <theg...@nospam.net> wrote: >> TheGist wrote: >>> Let (a_n,b_n) be a sequence of nested open intervals. Assume >>> that intersection of all (a_n, b_n) is empty. >>> Let A=Sup a_n=Inf b_n >>> Show that there exists an n_0 such that for all n > n_0 either a_n=A or >>> b_n=A. >>> Proof: >>> If r,s >= n_0 the x_r, x_s both belong to >>> (a_n0, b_n0) . therefore |x_r - x_s| <= b_n0 - a_n0 >>> since lim(b_n - a_n)=0 . Furthermore, since b_n - a_n =0 >>> it follows that x_r is Cauchy and thus converges so x_r -> x. >>> We know that A=Sup a_n=Inf b_n thus the subsequence x_r in (a_n0, b_n0) >>> converges to the single value A for all n > n_0 and thus so do a_n and >>> b_n . That is, either a_n=A or b_n=A. Q.E.D >>> Problems? Am I close or is it all wrong? >> I am afraid I am very stuck on this one. >> I need to start at square one again. >> Can anyone provide me with a hint to get started in the >> right direction?- Hide quoted text -
>> - Show quoted text -
> No problem. You know that a_1 <= a_2 <= ... and b_1 >- b_2 >= ... . > You also know (previous post) that sup a_n = sup b_n. Now if one > either of these sequences becomes stationary, say a_56 is the final > a, then there is nothing in the interesection because the b's march > all the way down to a_56, sweeping past every number in their path on > the way to their goal of a_56. On the other hand, if neither the a's > nor the b's becomes stationary, can this happen, or must there be an > intersection?
So, let me try and formalize this... Would you advise a proof by contradiction?
Suppose no such n_0 exists. Then neither a_n and b_n reach a limit. Then for a_n increasing and b_n decreasing eventually the intersection of all (a_n, b_n) is not null which is a contraditcion. Thus either a_n=A or b_n=A. Is this the argument you suggest? If so, I will need to more formally show that there is an eventual intersection I think. that there
>>>> Let (a_n,b_n) be a sequence of nested open intervals. Assume >>>> that intersection of all (a_n, b_n) is empty. >>>> Let A=Sup a_n=Inf b_n >>>> Show that there exists an n_0 such that for all n > n_0 either a_n=A or >>>> b_n=A. >>>> Proof: >>>> If r,s >= n_0 the x_r, x_s both belong to >>>> (a_n0, b_n0) . therefore |x_r - x_s| <= b_n0 - a_n0 >>>> since lim(b_n - a_n)=0 . Furthermore, since b_n - a_n =0 >>>> it follows that x_r is Cauchy and thus converges so x_r -> x. >>>> We know that A=Sup a_n=Inf b_n thus the subsequence x_r in (a_n0, b_n0) >>>> converges to the single value A for all n > n_0 and thus so do a_n and >>>> b_n . That is, either a_n=A or b_n=A. Q.E.D >>>> Problems? Am I close or is it all wrong? >>> I am afraid I am very stuck on this one. >>> I need to start at square one again. >>> Can anyone provide me with a hint to get started in the >>> right direction?- Hide quoted text -
>>> - Show quoted text -
>> No problem. You know that a_1 <= a_2 <= ... and b_1 >- b_2 >= ... . >> You also know (previous post) that sup a_n = sup b_n. Now if one >> either of these sequences becomes stationary, say a_56 is the final >> a, then there is nothing in the interesection because the b's march >> all the way down to a_56, sweeping past every number in their path on >> the way to their goal of a_56. On the other hand, if neither the a's >> nor the b's becomes stationary, can this happen, or must there be an >> intersection?
> So, let me try and formalize this... > Would you advise a proof by contradiction?
> Suppose no such n_0 exists. Then neither a_n and b_n reach a limit. > Then for a_n increasing and b_n decreasing eventually the intersection > of all (a_n, b_n) is not null
What do you mean by "eventually"? The intersection of all intervals either is empty (not "null") or it is not.
> >>>> Let (a_n,b_n) be a sequence of nested open intervals. Assume > >>>> that intersection of all (a_n, b_n) is empty. > >>>> Let A=Sup a_n=Inf b_n > >>>> Show that there exists an n_0 such that for all n > n_0 either > >>>> a_n=A or b_n=A. > >> No problem. You know that a_1 <= a_2 <= ... and b_1 >- b_2 >= ... . > >> You also know (previous post) that sup a_n = sup b_n. Now if one > >> either of these sequences becomes stationary, say a_56 is the final > >> a, then there is nothing in the interesection because the b's march > >> all the way down to a_56, sweeping past every number in their path on > >> the way to their goal of a_56. On the other hand, if neither the a's > >> nor the b's becomes stationary, can this happen, or must there be an > >> intersection?
> > Suppose no such n_0 exists. Then neither a_n and b_n reach a limit. > > Then for a_n increasing and b_n decreasing eventually the intersection > > of all (a_n, b_n) is not null
> What do you mean by "eventually"? The intersection of all intervals > either is empty (not "null") or it is not.
> > which is a contraditcion.
> Yes, but why is it not empty?
Because it'll catch A. Thus sup = inf doen't imply empty intersection.
If intersection is empty then
B \/ D = R
B = \/_j (-oo,aj]; D = \/_j [bj,oo)
If { aj | j in N } is finite, then some n with B = (-oo,a_n] otherwise B = (-oo, sup_j aj)
Likewise with { bj }. Thus four cases:
(-oo, sup_j aj) \/ (inf_j aj, oo) = R (-oo, a_n] \/ (inf_j aj, oo) = R (-oo, sup_j aj) \/ [b_m, oo) = R (-oo, a_n] \/ [b_m, oo) = R
etc.
-- Don't pass the buck. Outsource!
What ya have todo these days ta make a buck? Buck the system?
>>>>>> Let (a_n,b_n) be a sequence of nested open intervals. Assume >>>>>> that intersection of all (a_n, b_n) is empty. >>>>>> Let A=Sup a_n=Inf b_n
>>>>>> Show that there exists an n_0 such that for all n > n_0 either >>>>>> a_n=A or b_n=A.
>>>> No problem. You know that a_1 <= a_2 <= ... and b_1 >- b_2 >= ... . >>>> You also know (previous post) that sup a_n = sup b_n. Now if one >>>> either of these sequences becomes stationary, say a_56 is the final >>>> a, then there is nothing in the interesection because the b's march >>>> all the way down to a_56, sweeping past every number in their path on >>>> the way to their goal of a_56. On the other hand, if neither the a's >>>> nor the b's becomes stationary, can this happen, or must there be an >>>> intersection? >>> Suppose no such n_0 exists. Then neither a_n and b_n reach a limit. >>> Then for a_n increasing and b_n decreasing eventually the intersection >>> of all (a_n, b_n) is not null >> What do you mean by "eventually"? The intersection of all intervals >> either is empty (not "null") or it is not.
>>> which is a contraditcion. >> Yes, but why is it not empty?
> Because it'll catch A. > Thus sup = inf doen't imply empty intersection.
The question was not meant for you. I was trying to help th OP to get to the proof. If I wanted him to know, without effort, why is it that the intersection is empty, I would have told him.
> > On Oct 6, 5:58 pm, TheGist <theg...@nospam.net> wrote: > >> TheGist wrote: > >>> Let (a_n,b_n) be a sequence of nested open intervals. Assume > >>> that intersection of all (a_n, b_n) is empty. > >>> Let A=Sup a_n=Inf b_n > >>> Show that there exists an n_0 such that for all n > n_0 either a_n=A or > >>> b_n=A. > >>> Proof: > >>> If r,s >= n_0 the x_r, x_s both belong to > >>> (a_n0, b_n0) . therefore |x_r - x_s| <= b_n0 - a_n0 > >>> since lim(b_n - a_n)=0 . Furthermore, since b_n - a_n =0 > >>> it follows that x_r is Cauchy and thus converges so x_r -> x. > >>> We know that A=Sup a_n=Inf b_n thus the subsequence x_r in (a_n0, b_n0) > >>> converges to the single value A for all n > n_0 and thus so do a_n and > >>> b_n . That is, either a_n=A or b_n=A. Q.E.D > >>> Problems? Am I close or is it all wrong? > >> I am afraid I am very stuck on this one. > >> I need to start at square one again. > >> Can anyone provide me with a hint to get started in the > >> right direction?- Hide quoted text -
> >> - Show quoted text -
> > No problem. You know that a_1 <= a_2 <= ... and b_1 >- b_2 >= ... . > > You also know (previous post) that sup a_n = sup b_n. Now if one > > either of these sequences becomes stationary, say a_56 is the final > > a, then there is nothing in the interesection because the b's march > > all the way down to a_56, sweeping past every number in their path on > > the way to their goal of a_56. On the other hand, if neither the a's > > nor the b's becomes stationary, can this happen, or must there be an > > intersection?
> So, let me try and formalize this... > Would you advise a proof by contradiction?
> Suppose no such n_0 exists. Then neither a_n and b_n reach a limit. > Then for a_n increasing and b_n decreasing eventually the intersection > of all (a_n, b_n) is not null which is a contraditcion. Thus either > a_n=A or b_n=A. > Is this the argument you suggest? If so, I will need to more formally > show that there is an eventual intersection I think. > that there- >
Jumping in here after several other posters have spoken. Yes that is the general idea, but you have to formalize. You have to actually write a proof of your claims, such that the intersection is non-empty when you say it is non-empty and that it is empty if you say it is empty. Learning to formalize a proof is an important art that we tend to forget took a lot of learning once we have done it a few thousand times.
