Let (a_n,b_n) be a sequence of nested open intervals. Assume that intersection of all (a_n, b_n) is empty. Show that Sup a_n=Inf b_n
Proof: {a_n} is an increasing sequence bounded above and {b_n} is a decreasing sequence bounded below. We know from previous theorems that since {a_n} and {b_n} are monotonically increasing or decreasing and lim b_n=inf b_n and lim a_n=sup a_n Need to show lim(a_n - b_n)=0 To show this... b - a = (b - b_n) + (b_n - a_n) + (a_n - a) |b - a| <= |b - b_n| + |b_n - a_n| + |a_n - a| for any e > 0 we can find N such that for all n > N |b - b_e| < e/3 |b_n - a_n| < e/3 |a_n - a| < e/3 and so |b - a| < e. Thus b - a=0 and a=b.
> We know from previous theorems that since {a_n} and {b_n} > are monotonically increasing or decreasing > and lim b_n=inf b_n > and lim a_n=sup a_n > Need to show lim(a_n - b_n)=0 > To show this... > b - a = (b - b_n) + (b_n - a_n) + (a_n - a) > |b - a| <= |b - b_n| + |b_n - a_n| + |a_n - a| > for any e > 0 we can find N such that for all n > N > |b - b_e| < e/3 > |b_n - a_n| < e/3 > |a_n - a| < e/3 > and so |b - a| < e. Thus b - a=0 and a=b.
rabbits77 wrote: > Let (a_n,b_n) be a sequence of nested open intervals. Assume > that intersection of all (a_n, b_n) is empty. > Show that Sup a_n=Inf b_n
I think I previously made things harder and more awkward than necessary. I think we can say that by the definition of nested intervals lim(b_n - a_n)=0 since the distance between a_n,b_n approaches zero as n-> oo. Proof: {a_n} is an increasing sequence bounded above and {b_n} is a decreasing sequence bounded below. We know from previous theorems that since {a_n} and {b_n} are monotonically increasing or decreasing then lim b_n=inf b_n and lim a_n=sup a_n Since lim(b_n - a_n)=0 we have b - a = lim b_n - lim a _n = lim(b_n - a_n)=0 . Thus inf b_n = sup a_n .
I think this is a pretty straightforward and simple proof now. Unless I made nay errors. Have I?
Here is another attempt... Let (a_n,b_n) be a sequence of nested open intervals. Assume that intersection of all (a_n, b_n) is empty. Show that Sup a_n=Inf b_n
Proof: {a_n} is an increasing sequence bounded above and {b_n} is a decreasing sequence bounded below. We know from previous theorems that since {a_n} and {b_n} are monotonically increasing or decreasing then lim b_n=inf b_n and lim a_n=sup a_n lim b_n - lim a_n=lim(b_n - a_n)= lim L/2^(n-1)=0 where L is the distance of I_0 and the length of I_n is L/2^(n-1). Since, for example, we know(from a previous theorem) that for 0 < a < 1 we have a^n -> 0 Also, we know that this distance approaches zero as (a_n,b_n) becomes a smaller and smaller interval as a_n and b_n get closer and closer for each successive nested interval.
Is that correct?
If so then it follows that lim b_n - lim a _n = lim(b_n - a_n)=0 . Thus inf b_n = sup a_n .
rabbits77 wrote: > Here is another attempt... > Let (a_n,b_n) be a sequence of nested open intervals. Assume > that intersection of all (a_n, b_n) is empty. > Show that Sup a_n=Inf b_n
> Proof: > {a_n} is an increasing sequence bounded above > and {b_n} is a decreasing sequence bounded below. > We know from previous theorems that since {a_n} and {b_n} > are monotonically increasing or decreasing > then lim b_n=inf b_n > and lim a_n=sup a_n > lim b_n - lim a_n=lim(b_n - a_n)= lim L/2^(n-1)=0 > where L is the distance of I_0 and the length of I_n is L/2^(n-1).
I say this because if I_0 =(-L,L) and we bisect it repeatedly so that I_1=(0,L) and so forth.
> Since, for example, we know(from a previous theorem) that for 0 < a < 1 > we have a^n -> 0 > Also, we know that this distance > approaches zero as (a_n,b_n) becomes a smaller and smaller interval as > a_n and b_n get closer and closer for each successive nested interval.
> Is that correct?
> If so then it follows that lim b_n - lim a _n = lim(b_n - a_n)=0 . Thus > inf b_n = sup a_n .
William Elliot wrote: > On Mon, 6 Oct 2008, rabbits77 wrote:
>> Let (a_n,b_n) be a sequence of nested open intervals. Assume
> Of course, they are all not empty, ie no j with bj <= aj.
>> that intersection of all (a_n, b_n) is empty. >> Show that Sup a_n=Inf b_n
> a = sup_j aj <= b = inf b_j. > If a < b, then > nonnul (a,b) subset /\_j (aj, bj).
I am sorry(this may be a minor confusion) but I am not sure what you mean by this? If a < b then the open non-empty interval (a,b) is a subset of the intersection of all (a_j,b_j) but since we know that this is empty then we have a contradiction and we are done?
>> Proof: >> {a_n} is an increasing sequence bounded above >> and {b_n} is a decreasing sequence bounded below.
> Generalize to an uncountable nest of open intervals.
>> We know from previous theorems that since {a_n} and {b_n} >> are monotonically increasing or decreasing >> and lim b_n=inf b_n >> and lim a_n=sup a_n >> Need to show lim(a_n - b_n)=0 >> To show this... >> b - a = (b - b_n) + (b_n - a_n) + (a_n - a) >> |b - a| <= |b - b_n| + |b_n - a_n| + |a_n - a| >> for any e > 0 we can find N such that for all n > N >> |b - b_e| < e/3 >> |b_n - a_n| < e/3 >> |a_n - a| < e/3 >> and so |b - a| < e. Thus b - a=0 and a=b.
On Tue, 7 Oct 2008, rabbits77 wrote: > William Elliot wrote: > > On Mon, 6 Oct 2008, rabbits77 wrote:
> >> Let (a_n,b_n) be a sequence of nested open intervals. Assume
> > Of course, they are all not empty, ie no j with bj <= aj.
> >> that intersection of all (a_n, b_n) is empty. > >> Show that Sup a_n=Inf b_n
> > a = sup_j aj <= b = inf b_j. > > If a < b, then > > nonnul (a,b) subset /\_j (aj, bj).
> I am sorry(this may be a minor confusion) but I am not sure what you > mean by this?
In this senctece you are speaking English, not math. Thus a blank space is required to be between 'sorry' and '(". So don't be confused about punctuation.
> If a < b then the open non-empty interval (a,b) is a subset of > the intersection of all (a_j,b_j) but since we know that this > is empty then we have a contradiction and we are done?
In article <gcen6v$19...@aioe.org>, rabbits77 <rabbit...@my-deja.com> wrote:
> Here is another attempt... > Let (a_n,b_n) be a sequence of nested open intervals. Assume > that intersection of all (a_n, b_n) is empty. > Show that Sup a_n=Inf b_n
Here's an interesting supplemental problem that the above problem suggests, that you might find interesting to have a go at:
Show that one of the sets {a_i}, {b_i} must be finite, and the other one must be infinite.
If you are getting a good grasp of the material you are studying, this should be a short and easy proof.
