- #1

Jaquis2345

- 6

- 1

- Homework Statement
- Prove that if p1,p2,p3,... is a non-decreasing sequence and there is a point x right of every point of the sequence, then the sequence converges to x.

- Relevant Equations
- Def: The sequence p1, p2, p3, ... is non-decreasing if for each

positive integer n, pn<=pn+1:

Def: If M is a set and there is a point to

the right of every point of M, then there is either a rightmost point of M or

a first point to the right of M.

So far this is what I have.

Proof:

Let p1, p2, p3 be a non-decreasing sequence. Assume that not all points of the sequence p1,p2,p3,... are equal.

If the sequence p1,p2,p3,... converges to x then for every open interval S containing x there is a positive integer N s.t. if n is a positive integer and n>=N, then pn is contained by S (definition we use in class).

Let S = (a,b) be an open interval containing x.

Let N be a positive integer and let n be a positive integer s.t. n>=N.

Let M be a set s.t. M contains all points of the sequence pn.

Since M is a set there exists either a rightmost point of M or a first-point to the right of M.

Since pn is a non-decreasing sequence s.t. not all points of pn are equal, there exists pn+1 points in the set M. Thus, the set is infinite and has no rightmost point. So, M has a 1st point to the right.

Since x is to the right of every point in M, x is the first point to the right of M.

Thus, there exists no point q s.t. pn<q<x.

Since pn<x, pn<b.

Since a<x and x is the first-point to the right of M, a cannot exist between pn and x. So, a<x and pn is an element of S.

Since pn is an element of S, the sequence p1,p2,p3,... converges to x.

I think I have the right idea but I am unsure of my logic when it comes to the rightmost point of M and x being the first point to the right of M.

Proof:

Let p1, p2, p3 be a non-decreasing sequence. Assume that not all points of the sequence p1,p2,p3,... are equal.

If the sequence p1,p2,p3,... converges to x then for every open interval S containing x there is a positive integer N s.t. if n is a positive integer and n>=N, then pn is contained by S (definition we use in class).

Let S = (a,b) be an open interval containing x.

Let N be a positive integer and let n be a positive integer s.t. n>=N.

Let M be a set s.t. M contains all points of the sequence pn.

Since M is a set there exists either a rightmost point of M or a first-point to the right of M.

Since pn is a non-decreasing sequence s.t. not all points of pn are equal, there exists pn+1 points in the set M. Thus, the set is infinite and has no rightmost point. So, M has a 1st point to the right.

Since x is to the right of every point in M, x is the first point to the right of M.

Thus, there exists no point q s.t. pn<q<x.

Since pn<x, pn<b.

Since a<x and x is the first-point to the right of M, a cannot exist between pn and x. So, a<x and pn is an element of S.

Since pn is an element of S, the sequence p1,p2,p3,... converges to x.

I think I have the right idea but I am unsure of my logic when it comes to the rightmost point of M and x being the first point to the right of M.