Bounded non-decreasing sequence is convergent

[Jaquis2345]

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.

 

[FactChecker]

This seems wrong. The problem statement needs to say more about x. Consider the sequence 0.9, 0.99, 0.999, 0.9999, ..., which obviously converges to 1, and the point x=2.

 

[Jaquis2345]

This seems wrong. The problem statement needs to say more about x. Consider the sequence 0.9, 0.99, 0.999, 0.9999, ..., which obviously converges to 1, and the point x=2.

After emailing my professor it seems there was a typo in the problem we were given. I just need to show that the sequence converges not that it converges to x.

 

[PeroK]

After emailing my professor it seems there was a typo in the problem we were given. I just need to show that the sequence converges not that it converges to x.

Why do you think this statement is true?

 

[Jaquis2345]

Why do you think this statement is true?

After working on the problem for a while this is what I got (roughly). Since the point x is to the right of every point in the sequence, then from the completeness axiom there is either a rightmost point of the set contains all points of the sequence or a first point to the right of that set. I then went by cases. The first case I assumed that some arbitrary point s was contained by an open interval (a,b). And s was the right most point of the set containing all points of the sequence. Since it was the rightmost point it was obviously contained in the set. And since it was non-decreasing some point in that set had to equal s. Therefore my sequence was contained by (a,b). My second case dealt with the first point to the right but I think I got it.

 

[PeroK]

From the Completeness axiom, every set bounded above has a supremum is what you are trying to say?