# Thread: Question on monotone convergence theorem problem

1. ## Question on monotone convergence theorem problem

Hey guys,

Can you check whether my proof of the problem?

Show that a monotone sequence has at most one cluster point.

pf)

i) {a_n} is monotone increasing and bounded above.

By monotone convergence theorem, {a_n} -> x

If {a_n} is monotone decreasing and bounded below

b_n = -a_n and apply the same as above.

a_n -> x

Since a_n converges to x, a_n only has x as its cluster point.

ii) {a_n} is monotone increasing and not bounded.

There is N in N such that |a_N| $\displaystyle \geq M, \forall M>0$

Assume a_n -> x.

For any epsilon > 0 There is a_N such that |a_N - x | < 1, when n>N.

|a_N| - |x| $\displaystyle \leq |a_N - x| < 1$

M = Max { |a_1|, |a_2|, ... |a_N|, |a_n| +1}

|a_n| $\displaystyle \leq M$

Since a_n is unbounded, contradiction.

Since a_n is unbounded and diverges, a_n has no cluster point.

If a_n is monotone decreasing and unbounded, let b_n = - a_n and apply the same procedure.

2. ## Re: Question on monotone convergence theorem problem Originally Posted by rokman54 Can you check whether my proof of the problem?

Show that a monotone sequence has at most one cluster point.

Frankly, I follow little of what you posted. Why are you doing so may different things.
Look carefully at what you are asked to show.
Suppose that $\displaystyle (a_n)$ is a a monotone sequence.
Without loss of generality (wlog) we can assume the sequence is non-decreasing.

For proof by contradiction, suppose the sequence has two cluster points, $\displaystyle p~\&~q$.
Again wlog we can assume $\displaystyle p<q$. Let $\displaystyle c=\frac{q-p}{2}>0$.

The open intervals $\displaystyle (p-c,p+c)\cap(q-c,q+c)=\emptyset$.

By definition of cluster point, each of those intervals must contain infinitely many points of $\displaystyle (a_n)$.
Is the possible for a non-decreasing sequence?

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