# Every convergent sequence has a monotoic subsequence

• Mr Davis 97
In summary, to prove that every convergent sequence has a monotone subsequence, we define the limit L of the sequence and note that any epsilon-ball around L contains infinitely many points. We then choose a subsequence by selecting elements from the set (L, infinity) or (-infinity, L). To ensure monotonicity, we select elements such that each subsequent element is contained within the previous element and there are infinitely many elements to choose from. To make this proof more rigorous, additional supporting details and descriptions are needed, such as clarifying the direction of the sequence and addressing the possibility of repeated elements in the subsequence.

## Homework Statement

Prove that every convergent sequence has a monotone subsequence.

## The Attempt at a Solution

Define ##L## to be the limit of ##(a_n)##. Then every ##\epsilon##-ball about L contains infinitely many points. Note that ##(L, \infty)## or ##(-\infty, L)## (or both) has infinitely many elements. Suppose that ##(L, \infty)## has infinitely many elements. Choose ##n_1## such that ##a_{n_1} \in (L, \infty)##. Choose ##n_2 \in (L,a_{n_1})##. In general, choose ##n_{k+1}## such that ##a_{n_{k+1}} \in (L, a_{n_{k}})##. This assignment can always be made since there are infinitely many elements of ##(a_n)## in ##(L, a_{n_{k}})## to choose from such that ##a_{n_{k+1}} \in (L, a_{n_{k}})##.

Do you have a question about this? I think it is a good start and only needs more supporting detailed statements to make it a rigorous proof.

FactChecker said:
Do you have a question about this? I think it is a good start and only needs more supporting detailed statements to make it a rigorous proof.
I guess my question then would be what supporting detailed statements do I need? As of now this is the best I can do, so I'm trying to see what I'm missing in terms of rigor.

Just add more description. The words do not cost you anything. What about the lower interval? Is the sequence increasing or decreasing? Monitone because?

Mr Davis 97 said:
Then every ##\epsilon##-ball about L contains infinitely many points.

It need not contain infinitely many distinct points. The sequence 2,2,2,2,... converges to 2.

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