# Constructing a Sequence to Show the Existence of a Limit

• Anoonumos
In summary, the problem is to show that there exists a sequence (v_n)_{n \geq 0} in a non-empty, upper bounded subset V of R such that 1) v_0 \leq v_1 \leq ... and 2) the limit of the sequence is sup V. A recursive construction can be used to construct the sequence, where v_0 is any member of V and for each v_i, there exists v_{i+1} in V such that (v_i+ sup(V))/2 < v_{i+1}. This ensures that the distance from the previous term to sup(V) is cut in half on each step. It is not sufficient to just state that
Anoonumos
Hi,

## Homework Statement

V is a non-empty, upper bounded subset of R. Show that a sequence $$(v_n)_{n \geq 0}$$ in V exists such that: 1)$$v_0 \leq v_1 \leq ...$$ and 2) the limit of the sequence is sup V. (Hint: use a recursive construction)

## The Attempt at a Solution

V is non-empty and upper bounded so sup V exists.
Suppose $$sup V \in V$$ Construct the sequence: sup V = v0 = v1...
And I know how to go from there.

I'm having problems with the case: sup V is not in V.
I was thinking of constructing the sequence:
$$v_0 \in V$$
$$(v_n, supV) \subset (v_{n-1}, supV)$$

But I'm not sure if this is right or how to proof that the limit of this sequence is sup V.
Any ideas?

Let $v_0$ be any member of V. We can do this because V is non-empty.

The recursion step: for any $v_i$, $(sup V+ v_i)/2< sup(V)$ and so is not an upper bound on V/. There must exist some $v_{i+1}$ in V such that $(v_i+ sup(V))/2< v_{i+1}$. That cuts the distance from the previous term to sup(V) in half on each step.

Thanks!

Sorry for the double post, but I came across another problem when writing it down. v(i+1) doesn't have to exist right? And there doesn't have to be a smallest v(i+1). So how would one formulate the sequence?

$$v_{i+1} \in [v_1, sup V)$$ perhaps?

Or is it enough to state v(i+1) exists and v(i+1) ? vi and v(i+1) is in V, to formulate a sequence?

Last edited:
No, $v_{i+1}$ by this method must exist: Since in this case we are assuming that sup(V) is not in V itself, our $(v_i+ sup(V))/2$ is NOT sup(V) and so is not an upper bound for V. There must exist at least one member of V larger than $(v_i+ sup(V))/2$. I said nothing about $v_{i+1}$ being the smallest such member of V- choose any of them.

But it is not sufficient just to say that $v_{i+1}$ is some number in V larger than $v_i$. That sequence would not necessarily converge to sup(V).

I understand. I have one final question.
Nevermind, thanks :)

Last edited:

## 1. What is a sequence?

A sequence is a list of numbers in a specific order, typically with a pattern that can be identified. The order of the numbers is important and determines the sequence.

## 2. How is a sequence constructed?

A sequence is constructed by following a specific rule or pattern. This can involve adding or subtracting a certain number, multiplying or dividing by a constant, or using a more complex formula. The rule is applied to each number in the sequence to generate the next number.

## 3. What are the different types of sequences?

There are many different types of sequences, but some common types include arithmetic sequences, geometric sequences, and Fibonacci sequences. Arithmetic sequences involve adding or subtracting a constant number to each term, geometric sequences involve multiplying or dividing by a constant ratio, and Fibonacci sequences involve adding the two previous terms together to generate the next term.

## 4. Why are sequences important in science?

Sequences are important in science because they can help us understand and predict patterns and relationships in natural phenomena. They are also used in many mathematical and scientific models to make predictions and solve problems.

## 5. How can sequences be extended or continued?

Sequences can be extended or continued by using the same rule or pattern to generate additional terms. This allows us to predict future terms in the sequence and better understand the overall pattern and behavior of the sequence.

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