Proof: every convergent sequence is bounded

[SpY]]

Homework Statement

Prove that every convergent sequence is bounded.

Homework Equations

Definition of \lim_{n \to +\infty} a_n = L
\forall \epsilon > 0, \exists k \in \mathbb{R} \; s.t \; \forall n \in \mathbb{N}, n \geq k, \; |a_n - L| < \epsilon

Definition of a bounded sequence: A sequence is bounded iff it is bounded above and below, ie. \exists m \in \mathbb{R} \; s.t. a_n \geq m \; \forall n and similarly a_n \leq M

2. The attempt at a solution
Suppose a sequence a_n converges to some limit L.

Then by definition of the limit \forall \epsilon > 0, \exists k \in \mathbb{R} \; s.t \; \forall n \in \mathbb{N}, n \geq k, \; |a_n - L| < \epsilon

Rewriting the absolute value, L - \epsilon < a_n < L + \epsilon

Since L, \epsilon \in \mathBB{R}, L + \epsilon > a_n \; \text{and} L - \epsilon < a_n. So the sequence is bounded above and below, hence bounded.
...

In my lecture notes, the given proof chooses \epsilon = 1 but does this affect the proof since \epsilon is arbitrary? It is also written as \left \{ a_n : n \leq k \right \} \subset (L-1, L+1) but my notation is equivalent?

 

[jbunniii]

In my lecture notes, the given proof chooses \epsilon = 1 but does this affect the proof since \epsilon is arbitrary? It is also written as \left \{ a_n : n \leq k \right \} \subset (L-1, L+1) but my notation is equivalent?

You can choose any \epsilon you like. The resulting k will be different depending on which \epsilon you choose, but that doesn't matter. All you need is a specific pair of \epsilon and k that work.

Yes, these two notations mean the same thing:

\{a_n : n \geq k\} \subset (L - 1, L + 1)

and

\forall n \in \mathbb{N}, n \geq k, \; |a_n - L| < 1

 

[Stephen Tashi]

The basic problem with your proof is that you aren't using logical quantifiers properly. In statements about variables, a sentence at the top of the page may say something about \epsilon and later sentence may mention \epsilon, but the two \epsilon's need not have anything to do with each other. Each use of a variable should occur with the "scope" of a certain quantifier.

You go from this statement:

Then by definition of the limit \forall \epsilon > 0, \exists k \in \mathbb{R} \; s.t \; \forall n \in \mathbb{N}, n \geq k, \; |a_n - L| < \epsilon

To

Rewriting the absolute value, L - \epsilon < a_n < L + \epsilon

Without making it clear that you wish to pick a particular \epsilon from among the possibly thousands of different ones that exist. There is actually a name for this procedure. As I recall, the author Irving Copi called it something like "existential instantiation". It says if we know "there exists" a variable with certain properties, we may introduce a variable that is one particular instance.

You also aren't doing a good job with n. In the definition of bounded the quantifier is "for all" n and in the definition of limit the condition is "for all n \ge k". The two statements are not about the same n.

Once you pick a particular \epsilon, the definition of limit is going to get you
L - \epsilon < a_n < L + \epsilon for all n \ge k , but not "for all n ".

For example, L + \epsilon doesn't work for a upper bound because for some j < k we can have a_j > L + \epsilon. So you should define M as the larger of: max \{a_j: j < k \} and L + \epsilon.

 

[hunt_mat]

Note that a_{n}=a_{n}-a+a, and so that:
<br /> |a_{n}|\leqslant|a_{n}-a|+|a|<br />
From here, I think that you can show that all convergent sequences are bounded.

 

[SpY]]

Although I don't really see the need to choose an \epsilon, I understand why for clarity it makes sense to chose one. I thought that proving that there exists some k for which an is bounded would be sufficient, but I see now one has to consider a set of {a_n : n &lt; k} then use the fact that the union of bounded sets is bounded.

Thanks

 

[Maxter]

This is that quantifier problem. The epsilon interval only gives you a bound for some of the terms.

Also, at some initial point, you must *clearly fix* an epsilon. It's nice to make that concrete, particularly if you have to use it to prove something further down the road, hence choose it to be simply 1.

 

[Stephen Tashi]

Although I don't really see the need to choose an \epsilon

I agree that it isn't necessary to state a numerical value for \epsilon. It is necessary to state that a specific \epsilon is being chosen, even if it is represented only as a symbol.

If you don't pick a specific \epsilon then what would the expression L + \epsilon stand for? Would we interpret it as an infinite set of numbers formed by adding L to each possible \epsilon ? If we interpret it that way then it's doesn't help us find an upper bound since an upper bound is supposed be a number, not an infinite set of numbers.

 

[gb7nash]

If you don't pick a specific \epsilon then what would the expression L + \epsilon stand for? Would we interpret it as an infinite set of numbers formed by adding L to each possible \epsilon ?

No. Treat the epsilon as an arbitrary fixed number greater than 0. Similary, you can represent an even number a as a = 2k, where k is an integer. We have a fixed value of k. We don't treat this as all integers k.

 

[Stephen Tashi]

No. Treat the epsilon as an arbitrary fixed number greater than 0.

"Fixed" is the same as saying that \epsilon has been chosen to be specific.