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:
$$|a_{n}|\leqslant|a_{n}-a|+|a|$$
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 ${a_n : n < k}$ 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.