When proving that every convergent sequence is bounded, do we have to fix $\epsilon$ to a concrete number in the proof, or can we let it be arbitrary?

Question

My Calculus instructor presented the following proof for the claim that every convergent sequence is bounded:

We assume the sequence $\displaystyle \big\{ a_n \big\}_{n = 0}^\infty $ is convergent. Let $L$ be the limit.

We choose $\epsilon = 1$ in the definition of $ \displaystyle L = \lim_{n \to \infty} a_n $.

Thus we know $\exists n_0 \in \mathbb{N}, $ $$ \forall n \in \mathbb{N}, n \geq n_0 \Longrightarrow L - 1 < a_n < L + 1 \tag{1} $$ We take $$A = \min\{ L - 1, a_0, a_1, a_2, \dots, a_{n_0 -1} \} \\ B = \max\{ L + 1, a_0, a_1, a_2, \dots, a_{n_0 -1} \} $$

We will show that $\forall n \in \mathbb{N}, A \leq a_n \leq B$.

Let $n \in \mathbb{N}$.

If $n \geq n_0$, then from $(1)$, $A \leq L - 1 < a_n < L + 1 \leq B$.

If $n < n_0$, then by definition of $A$ and $B$, $A \leq a_n \leq B$.

$\blacksquare$

My question is, do we have to choose $\epsilon = 1$ in the proof?

Or could we say something more generic, such as "let $\epsilon > 0$", and work with $\epsilon$ throughout the proof?

If we can't use "let $\epsilon > 0$" instead, why can't we?

Answer 1

Assume $\{a_n\}_{n=1}^\infty$ is a convergent sequence such that $\lim\limits_{n \to \infty} a_n = L.$

Let $\varepsilon>0$ be given. By definition, there is $N \in \mathbb N$ so that $$L-\varepsilon<a_n<L+\varepsilon \, \text{ whenever } \, n \geq N.$$

Take $$A = \min\left\{ L-\varepsilon, a_0, a_1, \ldots, a_{n-1} \right\},$$ $$B = \max\left\{ L+\varepsilon, a_0, a_1, \ldots, a_{n-1} \right\}.$$

Now suppose $n \in \mathbb{N}$.

If $n \geq N$, then $$ A\leq L-\varepsilon<a_n < L+\varepsilon \leq B .$$

Otherwise, $n \leq N$ and we clearly have that $A \leq a_n \leq B$.

$\blacksquare$

Basically if we are trying to prove that a sequence converges, then we have to show that the definition of convergent sequence is satisfied for any $\varepsilon>0$. But if we are trying to prove something about a convergent sequence by using the fact that a sequence converges, then we may take any $\varepsilon>0$ (so long as it works for the argument at hand). For this problem, how we take $\varepsilon>0$ doesn't matter but probably most will take $\varepsilon=1$ out of simplicity and convention.

Answer 2

Any $\varepsilon>0$ will do. You can take, say $\varepsilon=\frac\pi3$, or $\varepsilon=\sqrt{\frac12}$. But $\varepsilon=1$ is the natural choice.

I suppose that what is confusing you is the fact that definition of convergent sequence begins with “$(\forall\varepsilon>0)$”, but that only means that if you want to prove (by the definition) that a sequence converges, then you can impose no conditions whatsoever on $\varepsilon$ (besides the fact that $\varepsilon>0$). But if you are using the fact that a sequence converge, then you can take any $\varepsilon>0$ that you want.

Answer 3

You can use any fixed value of $\epsilon$. But if you write it with a particular value - perhaps $1$ or $1000$, your proof will better convey the idea behind your thinking. If you begin "Let $\epsilon > 0$ ..." someone reading quickly might think you are just cutting and pasting from other proofs that start that way.

Answer 4

The idea of this proof is that the list $a_0,a_1,\dots, a_{n_0-1}$ is finite. This is true for any fixed $\epsilon\gt0$ but we choose $\epsilon=1$ by convention. For example we could equally take $\epsilon=k\gt0$ then for $n_0\in\mathbb{N}$ such that $n\ge n_0\implies |a_n-L|\lt k$ we have the bounds $$a_n\le \max{(L+k,a_0,\dots,a_{n_0-1})}$$ $$a_n\ge \min{(L-k,a_0,\dots,a_{n_0-1})}$$