Consequences of replacing $|a_n-a|<\epsilon $ with other assertions involving $\epsilon$ on the convergence of real sequences

Question

We are given a sequence $(a_n)$ of real numbers and a real number $a$. I want to show that the following are equivalent:

(i) for each $\epsilon>0$ there exists a $n_0 \in \mathbb{N}$ such that $|a_n-a|<\epsilon$ for each $n \geq n_0$,

(ii) for each $\epsilon>0$ there exists a $n_0 \in \mathbb{N}$ such that $|a_n-a|<2 \epsilon$ for each $n \geq n_0$,

(iii) for each $\epsilon \in (0,1)$ there exists a $n_0 \in \mathbb{N}$ such that $|a_n-a|<\epsilon$ for each $n \geq n_0$,

(iv) for each $\epsilon>0$ there exists a $n_0 \in \mathbb{N}$ such that $|a_n-a|<\frac{\epsilon}{2}$ for each $n \geq n_0$.

I have tried the following:

$(i) \Rightarrow (ii)$:

Since $(i)$ holds, we have that $\forall \epsilon'>0 \ \exists n_0 \in \mathbb{N}$ such that $|a_n-a|<\epsilon'$, $\forall n \geq n_0$.

We pick $\epsilon'=2 \epsilon$. Then we have that $\forall \epsilon>0 \ \exists n_0 \in \mathbb{N}$ such that $|a_n-a|<2 \epsilon$, $\forall n \geq n_0$.

$(ii) \Rightarrow (iii)$:

Since $(ii)$ holds, we have that $\forall \epsilon'>0 \ \exists n_0 \in \mathbb{N}$ such that $|a_n-a|<2 \epsilon'$, $\forall n \geq n_0$. We pick $0<\epsilon'<\frac{\epsilon}{2}<\frac{1}{2}$. Then we get that $\forall \epsilon \in (0,1)$ $\exists n_0 \in \mathbb{N}$ such that $|a_n-a|< \epsilon, \forall n \geq n_0$.

How does $(iii)$ imply $(iv)$ ?

$(iv) \Rightarrow (i)$:

Since $(iv)$ holds, we have that $\forall \epsilon'>0, \exists n_0 \in \mathbb{N}$ such that $|a_n-a|<\frac{\epsilon'}{2}, \forall n \geq n_0$. We pick $\epsilon'=2 \epsilon$. So we have that $\forall \epsilon>0$, $\exists n_0 \in \mathbb{N}$ such that $|a_n-a|<\epsilon$, $\forall n \geq n_0$.

Is everything that I did correct? If we show that $(iii)$ implies $(iv)$, we will have shown that the above statements are equivalent, right?

Answer 1

The crux of the $(iii)\Rightarrow (iv)$ proof is that when you have arbitrary $\epsilon' > 0$ you can always find $\epsilon \in (0,1)$ such that $\epsilon \leq \epsilon'$:

$$ |a_n - a| < \epsilon \leq \epsilon' \Rightarrow |a_n - a| < \epsilon' $$ I'm sure you can now write a full proof yourself.

The proofs that you've written are clean and good. I would make only one small adjustment: when you write

We pick $ϵ′=2ϵ$. Then we have that $∀ϵ>0\; ∃n_0∈\mathbb{N}$ such that $|a_n−a|<2ϵ$, $∀n≥n_0$.

I would write that quantifier $\forall \epsilon>0$ before the whole sentence, like that:

Let $\epsilon > 0$. We pick $ϵ′=2ϵ$. Then we have that $∃n_0∈\mathbb{N}$ such that $|a_n−a|<2ϵ$, $∀n≥n_0$.

($\forall$ and "Let..." are synonymous in this context.)

In order to write $\epsilon' = 2\epsilon$ it has to be specified what $\epsilon$ is first. That's why almost all proofs of sentences beginning with $\forall x \dots$ start with "Let $x$ be arbitrary.". (The other ones start with a lemma.)