The equivalence of two situations about convergence of the sequence.

Question

Let $\{a_n\}$ be sequence.

Consider the two situations $(E_1)$ and $(E_2)$.

$(E_1)\ $ $\forall \epsilon >0, \exists N \in \mathbb N$ s.t. $n \geqq N \Rightarrow |a_n-a|<\epsilon.$

$(E_2)\ $ $\forall \epsilon \in (0, l), \exists N \in \mathbb N$ s.t. $n \geqq N \Rightarrow |a_n-a|<\epsilon$ where $l>0$ is a constant number that is independent of $\epsilon$.

I want to prove $(E_1)$ and $(E_2)$ are equivalent. $(E_1)\Rightarrow (E_2)$ is trivial. So I have to show $(E_2)\Rightarrow (E_1)$.

I wonder whether my proof is correct or not.

My proof

Let $\epsilon >0$.

If $0< \epsilon <l$, there exists $N_1 \in \mathbb N$ s.t. $n\geqq N_1 \Rightarrow |a_n-a|<\epsilon$ from $(E_2)$.

So, consider the case of $\epsilon \geqq l$.

Since $l>0,$ there exists $N_2 \in \mathbb N$ s.t. $n\geqq N_2 \Rightarrow |a_n-a|<l.$

Then, $n\geqq N_2 \Rightarrow |a_n-a|<l\leqq \epsilon.$

I wonder whether this is correct or not. $N_2$ depends on $l$ but I think that $N_2$ has to depend not on $l$ but on $\epsilon$.

Answer 1

This sentence is a little weird

Since l>0, there exists N2∈N s.t. n≧N2⇒|an−a|<l. since (E2) applies to every $\epsilon \in (0,l)$ and $l \notin (0,l)$.

Besides this, the idea is right but you could write it down a little more clearly, for example (second case only):

Let $\epsilon_0 > l > \frac{1}{2}l$.

By (E2) we have $\exists N\in \mathbb{N}\ $ s.t. $\ n \geq N \Rightarrow |a_n - a| < \frac{1}{2}l < l < \epsilon_0$.

Therefore (E2) $\Rightarrow$ (E1)