Undergrad Is ##\delta##-steady needed in this proof, given ##\epsilon##-steady

[yucheng]

TL;DR

Is $\delta$-steady needed in this proof, given $\epsilon$-steady for all $\epsilon>0$?

In Tao's Analysis 1, Lemma 5.3.6, he claims that "We know that $(a_n)_{n=1}^{\infty}$ is eventually $\delta$-steady for everyvalue of $\delta>0$. This implies that it is not only $\epsilon$-steady, $\forall\epsilon>0$, but also $\epsilon/ 2$-steady."

My question is, why do we need the statement on $\delta$ when we already have $\epsilon$-steady, $\forall\epsilon>0$, which immediately follows that the sequence is $\epsilon /2$-steady since $\epsilon>0 \implies \epsilon /2>0$? Is this just his style, or is it logically necessary?

 

[mfb]

when we already have $\epsilon$-steady

Don't you get that from $\delta$? I don't have the book so maybe the context is relevant, but you need to start with some knowledge to conclude anything.

 

[yucheng]

Don't you get that from $\delta$? I don't have the book so maybe the context is relevant, but you need to start with some knowledge to conclude anything.

Clarification: the 'knowledge' is $\forall\epsilon>0$... For whatever reason, the author used $\delta>0$... in the proof instead, then only brought $\epsilon$ in later.

Is it to distance the $\epsilon$, I mean the definition for a Cauchy sequence already uses $\epsilon$, so if I want to say $\epsilon /2$ fulfils the condition, I can reframe the definition in terms of $\delta >0$, then point out that $\epsilon >0$ also fulfils the condition, i.e. ($\epsilon\in \{x:x = \delta\}$) it, then $\epsilon /2>0$ also fulfils it, i.e. $\exists\epsilon /2: \epsilon /2\in \{x:x = \epsilon\} \subset \{x:x = \delta\}$?

I apologize if I am abusing, if not misusing notation!