General question about Epsilon-Delta proofs.

Question

If I have proved that for every $2\epsilon$ there exists a delta, is it sufficient to write that since there exists a $\epsilon$ for every $2\epsilon$ and there exists a $\delta$ for every $2\epsilon$, there exists a $\delta$ for every $\epsilon$?

Answer 1

You phrased your question a bit strangely but basically: yes.

If you were able to find an upper bound of $2\varepsilon$, for all $\varepsilon >0$, for some expression, then also $\varepsilon$ is an upper bound for this same expression.

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If you have something like: $$\forall\,\varepsilon_1 >0, \exists\,\delta_1>0 : \ldots \implies K < \color{red}{2}\varepsilon_1 \tag{1}$$ and you want something like: $$\forall\,\varepsilon_2 >0, \exists\,\delta_2>0 : \ldots \implies K < \varepsilon_2 \tag{2}$$ then you can always, for an arbitrary but fixed $\varepsilon_2$, apply $(1)$ by taking $\varepsilon_1 = \tfrac{\varepsilon_2}{2}$ and $(2)$ follows.

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In proofs this is often avoided by requiring some intermediate expressions to be bounded by e.g. $\tfrac{\varepsilon}{2}$ (rather than $\varepsilon$) so you immediately end up with something like $K < \tfrac{\varepsilon}{2}+\tfrac{\varepsilon}{2}=\varepsilon$.