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In epsilon-delta proofs, when specifying delta, is it correct to specify the delta is equal to a value or delta is less than or equal to a value?

Which statement if more correct and why? Given epsilon > 0, choose delta = (number) OR Given epsilon > 0, choose delta < or = (number) OR Both?

Why? I would think that it would be more correct to specify that delta is < or = to a value since any smaller value would also work in the proof, right?

Thanks!

bamajon1974
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  • Can you provide an example illustrating your question? – Ryszard Szwarc Aug 30 '23 at 19:12
  • Please see the example linked below. From the first line: Suppose ϵ>0 and δ=min{1,ϵ10}....is it more correct to write δ <=min{1,ϵ10} ? https://math.stackexchange.com/questions/1743385/epsilon-delta-proof?rq=1 – bamajon1974 Aug 30 '23 at 19:18
  • You only need to show one $\delta$ works, but if $\delta_0$ works, then any smaller $\delta<\delta_0$ works. – Thomas Andrews Aug 30 '23 at 19:25
  • As was mentioned in the answer if a number $\delta_0>0$ satisfies the requirements, than any smaller number $\delta$ is fine as well. So if you find one $\delta$ for a given $\varepsilon>0$ you are done. Another question is to find maximal $\delta>0$ for a given $\varepsilon.$ For example $\lim_{x\to 1}x^2=1$ In order to find the largest $\delta$ you have to solve the inequality $|x^2-1|<\varepsilon .$ For small $\varepsilon$ you will get two intervals and you choose the one containing $1.$ But that's painful. So if you are not after the largest $\delta,$ just find one. – Ryszard Szwarc Aug 30 '23 at 19:32

2 Answers2

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You are correct when you say that "any smaller value would also work", at least in most common problems like limits, continuity, etc. But if we look at the definition for continuity of a function $f:\mathbb{R} \rightarrow \mathbb{R}$ at a point $a$, $$∀\epsilon>0\space \exists\delta>0 \text{ such that}$$ $$|x-a|<\delta \implies |f(x)-f(a)|<\epsilon,$$ note that we are saying that a $\delta$ exists. So it kinda doesn't matter (again, in most common problems) if it works for any $\delta$ less or equal than some value, but rather if there is some delta that works, which is normally the value you find for $\delta$ and show that works.

Burno
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Let $A\subseteq \mathbb{R}, a\in A$ and $f:A\to \mathbb{R}$ be function. If you prove $$\forall \epsilon>0 \exists \delta>0 \forall x\in A (|x-a|<\delta \implies |f(x)-f(a)|<\epsilon) \ \ \ \ \ \ \ \ \ \ \ (1)$$ You can also prove that $$\forall \epsilon>0 \exists \delta>0 \forall x\in A (|x-a|\leq\delta \implies |f(x)-f(a)|<\epsilon) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ Because, If $(1)$ is true, there is a $\delta>0$ that satisfies $(1)$. If $\delta_{1}<\delta$, then $\delta_{1}$ satifies the $(1)$. Thus, we have found $\delta=\delta_{1}$ for $(2)$.

To show that $(2) \implies (1)$ is trivial one.

Hence, $(1) \iff (2)$.

endvc1
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