I've known the definition of the continuity like the below (Here the $I$ is a interval on $\mathbb R$)
$(1)$ $\exists \lim_{x \to a}f(x) = L$ for $a \in I$(Existence of the limits)
$\forall \epsilon > 0, \exists \delta > 0$ $s.t$ $0 < |x-a| < \delta \implies|f(x)-L| < \epsilon$ for $\forall x \in (a-\delta, a+\delta)$
[When the $f$ is a continuous at $a$ ; $" \forall \epsilon > 0, \exists \delta > 0$ $s.t$ $|x-a| < \delta \implies|f(x)-L| < \epsilon$ for $\forall x \in (a-\delta, a+\delta) "$]
$(2)$ $f: I\to \mathbb R$ is continuous on $I$
$\forall \epsilon >0, \forall x,y \in I;\exists \delta >0 s.t. |x-y|<\delta \implies |f(x)-f(y)|<\epsilon$
$(3)$ $f:I\to\mathbb R$ is uniformly continuous On $I$
$\forall \epsilon >0;\exists \delta >0, \forall x,y \in I s.t. |x-y|<\delta \implies |f(x)-f(y)|<\epsilon$
Let's consider the $(1)$ and $(2)$, To show either the continuity or the existence limits by definition, I used these methods.
Firstly taking a "$\epsilon$". Secondly, find the $\delta $ corresponding the $\epsilon$ and $x \in I$ like the exercise in the textbook. So I regard the value delta as having two variables $\epsilon$ and a point.
I.e. $\delta= \delta(a,\epsilon)$[case $(1)$] or $\delta(x,\epsilon)$[case$(2)$].
But There is a claim, $\delta$ can only depend on $\epsilon$(I.e. independent of the $x \in I$, $\delta= \delta(\epsilon)$] Because delta only depend on the things before the $\delta$ is introduced. I can't totally understand this claim, because when we find the delta for $(1)$ & $(2)$, should be the epsilon and $x\in I$ considered. So my thought his claim is incorrect.
Secondly Let's consider the $(3)$, To show the uniformly continuous, I used the method similar with $(1)$ & $(2)$ but slightly different. Firstly, taking a "$\epsilon$". Secondly, find the $\delta $ corresponding the $\epsilon$ like a exercise in the textbook. Hence I've regarded the value delta as having single variable, $\epsilon$ (I.e. $\delta =\delta (\epsilon)$ [independent for $x \in I$])
But the before who claimed $\delta= \delta(\epsilon)$ for $(1)$ and $(2)$ said I'm incorrect. He claimed when considering the $(3)$, delta should be independent for epsilon. Why does it have to be? Then What is the meaning for find the $\delta = \delta(\epsilon)$ for showing uniformly continuity?
So my question is please make me sure which claim is right between me and his. I'm very confused the meaning of the delta for 3 cases. Is my thought right?
Thank you.
P.s.) Why does he consider delta should be independent for epsilon for (3)
$Q)$ The function $f,g : (0, \infty) \to \mathbb{R}$ satifies the $(1)$ & $(2)$. Show $\exists \lim \limits_{x \to 0^+} f(x)$
$(1)$ $\exists \lim \limits_{x \to 0^+} g(x)$
$(2)$ $\forall a,b \in (0,\infty) $, $\vert f(b) - f(a) \vert \leq \vert g(b) - g(a)\vert $
Here is my attempt.
Say $\exists \lim \limits_{x \to 0^+} g(x) = L$. Firstly, take a $\epsilon$. And then, $\exists \delta(=\delta(\epsilon))>0$ $s.t.$ $0 <x <\delta \Rightarrow \vert g(x) -L \vert < {\epsilon \over 2}$
Consider $0 <x,y <\delta(=\delta(\epsilon))$ Then, $\vert g(x)-g(y)\vert \leq \vert g(x)-L \vert + \vert g(y)-L \vert < \epsilon$
Hence By $(2)$, $\vert f(x) - f(y) \vert \leq \vert g(x) - g(y)\vert < \epsilon $.
I concluded the $f$ is uniformly continuous on $(0, \delta)$. Therefore, $\exists \lim \limits_{x \to 0^+} f(x)$
But he said "$\delta $ is depend on the $\epsilon$. So this is not definition of the uniformly continuity." But I'm not agree with the his claim. Because Hence under the condition taking the epsilon for limit for $g(x)$, then automatically $∀x,y∈(0,δ(ϵ))⇒|f(x)−f(y)|<ϵ$ Also lecture said my friend is right. I can't understand which point I was wrong. :(