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. :(


1 Answer 1


Your definition of (2) "continuity on $I$" is incorrect (you may have meant the right thing but what you wrote doesn't convey what you probably meant). Also, in (1) if you want to be more precise, you should avoid using quantifiers at the end of a mathematical statement. I know this is common practice in everyday language and also in many mathematical texts, and I'm also guilty of this from time to time but you should always use quantifiers first (in the correct order), and only then write out the conditions (also I prefer to write out in natural language rather than in symbols... but that's just me). So, for example:

  1. (i) "Existence of limit at $a$":

there exists $L\in \Bbb{R}$ such that for every $\epsilon>0$, there is a $\delta>0$ such that for all $x\in I$, if $0< |x-a|<\delta$ then $|f(x)-L|<\epsilon$.

  1. (ii) "Continuity of $f:I\to \Bbb{R}$ at a point $a\in I$":

for every $\epsilon>0$, there is a $\delta>0$ such that for all $x\in I$, if $|x-a|<\delta$ then $|f(x)-f(a)|<\epsilon$.

In both these cases, $\delta$ depends on the point $a$ and $\epsilon$.

Next, we have

  1. "Continuity of $f$ on $I$":

for every $x\in I$, for every $\epsilon>0$, there exists a $\delta>0$ such that for all $y\in I$, if $|y-x|<\delta$ then $|f(y)-f(x)|<\epsilon$.

Here, $\delta$ depends on $x$ and $\epsilon$. This is of course completely reasonable because saying "$f$ is continuous on $I$" is taken to mean "for every $x\in I$, $f$ is continuous at $x$", which is why the definition is so similar to 1(ii); we just tack on a "for all $x\in I$" in front of the whole thing.

Finally, we have

  1. "$f$ is uniformly continuous on $I$":

For every $\epsilon>0$, there is a $\delta>0$ such that for all $x,y\in I$, if $|x-y|<\delta$ then $|f(x)-f(y)|< \epsilon$.

In this case $\delta$ depends only on $\epsilon$ (compare with (2) where $\delta$ depended on $\epsilon$ AND $x$).

You then say

He claimed when considering the (3), delta should be independent for epsilon.

I'm not sure what you mean here. $\delta$ in every case most certainly depends on $\epsilon$ (because if $I=[\alpha,\beta]$ is an interval then the $\delta$ doesn't depend on $\epsilon$ if and only if $f$ is constant on $I$).

By the way, note that in all of these cases, the $\delta$ also depends on the function $f$, so if you really wanted in 1(ii), you could write $\delta_{f,a,\epsilon}$. Of course no one goes this far; once you understand the gist of what depends on what, we omit this from the notation (and also because the way it is written is already unambiguous, one just needs practice with "reading" mathematical statements).

  • $\begingroup$ Ah Thanks for clarify the notation. I added the post why does he thought delta should be independent for epsilon. $\endgroup$ Commented Sep 19, 2020 at 2:53
  • $\begingroup$ This is good, and complete. I can’t imagine anybody bettering it. $\endgroup$
    – Lubin
    Commented Sep 19, 2020 at 3:02

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