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I understand the geometric differences between continuity and uniform continuity, but I don't quite see how the differences between those two are apparent from their definitions. For example, my book defines continuity as:

Definition 4.3.1. A function $f:A \to \mathbb R$ is continuous at a point $c \in A$ if, for all $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $|x-c| < \delta$ (and $x \in A$) it follows that $|f(x)-f(c)| < \epsilon$.

Uniform continuity is defined as:

Definition 4.4.5. A function $f:A \to \mathbb R$ is uniformly continuous on $A$ if for every $\epsilon > 0$ there exists a $\delta > 0$ such that $|x-y| < \delta$ implies $|f(x)-f(y)| < \epsilon$.

I know that in Definition 4.3.1, $\delta$ can depend on $c$, while in definition 4.4.5, $\delta$ cannot depend on $x$ or $y$, but how is this apparent from the definition? From what appears to me, it just seems like the only difference between Definition 4.3.1 and Definition 4.4.5 is that the letter $c$ was changed to a $y$.

My guess is that the first definition treats $c$ as a fixed point and it is only $x$ that varies, so in this case, $\delta$ can depend on $c$ since $c$ doesn't change. Whereas for the second definition, neither $x$ or $y$ are fixed, rather they can take on values across the whole domain, $A$. In this case, if we set a $\delta$ such that it depended on $y$, then when we pick a different $y$, the same $\delta$ may not work anymore. Is this somewhat a correct interpretation?

Anymore clarifications, examples, would be appreciated.

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First of all, continuity is defined at a point $c$, whereas uniform continuity is defined on a set $A$. That makes a big difference. But your interpretation is rather correct: the point $c$ is part of the data, and is kept fixed as, for instance, $f$ itself. Roughly speaking, uniform continuity requires the existence of a single $\delta>0$ that works for the whole set $A$, and not near the single point $c$.

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The difference is in the ordering of the quantifiers.

  • Continuity:

For all $x$, for all $\varepsilon$, there exist such a $\delta$ that something something.

  • Uniform continuity:

For all $\varepsilon$, there exists such a $\delta$ that for all $x$ something something.

For something to be continuous, you can check "one $x$ at a time", so for each $x$, you pick a $\varepsilon$ and then find some $\delta$ that depends on both $x$ and $\varepsilon$ so that $|f(x)-f(y)|<\varepsilon$ if $|x-y|<\delta$. As you can see if you try it on $f(x)=1/x$ on $(0,1)$, you can find such a $\delta$ for every $x$ and $\varepsilon$. However, if you fix $\varepsilon$, the values for $\delta$ that you need become arbitrarily small as $x$ approaches $0$.

If you want uniform continuity, you need to pick a $\varepsilon$, then find a $\delta$ which is good for ALL the $x$ values you might have. As you see, for $f(x)=1/x$, such a $\delta$ does not exist.

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    $\begingroup$ The language you use, and therefore the order you use the language is not found in the definitions provided from the OP's textbook, and therefore isn't very helpful. Can you please reword your answer to fit the question: "I know that in Definition 4.3.1, δ can depend on c, while in definition 4.4.5, δ cannot depend on x or y, but how is this apparent from the definition?" I ask, because I have the same question and the same textbook definition. I know that your answer is correct but why is it apparent from the GIVEN definition? $\endgroup$ – rocksNwaves Apr 30 '18 at 23:56
  • $\begingroup$ @rocksNwaves, indeed, in the script we use, the order is the same as in the uniform continuity: $$$$ continuity at a point $c\in I\subseteq\Bbb R:$ $\\(\forall\varepsilon>0)(\exists\delta>0)(\forall x\in I)((|x-c|<\delta)\implies (|f(x)-f(c)|<\varepsilon))\\$ $$$$ whereas the uniform continuity on an open interval $I\subseteq\Bbb R:$ $$$$$\\(\forall\varepsilon>0)(\exists\delta>0)(\forall x',x''\in I)((|x'-x''|<\delta)\implies(|f(x')-f(x'')|<\varepsilon))\\$ $\endgroup$ – Invisible Apr 24 at 22:31
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The subtle difference between these two definitions became more clear to me when I read their equivalent sequence definitions. First take the definition of a continuous function.

Definition A function $f: D\to\mathbb{R}$ is said to be continuous at the point $x_0$ in $D$ provided that for every sequence $\{x_n\}$ in $D$ that converges to $x_0$, the image sequence $\{f(x_n)\}$ converges to $f(x_0)$.

Now compare this to a uniformly continuous function.

Definition A function $f: D\to\mathbb{R}$ is said to be uniformly continuous provided that for any two sequences $\{y_n\}$ and $\{x_n\}$ in $D$ have the property $$\lim_{n\to\infty}(y_n-x_n)=0,$$ then $$\lim_{n\to\infty}(f(y_n)-f(x_n))=0$$

Notice how the second definition mentions no convergence to a point, but that two sequences are tending toward the same value and at the same rate. These sequences can both be divergent sequences when alone, but their terms can become arbitrarily close to each other.

The classic example is $f:\mathbb{R}\to\mathbb{R}, f(x)=x^2$ is continuous but not uniformly continuous. Take the two sequences $\{y_n\}=\{\sqrt{n^2+1}\}$ and $\{x_n\}=\{n\}$. (Note, both sequences diverge). Take the $\lim_{n\to\infty}{y_n-x_n}$, and solve by multiplying numerator and denominator by its conjugate. $$\lim_{n\to\infty}(\sqrt{n^2+1}-n)=\lim_{n\to\infty}\frac{n^2 +1-n^2 }{\sqrt{n^2+1}+n}=\lim_{n\to\infty}\frac{1}{\sqrt{n^2+1}+n}=0.$$ Now, looking at $\lim_{n\to\infty}{f(y_n)-f(x_n)}$ we get the following $$\lim_{n\to\infty}{(\sqrt{n^2+1})^2-n^2}=\lim_{n\to\infty}{n^2+1-n^2}=1$$ So this goes against the definition of uniform continuity. We need the difference of function values to also go to $0$, as well, in order for it to be uniformly continuous.

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  • $\begingroup$ Clint.Have you seen the proof of the equivalence of your sequence definition of uniform continuity with Definition 4.4.5 in the OP's question? Thanks. $\endgroup$ – Peter Szilas Aug 27 '19 at 16:57
  • $\begingroup$ @PeterSzilas I have; though admittedly, it has been some time. I remember bridging the two through proof in an assignment for a class. For purposes of this comment, I copied the definition from Advanced Calculus, Second Edition, by Patrick M. Fitzpatrick. The definition was found in section 3.4: Uniform Continuity. $\endgroup$ – Clint Chelak Aug 30 '19 at 19:30
  • $\begingroup$ @PeterSzilas To help in starting the proof, see Lemma 2.9 in the same book, called the Comparison Lemma. The proof gives strategies to tie together sequences and delta-epsilon proofs. I know somewhere in that book it talks about the equivalence of limit definitions and delta-epsilon definitions and its reasoning for keeping to the limit definitions, but I couldn't find it in my brief search. If time permits, I'll create and link a new thread with the proof. $\endgroup$ – Clint Chelak Aug 30 '19 at 19:53
  • $\begingroup$ Clint.Thanks.The above theorem is at times quite useful. Even proving $\epsilon -\delta$ (simple) continuity starting from the sequential continuity definition (by contradiction) is a bit tricky. Let you know if I come across something interesting.Greetings. $\endgroup$ – Peter Szilas Aug 31 '19 at 7:35
  • $\begingroup$ didn't get that sequence part $\endgroup$ – Holy Answerer Apr 10 '20 at 6:44
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Let me focus on this part of the question:

"I know that in Definition 4.3.1, $\delta$ can depend on $c$, while in definition 4.4.5, $\delta$ cannot depend on $x$ or $y$, but how is this apparent from the definition?"

This is apparent from the order of the quantifiers. When we write out these two statements into "Prenex normal form", we have that:

$$ \forall c \in A,\forall \epsilon >0, \exists \delta, \forall x \in A \:( |x-c|<\delta \implies |f(x)-f(c)|<\epsilon) $$

$$\forall \epsilon >0, \exists \delta, \forall x,c\in A \: (|x-c|<\delta \implies |f(x)-f(c)|<\epsilon) $$

In the first statement, note that the universal ($\forall$) quantifier $\forall c$ precedes the existential ($\exists$) quantifier $\exists \delta$, and the universal quantifier $\forall x$ follows the existential quantifier $\exists \delta$.

Note that in the second definition, the universal quantifier $\forall c$ now also follows the existential quantifier $\exists \delta$.

To see the significance of the quantifier order, consider the following, where C is the set of cars, P is the set of people, and R is a relation such that cRp means c is owned by p.

$$ \forall c\in C, \exists p \in P: cRp $$ $$ \exists p\in P, \forall c \in C: cRp $$

Observe that in the first statement of the example, the universal quantifier precedes the existential quantifier. This statement means each car $c$ has an owner $p$. Observe that the person p depends on the car. In the second statement, the universal quantifier follows the existential quantifier. This statement means there is some person $p$ who owns EVERY car. Thus this person doesn't depend on the car (since he has all of them, or in other words; given every car, he has it).

To conclude, for any variables $x,y$, $y$ can depend on $x$ if and only if the universal quantifier for $\forall x$ precedes the existential quantifier for $\exists y$.

Applying this theorem to your definitions, we see that in the definition of continuity, the universal quantifiers $\forall x\in\mathbb{R}$ and $\forall\epsilon$ precede $\exists\delta$ . Thus here $\delta$ may depend on both $x,\epsilon$ . However, in the definition of uniform continuity, the only universal quantifier that precedes $\delta$ is $\forall\epsilon$ . Thus delta may only depend on $\epsilon$ and not $x$ .

In the definition of uniform continuity, $\exists \delta $ precedes neither $x$ nor $c$, therefore it can depend on neither of them, but only on $\epsilon$.

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To understand the difference between continuity and uniform continuity, it is useful to think of a particular example of a function that's continuous on $\mathbb R$ but not uniformly continuous on $\mathbb R$. An example of such a function is $f(x)=x^2$. Here to understand the failure of uniform continuity of the function, it is particularly useful to exploit the hyperreals. Note that if $H$ is an infinite number and we choose an infinitesimal $\epsilon=\frac{1}{H}$, then the values of $f$ at the infinitely close points $H$ and $H+\epsilon$ are themselves not infinitely close. This violation of the property of microcontinuity of $f$ at $H$ captures the essence of the failure of uniform continuity of $f$ on $\mathbb R$.

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enter image description here

This intuitive GIF image from Wikepedia helped me most.

$f(x)=\frac{1}{x}$ as shown in the image is continuous but not uniformly continuous, because obviously if, for instance when $x_1=0.1$ we can see that $|f(x_1)-f(x_1+0.2)| \gt 0.5$; while $g(x)=\sqrt x$ is both continuous and uniformly continuous since we can find a number, for instance 0.5 bellow, such that $|f(x_1)-f(x_1+0.2)| \lt 0.5$ for every $x_1$. Here 0.2 and 0.5 should be numbers in R which just exist for the given function.

Hope this will be of any help to you.

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I find it interesting that each answer here throws in something different to the understanding of uniform continuity. And I have something different to add.

As observed by Siminore, continuity can be expressed at a point and on a set whereas uniform continuity can only be expressed on a set. Reflecting on the definition of continuity on a set, one should observe that continuity on a set is merely defined as the veracity of continuity at several distinct points. In other words, continuity on a set is the "union" of continuity at several distinct points. Reformulated one last time, continuity on a set is the "union" of several local points of view.

Uniform continuity, in contrast, takes a global view---and only a global view (there is no uniform continuity at a point)---of the metric space in question.

These different points of view determine what kind of information that one can use to determine continuity and uniform continuity. To verify continuity, one can look at a single point $x$ and use local information about $x$ (in particular, $x$ itself) and local information about how $f$ behaves near $x$. For example, if you know that $f$ is bounded on a neighborhood of $x$, that is fair game to use in your recovery of $\delta$. Also, any inequality that $x$ or $f(x)$ satisfies on a tiny neighborhood near $x$ is fair game to use as well. You can even use $f(x)$ to define $\delta$.

However to verify uniform continuity, you can't zoom in on any particular point. You can only use global information about the metric space and global information about the function $f$; i.e. a priori pieces of information independent of any particular point in the metric space. For example, any inequality that every point of $X$ satisfies is fair game to use to recover $\delta$. If $f$ is Lipschitz, any Lipschitz constant is fair to use in your recovery of $\delta$.

There are two propositions which I think exemplify the difference between continuity and uniform continuity:

Let $X$ and $Y$ denote two metric spaces, and let $f$ map $X$ to $Y$.

  • $f$ is continuous on $X$ if and only if for every $x$ in $X$ and for every $\epsilon>0$ there is a $\delta>0$ such that $$\text{diam}\, f(B_{\delta/2}(x))<\epsilon\,.$$

  • $f$ is uniformly continuous on $X$ if and only if for every $\epsilon>0$ there is a $\delta>0$ such that $$\text{diam}\,f(E)<\epsilon$$ for every subset $E$ of $X$ that satisfies $\text{diam}\,E<\delta$.

Thus continuity in a certain sense only worries about the diameter of a set around a given point. Whereas uniform continuity worries about the diameters of all subsets of a metric space simultaneously.

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Exactly, the delta of uniform continuity is not changeable, which decides the ball of x, y. However, the delta of continuity is decided by the point c, it varies due to the change of c.

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