# Slightly changing the formal definition of continuity of $f: \mathbb{R} \to \mathbb{R}$?

I'm curious for some perspectives on why it would be wrong to change the definition of continuity of $f: \Bbb R \to \Bbb R$ in the following way:

Original definition. $f : \Bbb R \to \Bbb R$ is said to be continuous at $x \in \Bbb R$ if $\forall \epsilon > 0$ $\exists \delta > 0$ such that $|x - a| < \delta \implies |f(x) - f(a)| < \epsilon$.

Altered definition. $f : \Bbb R \to \Bbb R$ is said to be continuous at $x \in \Bbb R$ if $\forall \delta > 0$ $\exists \epsilon > 0$ such that $|x - a| < \delta \implies |f(x) - f(a)| < \epsilon$.

The altered definition is more in line with what I think when I think about continuity intuitively: nearby points are sent to nearby points. It only makes sense to me to be able to choose "nearness" in the domain (i.e., $\forall \delta > 0$) and show there is nearness in the codomain (i.e., $\exists \epsilon > 0$) to prove intuitively that "nearby points are sent to nearby points".

Similarly, if $X, Y$ are topological spaces, we say $f: X \to Y$ is continuous if the preimages of open sets are open. What would be wrong about changing the definition to say that a map is continuous if the images of open sets are open (i.e., $f$ is continuous if it is an open map)? This is more inline with the intuitive idea of "nearby points being sent to nearby points" -- you pick nearness in the domain (i.e., an arbitrary open set) an show nearness in the codomain (i.e., the image is open).

Does anyone have any useful remarks?

• Your definition is very weak - for example, always taking the witnessing $\epsilon$ to be 200 will show that all functions with range $\subseteq [-100,100]$ are continuous... – Alex Kruckman Mar 20 '15 at 19:44
• Then all bounded functions are going to be "continuous" by your definition. – A.Γ. Nov 2 '16 at 13:51
• @MauroALLEGRANZA But the positions of $\epsilon$ and $\delta$ have changed only in the quantifiers, so it doesn't look like it is about changing names. – GoodDeeds Nov 2 '16 at 13:51
• The definition you propose is the definition of $f$ being bounded on every interval containing $x_0$. But there are functions that are bounded that certainly aren't continuous (see GoodDeeds' answer), but also some functions that are continuous at $x_0$ but not bounded on every interval containing $x_0$ (see Nick's answer). So your definition does not capture continuous functions, because there are discontinuous functions that satisfy it, but also continuous functions that don't. – kccu Nov 2 '16 at 13:54
• This is a really good question since it forces you to think about the meanings of "there exists" and "for all" and the order in which those quantifiers occur. You should be able to learn a lot from the answers. – Ethan Bolker Nov 2 '16 at 13:58

Your confusion seems to lie in trying to translate the sentence "if two points are close, then their images under $f$ are close" into a formal mathematical statement. Since the sentence mentions the points in the domain first, it seems like the mathematical statement should also "start in the domain".

I'd like to suggest that a better gloss on the meaning of continuity is "if two points are close enough, then their images under $f$ are close". This is because for a continuous function like $f(x) = 100x$, points have to be much closer in the domain to guarantee a level of closeness in the range. That is, to guarantee that $|f(x) - f(y)| < \frac{1}{2}$, we must have $|x-y| < \frac{1}{200}$.

Now to translate. The issue is that "close" is a vague word. Formally, when we say "close", we need to specify how close. So let's change the first instance of "close" to "$\delta$-close" and express it as $|x-y|<\delta$ and change the second instance of "close" to "$\epsilon$-close" and express it as $|f(x) - f(y)|<\epsilon$. Then the sentence becomes $$|x-y|<\delta \implies |f(x) - f(y)|<\epsilon.$$

But we don't want the definition of continuity to depend on $\delta$ and $\epsilon$ - we want to quantify them out. How close should $f(x)$ and $f(y)$ be? Well, as close as we want. So we need to quantify over all $\epsilon > 0$. How close should $x$ and $y$ be? Well, close enough: as close as they need to be to satisfy the conclusion $|f(x) - f(y)|<\epsilon$. This makes it clear that the $\delta$ depends on the $\epsilon$.

• So saying that, intuitively, continuity means nearby points are mapped to nearby points is too vague. If I understand your answer, we should amend the intuitive statement to say "points that are close enough are mapped to close points", which is an intuitive statement that makes clear via the word "enough" that the "closeness" we are choosing is in the range, i.e., $\delta$ depends on the free variable $\epsilon$. – layman Mar 20 '15 at 21:03
• That's my claim, yes. I wouldn't call $\epsilon$ a free variable, since it's also quantified. But the point whatever $\epsilon$ you pick, you can find a $\delta$ (which depends on the $\epsilon$). – Alex Kruckman Mar 20 '15 at 21:08
• What's wrong with calling $\epsilon$ a free variable? – layman Mar 20 '15 at 21:09
• Free variables aren't bound by quantifiers, and $\epsilon$ is quantified in the definition of continuity. – Alex Kruckman Mar 20 '15 at 21:11
• In the statement $\forall x\forall y |x-y|<\delta \implies |f(x) - f(y)|<\epsilon$, $\epsilon$ and $\delta$ are free variables. So the statement could be true or false depending on the values of $\epsilon$ and $\delta$. If $f(x) = 100x$, then the statement is true if $\epsilon = 1$ and $\delta = \frac{1}{100}$, but false if $\epsilon = 1$ and $\delta = 1$. On the other hand, in the statement $\forall \epsilon \exists \delta \forall x \forall y |x-y|<\delta \implies |f(x) - f(y)|<\epsilon$, $\delta$ and $\epsilon$ are not free variables - for a given $f$ the statement is either true or false. – Alex Kruckman Mar 20 '15 at 21:20

Your proposed definition would force us to accept as "continuous" functions that are nowhere continuous. An example is

$$f(x) = \begin{cases} 0 & \text{ if x is rational} \\ 1 & \text{ if x is irrational} \end{cases}$$

Pick your favourite $\delta > 0$; we can now choose $\epsilon = 2$. Clearly, for any $x_0$ we have that if $|x-x_0| < \delta$ then $|f(x) - f(x_0)| < \epsilon$.

The definition says that $$\forall\epsilon\gt0\exists\delta\text{ such that } |x-x_0|\lt\delta\implies|f(x)-f(x_0)|\lt\epsilon$$

The version you mention says that

$$\forall\delta\gt0\exists\epsilon\text{ such that } |x-x_0|\lt\delta\implies|f(x)-f(x_0)|\lt\epsilon$$

The two are not the same. The first says that you can get $f(x)$ as close to $f(x_0)$ as you wish by getting $x$ sufficiently close to $x_0$. The second says that however close(or far)$x$ may beto $x_0$, you can find a positive number greater than $|f(x)-f(x_0)|$, which is always true, when $f(x)$ and $f(x_0)$ are finite.

For example, suppose you have the function $$f(x)=\begin{cases}0&x=0\\1&\text{ else}\end{cases}$$

You can prove the function to be continuous at $x=0$ using the second definition. For any $\delta$, if you take $\epsilon\gt1$, the statement holds.

• Many thanks for your clear answer. As kccu's comment says above, your answer and that of Nick are both excellent answers, and I didn't want to flag one and not the other as the best solution. Kindly accept my thanks in this comment instead. – user135626 Nov 2 '16 at 21:08
• This answer actually explains why the two are different instead of just giving an example. +1. – rogerl Nov 2 '16 at 21:59

Take some piecewise function, like:

$$f(x) = \begin{cases} \frac{1}{x} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}$$

Now let $x_0 = \frac{1}{10}$ (something really close to $x=0$). I think we would agree that this function is continuous at $x_0 = 1/10$. If your suggested alternative definition worked, then we could pick ANY $\delta$. So let's pick $\delta = 1$. Then your alternative definition would suggest we could find some $\varepsilon$ so that $f(x)$ would be within a distance $\varepsilon$ from $f(1/10) = 10$ for all $x$ within the interval $\left( -\frac{9}{10}, \frac{11}{10} \right)$. But this obviously can't happen (there can be no such $\varepsilon$) since $f(x)$ is unbounded as $x$ approaches $0$.

• Many thanks for your clear answer. As kccu's comment says above, your answer and that of GoodDeeds are both excellent answers, and I didn't want to flag one and not the other as the best solution. Kindly accept my thanks in this comment instead. – user135626 Nov 2 '16 at 21:08

It is a bit unconventional but you can do this (and it is even quite useful!). However, you need to have a condition on how to choose epsilon to make it work

Here is a $\delta$-$\epsilon$ definition of continuity of a function $f$ in $x_0$:

$\forall \delta>0, \exists \epsilon = \epsilon(\delta) > 0$ such that

1. $\forall x$ with $|x - x_0| \le \delta$ we have $|f(x) - f(x_0)| \le \epsilon(\delta)$
2. $\inf_{\delta > 0} \epsilon(\delta) = 0$

It is best to think of the $\forall \delta >0 \exists \epsilon(\delta) > 0$ part as the existence of a function $\epsilon(\delta)$ with certain properties. The second property, ensures that the above counterexamples do not work. By replacing $\epsilon(\delta)$ by $\epsilon'(\delta) = \inf_{\delta' \ge \delta} \epsilon(\delta)$ we can w.l.o.g. assume that

1. if $\delta_1 \le \delta_2$ we have $\epsilon(\delta_1) \le \epsilon(\delta_2)$

This makes it much clearer what is going on, and makes it an easy exercise to prove that the $\delta$-$\epsilon$ definition is equivalent to the $\epsilon$-$\delta$ definition (see below). The function $\epsilon(\delta)$ tells you not only that $f(x) \to f(x_0)$ as $x \to x_0$ but how fast this is happening, i.e. $\epsilon(\delta)$ controls how a deviation of $x$ from $x_0$ results in a (maximal) deviation of $f(x)$ from $f(x_0)$. The $\delta$-$\epsilon$ definition makes many standard proofs or explicit computations much more natural without "magic" choices like "choose $\delta = \epsilon^{100}/\pi^2$" often found in proofs using the $\epsilon$-$\delta$ definition. E.g. to show that the product of two functions is continuous suppose that $\epsilon_f(\delta)$ and $\epsilon_g(\delta)$ control the convergence of $f(x) \to f(x_0)$ and $g(x) \to g(x_0)$. Then for $|x - x_0| < \delta \le \delta_0$ (which is all that matters) we have

$|f(x)g(x) - f(x_0)g(x_0)| \le |f(x) - f(x_0)||g(x_0)| + |g(x) - g(x_0)||f(x)| \le |g(x_0)|\epsilon_f(\delta) + (|f(x_0)| + \epsilon_f(\delta_0))\epsilon_g(\delta)$

it is trivial that $\epsilon_{fg}(\delta) := |g(x_0)|\epsilon_f(\delta) + (|f(x_0)| + \epsilon(\delta_0))\epsilon_g(\delta)$ satisfies property 2 and 3 and so controls the convergence of $fg(x) \to fg(x_0)$, showing that $fg$ is continuous in $x_0$.

In fact while the $\delta$-$\epsilon$ is unconventional we often use it when we want stronger notions of continuity and stronger control over the convergence of $f(x) \to f(x_0)$

A function f is Lipschitz continuous in $x_0$ if 1. $\forall x$ with $|x - x_0| \le \delta$ we have $|f(x) - f(x_0)| \le C\delta$ for some constant $C > 0$.

Clearly property 2 and 3 are now trivially fulfilled. Often we want a $C$ to be uniform for the whole domain of $f$. If such a $C$ exists we call $f$ Lipschitz continuous.

likewise: A function is $\alpha$ Hölder continuous for $0 < \alpha$ if 1. $\forall x$ with $|x - x_0| \le \delta$ we have $|f(x) - f(x_0)| \le C\delta^\alpha$ for some constant $C > 0$.

Again property 2 and 3 are trivially fulfilled.

======

Proof that $\delta$-$\epsilon$ continuous $\iff$ $\epsilon$-$\delta$ continuous.

$\implies$: Choose $\epsilon_0 > 0$. By property 2, $\exists \delta_0$ such that $\epsilon(\delta_0) \le \epsilon_0$ hence $\forall x$ with $|x - x_0| \le \delta_0$ we have $|f(x) - f(x_0)| \le \epsilon_0$.

$\Leftarrow$ By the $\epsilon$-$\delta$ definition, there exists a function $\delta(\epsilon)$ such that $\forall x$ with $|x - x_0| \le \delta(\epsilon)$ we have $|f(x) - f(x_0)| \le \epsilon$

For $\delta_0 > 0$, let

$\epsilon(\delta_0) = \inf \{\epsilon >0, 0<\delta_0 \le \delta(\epsilon)\}$ [1].

Fix $\epsilon_0 > 0$ and assume $\delta_0 \le \delta(\epsilon_0)$. Then $\forall x$ with $|x- x_0| \le \delta_0\le \delta(\epsilon_0)$ we have $|f(x) - f(x_0)| \le \epsilon_0$. Now reversing points of view and taking any $\epsilon_0$ with $\delta(\epsilon_0) \ge \delta_0$ we see that on taking inf's we get $|f(x) - f(x_0)| \le \epsilon(\delta_0)$ as in property 1. We have chosen $\epsilon$ non decreasing as function of $\delta_0$ as in property 3 by construction. In particular for $\delta_0 \le \delta(\epsilon_0)$ we have $\epsilon(\delta_0) \le \epsilon(\delta(\epsilon_0))\le \epsilon_0$, so $0\le \inf_{\delta_0 > 0} \epsilon(\delta_0) \le \inf_{\epsilon_0 > 0} \epsilon(\delta(\epsilon_0)) \le \inf_{\epsilon_0 >0} \epsilon_0 = 0$ proving propery 2.

[1] We strictly speaking have to assume that $0 < \delta_0 \le \delta(1)$ say, to make sure that the inf is over a non empty set, but that does not matter, we only need control functions in a neighborhood of $\delta_0 = 0$.

• Thank you for this thorough reply. It is interesting to know that such alternative definitions exist. It is also slightly reassuring to know that my question might not have been too absurd/stupid, and that it has occured to others before in other forms. – user135626 Nov 2 '16 at 22:06
• In the general $\delta$-$\epsilon$ definition you gave above, my understanding of what you wrote is that you are trying to enforce a condition that we can find valid values of $\epsilon$ that are arbitrarily small at a given $\delta>0$, so that unbounded or discontinuous examples would not occur (such as the examples by Nick and GoodDeeds), but: (1) is stating the infimum in condition 2 like this enough to guarantee this enforcement, and why/how? and (2) could you please explain more why we needed to replace $\epsilon$ by $\epsilon'$ and how would that lead to condition 3? – user135626 Nov 2 '16 at 23:46
• I edited in a formal proof of the equivalence. Was a bit more involved than I remembered. – Rogier Brussee Nov 3 '16 at 12:12