# The epsilon-delta definition of continuity

As we know the epsilon-delta definition of continuity is:

For given $$\varepsilon > 0\ \exists \delta > 0\ \text{s.t. } 0 < |x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon$$

My question: Why wouldn't this work if the implication would be:

For given $$\varepsilon > 0\ \exists \delta > 0\ \text{s.t. } |f(x) - f(x_0)| < \varepsilon \implies 0 < |x - x_0| < \delta ?$$

No. The point is that some continuous functions do not have this property! So the two statements are not equivalent.

Consider any constant real function $f$: let $x,y$ be any two points and $\epsilon >0$. Then $|f(x)-f(y)|=0<\epsilon$, but, we can make $|x-y|$ "as large as we want". For instance, take $x=0$ and let $y$ go to infinity: no $\delta$, in this case, works.

There are three ways to interpret this question.

1. You could be asking "what is the difference between the two directions of implication?".

2. You could be asking "what, intuitively, is the property of continuous functions captured by the real definition that is not captured by the backwards one?"

3. You could be asking "what are the consequences of the backwards definition?"

The answer to question 1 is that they are unrelated. To understand how, practically, they are unrelated, is answered by 2 and 3, but from the logical perspective, there is really no connection between $A \implies B$ and $B \implies A$.

The answer to 2 is what everyone always says about continuity: it is supposed to be the property that "values of $f$ at close values of $x$ are close". Presumably you have seen the informal "derivation" of the $\epsilon\delta$ definition from this prescription, but here it is again.

• First, we have to express what it means that two numbers $x$ and $y$ are close; let's say that we have an "expectation" of closeness given by a number $e > 0$, and that their proximity "meets our expectation" if $|x - y| < e$.

• Now we have to ask how to express the relationship between the closeness of two inputs $x$ and $a$ and of their corresponding outputs $f(x)$ and $f(a)$. We should establish an expectation, $\epsilon > 0$, for the proximity of $f(x)$ and $f(a)$, and depending on $f$ and $a$ we develop an expectation for the required proximity, $\delta > 0$ of $x$ and $a$ to make it happen.

• Finally, we write out the chain of causality: we start with $\epsilon > 0$ and find $\delta > 0$, and if $x$ and $a$ meet the expectation established by $\delta$, then this (by intention) should result in $f(x)$ and $f(a)$ meeting the expectation established by $\epsilon$. Or in symbols: $$\forall \epsilon > 0, \exists \delta > 0 \text{ s.t. } |x - a| < \delta \implies |f(x) - f(a)| < \epsilon.$$

There are actually two places where the chain of causality affects the order of things in the symbolic expression:

• Starting with $\forall \epsilon$, or in words, starting with an expectation for the proximity of $f(x)$ and $f(a)$. The backwards version has this order also. However, to justify this, let me say that if you were to try to establish an expectation for the proximity of $x$ and $a$ first, you'd be trying to influence the proximity of the inputs of $x$ based on the proximity of its outputs, which is he opposite direction that functions work (input to output).

• Making $x$ and $a$ satisfy their proximity expectation before $f(x)$ and $f(a)$. This is reversed in the backwards version, and the reason this is wrong is that if we say first that $f(x)$ is close enough to $f(a)$ without having found $x$, then we are making a blanket statement about any $x$ with the same value under $f$. It's like saying that you hate olives without having tasted the ones you are being offered.

Hopefully, this sheds some light on why the formal definition is taken to be exactly the way it is, in order to match the intuition.

Finally, we come to question 3: what does the backwards definition actually describe? The answer: every function. Because no matter what $\epsilon$ is, so long as we have $|f(x) - f(a)| < \epsilon$ just choose $\delta = 2|x - a|$, and then we have $|x - a| < \delta$. (And of course, if we don't have $|f(x) - f(x)| < \epsilon$, then we don't need to bother verifying the implication at all!) Speaking informally, this is because we not only get to choose $\delta$ after $\epsilon$, we also get to use it afterwards, so the statement with $\epsilon$ in it has no effect. In logic terms, we have both $F \implies T$ and $T \implies T$, so if you get to choose the right-hand side of an implication, you can always arrange it to be true. It's the left-hand side that's hard.

• The reverse definition does not allow $\delta$ to depend upon $x$. The statement is that a single $\delta$ should work for all $x$. Any function with bounded range, for example, would fail the reverse condition. – Marcel Besixdouze May 9 '14 at 1:48
• True. That was pretty stupid. – Ryan Reich May 9 '14 at 4:43
• This answer is discussed here – Harry49 Dec 6 '18 at 14:07

Consider the implications of using this definition for any constant function (which should all be continuous, if any function is to be continuous).

• In particular, for $c \in \mathbb{R}$ consider the constant function $f(x) = c$. Given $x_0 \in \mathbb{R}$, taking $\varepsilon = 1$, note that for any $\delta > 0$ if $x = x_0 + \delta$ we have that

• $| f(x) - f(x_0) | = | c - c | = 0 < 1 = \varepsilon$; but
• $| x - x_0 | = |(x_0 + \delta ) - x_0 | = \delta \not< \delta$.

Therefore the implication $| f(x) - f(x_0) | < \varepsilon \rightarrow | x - x_0 | < \delta$ does not hold. It follows that the function $f$ does not satisfy the given property.

• If I may ask, does the given definition pose problems for any non-constant functions? – MathematicsStudent1122 Dec 26 '15 at 14:44
• @MathematicsStudent1122 Also functions like $f(x)= \frac{\sin (1/x)}{x}$ ($f(0)=0$) which takes the value $0=f(0)$ at points arbitrarily close to $x_0 = 0$. – user642796 Dec 26 '15 at 18:25

This is because you want a small difference in $x$ to create a small amount of change in $f(x)$.

Let us suppose that we have some function $f$. Suppose that in some practical application that we need the value of $f$ to be close to some value $y$ with a an error bound that we call $\epsilon$. Then we need $|f-y|<\epsilon$. We control the output of $f$ by varying the $x_0$ a bit. Now suppose that we know the exact value $f(x_0)=y$. But in practice it is difficult (probably impossible) to get $x=x_0$ on the nose. So we need to know that If we can control $x$ sufficiently well our error $|f-y|$ will not exceed epsilon, no matter what epsilon we choose. We also need to know how tight that control needs to be. The tightness of the control of $x$ to guarantee that we can control $f$ is exactly $\delta$. If the error in $x$ exceeds $\delta$ we might be Ok, but we also might be in trouble.

What you want is that $f(x)$ is close to $f(y)$ whenever $x$ is close to $y$. You do not really care about whether $f(x)$ is close to $f(y)$ if $x$ and $y$ are far away.

My question: Why wouldn't this work if the implication would be: $$\forall \varepsilon > 0\ \exists \delta > 0\ \text{s.t. } |f(x) - f(x_0)| < \varepsilon \implies 0 < |x - x_0| < \delta ?$$

That condition can never be satisfied by any function (take $x = x_0$), but this can be repaired by restricting the domain to $x \neq x_0$ or (equivalently) writing the condition as

$$\forall \varepsilon > 0\ \exists \delta > 0\ \text{s.t. } |f(x) - f(x_0)| < \varepsilon \implies |x - x_0| < \delta$$

which says that the function escapes any finite interval around $f(x_0)$ in finite time (thinking of $x$ as a time variable and $f(x)$ as position of a particle).

For example, if $f'(x) > 3$ for all $x$, then the escape property is true for $f$ at all $x_0$.

The value of function can be equal (approaching same value) for any two points in domain which are at any distance. We're not interested in "when the functional values are equal for two points...", we are interested in "when the two points of domain are arbitrarily close the functional values at those points..." when we talk about continuity.