# Continuity Rewritten: $\forall\delta>0,\exists\varepsilon>0\dots$

Take the following definition of the continuity of a function $f$ at some point $x_0$:

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

Why can I not rewrite it as follows?

$$\color{red}{\forall\delta>0,\exists\varepsilon>0}\text{ such that }|x_0-x|<\delta\implies|f(x_0)-f(x)|<\varepsilon.$$

I suppose that it is because it would not work for functions whose domains are not all of the reals. For example, take

$$f(x)=\ln x.$$

If I want to show that it is continuous at $x_0=1$ and take some $\delta=2>0$, then some $x$ will fall within $(-1,0]$, for which $f$ has no definition.

Is my train of thought sound? Or is there a more subtle reason?

• There are multiple reasons your proposed red rewrite is unsuited as an alternative definition for continuity. Your observation is as sound as any, and it seems succinct to me. Apr 24, 2015 at 18:58
• @BarryCipra, my diction was off: I meant to express "subtle," not "succinct." Thank you. Apr 24, 2015 at 19:04
• Actually I think your observation makes an admirably subtle point, namely that the proposed redefinition requires the function to be defined for all $x$. Apr 24, 2015 at 19:21

Consider the function

$$f(x) = \begin{cases} 0, \ x \leq 0 \\ 1, \ x > 0 \end{cases}$$

Then for all $\delta > 0$ there exists an $\epsilon$, say $\epsilon = 2$, such that

$$|x - 0| = |x| < \delta \ \Longrightarrow \ |f(x) - f(0)| = |f(x)| < \epsilon$$

Hence this definition doesn't correspond to our sense of continuity, intuitive or otherwise.

• +1 typo in your first line though Apr 24, 2015 at 18:45

I realize this is an old question, but I stumbled upon it and I felt that the answers don't address an important point in your question. Namely is your reasoning sound? Given that this was three years ago, you may be well aware of what I am going to say, but for the sake of anyone else reading the question and answers, I'll add this.

The answer to this is no, though perhaps it's a nitpick, but let me explain.

Your problem with the, let's call it $$\delta$$-$$\epsilon$$ definition you've come up with is that it seemed to imply that the function must be defined on all of $$\Bbb{R}$$. However there is a subtlety here, we must be careful about the quantification of $$x$$. While in high school calculus, we say things like the square root function is from $$\newcommand{\RR}{\Bbb{R}}\RR\to\RR$$, but its undefined when $$x<0$$. However in mathematics, functions are always defined on their entire domain. Thus we say that $$f(x)=\sqrt{x}$$ is a function $$f:\RR_{\ge 0}\to\RR$$. I.e. the square root function is a function defined only on positive reals. This is important because in the usual definition of continuity, $$x$$ is quantified over the domain of the function. I.e. we say a function from a subset $$D\subseteq\RR$$ to $$\RR$$ is continuous at $$x_0\in D$$ if $$\forall_{\epsilon >0}\exists_{\delta >0}\forall_{x\in D}|x_0-x|<\delta \implies |f(x_0)-f(x)|<\epsilon.$$

Thus in order for your definition to match the usual notion of continuity, it would have to read $$\forall_{\delta >0}\exists_{\epsilon >0}\forall_{x\in D}|x_0-x|<\delta \implies |f(x_0)-f(x)|<\epsilon.$$

Then because $$x$$ is required to be in the domain of the function by definition, the fact that we can choose $$\delta$$ as big as we want doesn't force any element of $$\RR$$ to be part of the domain of $$f$$.

As for why it's important that we restrict $$x$$ to be in the domain of $$f$$ for the usual definition of continuity, well for one thing, we'd like to say that the square root function is continuous at 0, which would be impossible if we allowed $$x$$ to be any real number less than a distance of $$\delta$$ from $$x_0$$, since for any $$\delta$$, $$-\delta/2$$ is a real number less than a distance of $$\delta$$ from 0, but $$-\delta/2$$ is negative, and hence not in the domain of $$\sqrt{x}$$.

Finally, just some comments on the $$\delta$$-$$\epsilon$$ definition you've described (modified so that $$x$$ has to be in the domain of the function). Let's say a function satisfying the property you've described is called bounded on finite intervals around $$x_0$$

Some properties:

1) If $$f$$ is bounded on finite intervals around $$x_0$$ for some particular $$x_0$$, it is bounded on finite intervals around $$x$$ for any $$x$$ in the domain, $$D$$, at all.

Proof: Suppose we are given $$\delta$$, let $$\delta'=\delta+|x-x_0|$$. Then since $$f$$ is bounded on finite intervals around $$x_0$$ it is bounded on $$(x_0-\delta',x_0+\delta')\supseteq (x-\delta,x+\delta)$$. Hence $$f$$ is bounded on finite intervals around $$x$$.

2) $$f$$ is bounded on finite intervals around $$x_0$$ for some particular $$x_0$$ if and only if $$f$$ is bounded on all bounded subsets of its domain. (Proof is similar to the last property)

3) If $$f$$ is continuous with closed domain then $$f$$ is bounded on all bounded subsets of its domain. Thus it is bounded on finite intervals around $$x_0$$ for any $$x_0$$ in its domain by 2 and 1.

Proof. Let $$B$$ be a bounded subset of $$D$$. Thus $$\bar{B}\subset D$$, and $$\bar{B}$$ is compact. Since $$f$$ is continuous it is bounded on compact sets, and hence on $$\bar{B}$$. Thus $$f$$ is bounded on $$B$$.

4) Not all continuous functions are bounded on bounded subsets of their domain though. As an example $$\frac{1}{x}$$ is unbounded on $$(0,1)$$.

Not quite you need to choose $\epsilon$ first, then find a $\delta$ that works. Because of the domain issues, we must have $0<\delta<1$ if it exists.

I think this is mainly a logical issue. For example take the claim: For any apple, there exists a tree such that the apple came from that tree.

You want to say this is equivalent to: For any tree, there exists an apple such that the apple came from that tree.

Clearly these mean different things.

• An implication is indeed not always equivalent to its converse. For this particular case, however, it was not readily obvious to me. Apr 24, 2015 at 18:59