# Is there a way to prove continuity without using epsilon-delta?

I'd like to know if there is an easier way of provining continuity instead of using the epsilon-delta criteria? I cannot understand it because it's way too complicated for me...

There is no workaround? Like converting the series to a function, then prove convergence on the function. If the converted function is continuous, the series will be continuous as well. Something like that would be possible?

Or there are other ways which are a bit easier?

Sorry for asking a question like that but I couldn't find anything on the internet. I really hope there is another way...

• I have never heard of "constant continuity of series" and searching online has brought me nothing using that phrase. Perhaps you mean uniform convergence? Please state the definition of constant continuity that you are using. – Will R Aug 11 '16 at 12:07
• Will R Oh yeah that's what I meant :P Got troubles translating some words correctly, especially in that context. – tenepolis Aug 11 '16 at 12:13
• Are you talking about a numerical series or a function series? Uniform convergence doesn't make sense in the context of a numerical series. – Wavelet Aug 11 '16 at 12:15
• Example would be $f:[0,\infty)\ni x\mapsto \frac{x^{2}}{x+1}\in \mathbb{R}$. Now I realize, I shouldn't have said series since this is a function.. Anyway, would like to know if it would be possible without epsilon delta. – tenepolis Aug 11 '16 at 12:21
• So you just want to know how to prove continuity without using $\epsilon$-$\delta$? – Wavelet Aug 11 '16 at 12:24

Based on @Dave L. Renfro's comment, I am inclined to include a detailed example, since that might help more than or at least supplement my explanation. This one is moderately tricky, if I remember and execute this correctly (it's been awhile).

Say $f(x)=x^{2}$ and we want to find a general $\delta$ that holds for all $x \in \mathbb{R}$ such that for alll $\epsilon>0$ we have that $\lvert x-a \rvert<\delta$ implies that $\lvert x^{2}-a^{2} \rvert<\epsilon$.

Now, to start, note that we can factor our inequality $$\lvert x^{2}-a^{2} \rvert=\lvert x-a\rvert\lvert x+a\rvert$$ now notice that we have $$\lvert x-a \rvert<\delta \text{ and } \lvert x-a\rvert\lvert x+a\rvert<\epsilon$$ so we can write $$\lvert x-a\rvert<\frac{\epsilon}{\lvert x+a\rvert}$$ and furthermore, say that we were to assume $\delta$ satisfies $$\lvert x-a\rvert<\delta \leq \frac{\epsilon}{\lvert x+a\rvert}$$ and $$\delta \leq \frac{\epsilon}{\lvert x+a\rvert} \implies \lvert x-a\rvert\lvert x+a\rvert<\epsilon$$ but notice that $\delta$ depends on $x$, and ideally, we want $\delta$ to be a function of $a$ and $\epsilon$ only. To sidestep this issue, we assume that $\delta <1$ then by choosing a $\delta \leq 1$ we have that $\lvert x-a \rvert < 1$ whenever $\lvert x-a\rvert<\delta$. Now note that $\lvert x-a \rvert \leq 1$ is equivalent to saying that $x$ is within a distance of $1$ of $a$ which implies that for $\lvert x+a\rvert$ we have $$\lvert x+a \rvert +\lvert x-a\rvert<1+\lvert x+a\rvert \implies \lvert x+a \rvert <\lvert 2a \rvert +1$$ so now we have a bound on $\lvert x+a \rvert$ which only depends on $a$ so we can write $$\lvert x-a\rvert < \delta=\min\left\{1,\frac{\epsilon}{\lvert 2a \rvert+1} \right\}<\frac{\epsilon}{\lvert x+a\rvert}$$ Which shows that $f(x)=x^{2}$ is continuous for any $a \in \mathbb{R}$.

• Kouba's THE CALCULUS PAGE PROBLEMS LIST has a lot of worked examples (see under "limit of a function using the precise epsilon/delta definition of limit"). I used to use this web site with classes I taught in the late 1990s --- I can't believe it's still around! – Dave L. Renfro Aug 11 '16 at 16:40
• A remark: this doesn't show that $x^2$ is not uniformly continuous (the last paragraph may seem confusing). – YoTengoUnLCD Aug 15 '16 at 21:50

This answer is about the continuity of functions, since the comments under the question indicate that the question was really about continuity of functions rather than convergence of series.

The continuity of a function $f:A\to\mathbb R$, where $A \subseteq \mathbb R$, is usually defined in terms of limits of the form $\lim_{x\to a} f(x)$. Limits of functions are usually defined using quantifiers over variables $\epsilon$ and $\delta$.

The reason the definition is important is that it applies to every function you could ever look at. It is possible to write the criteria for continuity differently so that they still apply to every function, either by proof or by definition. This other question gives an example of an alternative way to define continuity. But the application of this definition to an actual function is (I think) at least as complicated as the $\epsilon$-$\delta$ method.

I think if you want a method you can apply to any function, you need to have enough mathematical understanding to do $\epsilon$-$\delta$ proofs reliably. The alternatives will require just as much sophistication.

For many functions, however, you can avoid having to write out the $\epsilon$-$\delta$ method if you are allowed to apply a few known facts. For example, if $f$ and $g$ are both continuous functions over the domain $A$, then $f + g: x \mapsto f(x) + g(x)$ is also a continuous function over the domain $A$. There are several useful facts like this by which you can combine known continuous functions to show that other functions are continuous; page 2 of this document, for example, shows some of them.

Realize that someone had to apply $\epsilon$-$\delta$ proofs or similarly difficult methods in order to prove all of those facts in the first place. But once you know they are true, they make it very easy to show that (for example) the function $x \mapsto \frac{x^2}{x+1}$ on the domain $[0,\infty)$ is continuous.

• $\frac{x^2}{x+1}$ is not continuous at $x=-1$. Did you mean $\frac{x}{x^2+1}$? – Oriol Aug 11 '16 at 19:35
• @Oriol OP had mentioned that particular function in a comment, but restricted to the non-negative reals. I neglected to specify the domain (my mistake, sorry!); I think that's fixed now. – David K Aug 12 '16 at 0:24

You bet there is! To prove the continuity of $y=f(x)=\frac{x^2}{x+1}$ on positive reals, just show that every infinitesimal $x$-increment always leads to an infinitesimal change in $y$, which is the Cauchy-Robinson definition of continuity. Namely, if $\alpha>0$ is infinitesimal then $$f(x+\alpha)-f(x)= \frac{(x+\alpha)^2}{x+\alpha+1}-\frac{x^2}{x+1}=\frac{(x+\alpha)^2(x+1)-(x+1+\alpha)x^2}{ (x+1+\alpha)(x+1) }.$$ Multiplying out, we get $$f(x+\alpha)-f(x)= \frac{x^3+2x^2\alpha+\alpha^2x + x^2+2x\alpha+\alpha^2-x^3-x^2-\alpha x^2}{ (x+1+\alpha)(x+1)}.$$ Simplifying, we get $$f(x+\alpha)-f(x)= \frac{2x^2\alpha+\alpha^2x +2x\alpha+\alpha^2-\alpha x^2}{(x+1+\alpha)(x+1)}=\alpha\frac{2x^2+\alpha x +2x+\alpha- x^2}{(x+1+\alpha)(x+1)}.$$ This is a product of an infinitesimal $\alpha$ by a finite number $\frac{2x^2+\alpha x +2x+\alpha- x^2}{(x+1+\alpha)(x+1)}$ which is therefore infinitesimal. QED. For details on the rules governing manipulation with infinitesimals see Elementary calculus.

• This is very slick, but in the setting of a standard elementary analysis course, it’s not a proof of continuity. It’s proving something related in a rather different setting; and there are big transfer theorems that show that this implies continuity in the standard sense. But I wouldn’t accept this from a student as a proof of continuity in the usual setting, unless they could also demonstrate some understanding of the transfer theorems, since without those, this is a proof of something rather different. – Peter LeFanu Lumsdaine Aug 11 '16 at 18:54
• @PeterLeFanuLumsdaine, thanks for your comments. I have two remarks. (1) At my university, we teach first semester calculus to 130 students every year using infinitesimals a la Keisler, and this is indeed the definition of continuity. (2) You seem to be convinced that the epsilon-delta is the "true" definition but at any hate historically this is not accurate since the first definition was Cauchy's and it was the one I gave in my answer. The epsilon-delta definition is a paraphrase of Cauchy's definition due to Weierstrass. If you like I can send you a text containing a more detailed discu – Mikhail Katz Aug 12 '16 at 7:04
• Also, the OP may be engaged in self-study rather than taking a course at your university, in which case the infinitesimal approach is the more intuitive one that prepares the ground for facing the epsilon-delta paraphrases later. At any rate the OP specifically asked for a different approach, which is what I provided. – Mikhail Katz Aug 12 '16 at 7:06
• @MK: don’t worry, I’m certainly not advocating the idea that ε-δ is the one true definition of continuity — I’m a big fan of alternative approaches to the reals and analysis. But ε-δ (or its rephrasing in terms of open sets, or similar) is still by far the most standard approach, and it’s the one the OP mentions in their question — so I’m assuming that this is the setting they’re coming from. I read their question as wanting an alternative way of proving continuity (implicitly: in the standard setting), not as asking for an alternate definition of the reals/continuity. – Peter LeFanu Lumsdaine Aug 12 '16 at 13:33
• @YoTengoUnLCD, the transfer theorem is explained in detail with examples in the Elementary Calculus reference that I linked in my answer. For a higher level approach I can recommend additional textbooks if you are interested. – Mikhail Katz Aug 16 '16 at 9:09

Topologically you can argue that given two topological spaces $(A, \tau_{A}),(B, \tau_{B})$ a function is continuous if for $f:A \to B$ we have $U \in \tau_{B} \implies f^{-1}(U) \in \tau_{A}$.

Edit:

First of all, given an arbitrary space $X$, a metric is a function $d:X \times X \to [0,\infty)$ where together the space and the metric $(X,d)$ form a metric space (and also note that a metric space is a topological space since metrics induce topologies). To be a metric, $d$ must satisfy some intuitive properties namely that $$d(x_{1},x_{2})=0 \implies x_{1}=x_{2}$$ and $$d(x_{1},x_{2})=d(x_{2},x_{1})$$ and finally the triangle inequality $$d(x_{1},x_{3}) \leq d(x_{1},x_{2})+d(x_{2},x_{3})$$ Now we usually denote the metric on $\mathbb{R}$ by $\lvert \cdot \rvert$ which for $x_{1},x_{2} \in \mathbb{R}$ we have just $\lvert x_{1} \rvert =d(0,x_{1})$ and $\lvert x_{1}-x_{2} \rvert=d(x_{1},x_{2})$.

Now for $\epsilon-\delta$ proofs, we have the general statement in plain English that for a function $f:\mathbb{R} \to \mathbb{R}$ we have that $$\lim_{x \to x_{0}}f(x)=L$$ if for $\epsilon >0$ there exists a $\delta > 0$ such that $\lvert x- x_{0}\rvert <\delta$ imples that $\lvert f(x)-L \rvert<\epsilon$

Now what is really going on here? Well first note that we can choose our value of epsilon since it is arbitrary and if $f$ is continuous, then there will be a corresponding value of $\delta$ that guarantees that $\lvert x-x_{0} \rvert<\delta$. This means that $\delta$ IS A FUNCTION OF EPSILON, which is the part that so many teachers do not explicitly mention, thus leading to so much confusion. Again, just to make that as explicit as possible, $\delta$ is uniquely determined by our choice of $\epsilon$. Now what exactly is $\epsilon$? Well, recall from algebra that $$\lvert f(x)-L \rvert<\epsilon \text{ is equivalent to writing } L-\epsilon<f(x)<L+\epsilon$$ so $\epsilon$ quantifies the degree to which we "squeeze" the interval $(L-\epsilon,L+\epsilon)$ and $\delta$ quantifies the degree to which we squeeze the interval $(x_{0}-\delta,x_{0}+\delta)$ based on our choice of $\epsilon$. Now, this might seem backwards since $x$ is the independent variable and the value of $f(x)$ depends on our choice of $x$ but that is actually exactly why we need this framework. Let's say we reversed the statement and we squeezed $(x_{0}-\delta,x_{0}+\delta)$ by choosing $\delta$ instead of epsilon, then the fact that there is some value $\epsilon$ such that $\lvert f(x)-L\rvert <\epsilon$ is trivial for any (bounded) function. Basically, the intuition for continuity is that continuity implies that when two values $x_{1}$ and $x_{2}$ are "close" (an inherently topological notion) to each other then $f(x_{1})$ and $f(x_{2})$ have to be close to each other as well. Thus, by squeezing the interval $(L-\epsilon, L+\epsilon)$ then for a continuous function, we know that the interval $(x_{0}-\delta,x_{0}+\delta)$ will be squeezed to a distance of $\delta$ as a function of $\epsilon$.

Now see if you can "reverse engineer" the $\epsilon-\delta$ definition of continuity to arrive at the topological definition I provided above.

• Ok thank you but I think this is not easier, it seems a bit harder for me : / – tenepolis Aug 11 '16 at 12:28
• There is no other easier way? – tenepolis Aug 11 '16 at 12:29
• To rigorously prove continuity? I'm afraid not. Is there anything in particular that gives you so much difficulty understanding $\delta-\epsilon$ proofs? – Wavelet Aug 11 '16 at 12:30
• For function I don't have problem, just show for left side and right side limit, if same then continuous.. Why not for this : / epsilon delta work with metrics right? I don't understand anything there.. – tenepolis Aug 11 '16 at 12:34
• Yes with $\epsilon-\delta$ there is an implicit assumption that you are working in a metric space..I'll edit my question in a second and see if I can't break it down for you. – Wavelet Aug 11 '16 at 12:36

You could try reasoning within a framework which extends the reals with actual infinitesimals (I am aware of three such frameworks).

• Welcome! You should at least spell out the names of those three :-) – Singhal Aug 11 '16 at 15:11
• Done; see my answer. – Mikhail Katz Aug 11 '16 at 15:30

Usually, there are simple ways to show continuity.

# Composition of continuous functions is continuous

If you build your function out of continuous functions, it is automatically continuous. For example, consider the function:

$$f(x) = \sqrt{1 + x^2}$$

Constants, addition, squaring, and square roots are all continuous on their domain, and this implies $f$ is continuous.

# Compute the limit

The usual definition of continuity is not $\epsilon-\delta$: it is in terms of limits. You spend a lot of time in your calculus classes learning how to compute them.

For example, consider the function

$$\mathrm{sinc}(x) = \begin{cases} \frac{\sin x}{x} & x \neq 0 \\ 1 & x = 0 \end{cases}$$

By the method above, we can immediately see that $\mathrm{sinc}$ is continuous whenever $x \neq 0$. To show continuity at $x=0$, we just compute the limit.

Goal: prove $\mathrm{sinc}(0) = \lim_{x \to 0}\mathrm{sinc}(x)$

The left hand side is $1$. The right hand side can be computed by

$$\lim_{x \to 0} \mathrm{sinc}(x) = \lim_{x \to 0} \frac{\sin x}{x} = 1$$

Both sides are equal, and so $\mathrm{sinc}$ is continuous at zero as well, so it's continuous everywhere.