Proving that $f(x) = \frac{1}{x}$ is not uniformly continuous over $(0,1)$ - Approach to choosing the correct $x$ and $y$

Question

I've looked at the various other posts on this question and I've also worked on uniform continuity in the past. But right now as I work through Spivak's Calculus, I'm seeing that I haven't actually learned how to find choices for my $x$ and $y$ elements in a structured well thought out way. In the past what I thought was doing this was actually me just regurgitating what a professor did in their proof without any understanding of how $x$ and $y$ were chosen.

To illustrate where I'm stuck:

Let $\epsilon = 1$. Taking arbitrary $\delta > 0$. I encounter the following:

$$\bigg|\frac{1}{x} - \frac{1}{y} \bigg| > \epsilon = 1$$

I've seen prior choices such as letting $x = \delta$ and $y = \frac{\delta}{2}$ or letting $x < \delta$ and $y = \frac{x}{2}$. But I want to understand how these choices were made initially.

I know that $x(\delta)$ and $y(\delta)$ as well if I fiddle with the the expression above I have

$$\bigg|\frac{x-y}{xy}\bigg|$$

I also know that since $x,y \in (0,1)$ that $xy < 1$ (but this now won't help me when I want to extend the interval to $(0, \infty)$.

It appears I'm throwing a lot of stuff at the wall hoping for something to stick and this isn't very conducive. How should I be reasoning these steps out?

Answer 1

Simply let $ x = 2y \implies \left|\dfrac{x-y}{xy} \right| = \dfrac{1}{x} \ge \epsilon = 1$ when $0 < x < \delta < 1$.

Answer 2

Informally, uniform continuity means the (absolute value of the) slope of the function is bounded above. You should recognize that the derivative of $\frac 1x$, which is $\frac {-1}{x^2}$, can get as large as you want, so the function is not uniformly continuous. The problem occurs close to $0$, so that is where you should choose your points. You just have to find, based on the parameters you are given, how close you have to get.

You should not use $x$ for the variable in your function and again for one of the points, so I will use $y$ and $z$ as the two points.

The formal definition of uniform continuity says that for any $\epsilon \gt 0$ there is a $\delta \gt 0$ such that $|y-z| \lt \delta \implies |f(y)-f(z)| \lt \epsilon$. I like to think of quantified statements as a game. If somebody says the statement is true and I dispute that, I get to choose all the variables with a $\forall$ in my effort to make the statement false, while they get to choose all the variables with a $\exists$ in an effort to make the statement true. We play in order of the quantifiers, so I choose $\epsilon$ first and choose $1$. Now they have to choose $\delta$. Here is the difference with normal continuity. In the normal continuity definition, I have to choose $y$ before $\delta$ is chosen. For uniform continuity, $\delta$ is chosen before $y$. We assume $\delta$ is given to us. We need to find $y,z$ that are within $\delta$ of each other but $|\frac 1y - \frac 1z| \gt 1$. Here we can just take $z=y+\frac 12\delta$. That keeps them within $\delta$ and we know $\frac 1y \gt \frac 1z$ so we can get rid of the absolute value signs. Now we need $$\frac 1y - \frac 1{y+\frac 12\delta} \gt 1\\ y+\frac 12\delta -y \gt y(y+\frac 12 \delta)\\ \delta \gt 2y^2 + y\delta\gt 2y^2\\ y \lt \sqrt{\frac \delta 2}$$ So I can choose $y=\frac 12 \sqrt \delta, z=y+\frac 12\delta$ and the claim will fail.

Answer 3

Some rough steps :

• Visualize the graph of this function : since you are asked to prove that the function is not uniformly continuous, you must look for a place where the function is increasing really fast (or has steep slope) . (Note : a heuristic is that this can happen near a point where the function takes the value infinity : that point won't be there in the domain, but points close to it will be there)

• Why? Because, at these points, the $x$ coordinates are really close, but the $y$ coordinates are far apart. Remember uniform continuity says that the $y$ - coordinate difference between two points on the function's graph can be uniformly controlled by controlling their $x$-coordinate difference : so the points lying on these high-slope regions are most likely to provide a contradiction.

• After doing this, look for points $x,y$ for which the expression for $f(x)$ is very simple. For example, if $f(x) = \sin x$, then for $x$ being multiple of $\frac {\pi}4$, for example, the expression is simple.

• Sometimes, you will be able to do both : find a sequence of points in the slopy region, for which the function values are really easy to evaluate. Then you can compare these values and come to a contradiction.

---

Let us apply these principles for $f(x) = \frac 1x$. Drawing $\frac 1x$ in your head, you see that if $x = 0$ the expression is infinite. So points near $0$ is where you can try to look.

Note that if $x$ is the reciprocal of an integer, then $f(x) $ is that integer! So taking $x = \frac 1n$ for $n \in \mathbb N$ gives simple values of $f(x)$. Thus, we don't need to complicate matters by looking at the general expression $|f(x)-f(y)|$.

Finally, we notice that if $x_n = \frac{1}{n}$ and $y_n = \frac 1{n+1}$, then $|x_n-y_n|$ can be made small enough, but we always have $|f(x_n)-f(y_n)| = 1$. So, for a given $\delta$, we can easily take such large $N$ that $\frac 1N < \delta$ (and $|x_N - y_N| =x_N-y_N < x_N < \delta$) but $|f(x_n)-f(y_n)| = 1$.

Thus uniform continuity is contradicted with $\epsilon = 1$, because we can find for any $\delta$ a pair of points differing by less than $\delta$ but whose function values have difference $1$.

You can keep these points in mind while approaching "non-uniform continuity" questions.