08 Apr 2019  0

We look into a weird consequence of the definition of continuity given by Karl Weierstrass.

We know that there are certain definitions and results in mathematics that lead to unintuitive consequences, some mild, which just requires a bit of tweak in perspective, and others straight up mind-boggling. I came across one of the former today, See MathSE. which initially shoke me so much I felt like a truck hit me and had almost sent me down a very badly-timed existential crisis This is a few days right before an important exam. .

I will not review what Weierstrass continuity means first, but hint that it is the definition of continuity that uses $\epsilon$’s and $\delta$’s. If you are not familiar with it, this post will be uninteresting if you follow the sequence of the writing, so look downwards for the definition of Weierstrass continuity. If you are aware of the definition, follow along.

Graph of f(x)
Consider the function $f : [0, 1) \cup [2, 3] \to [0, 2]$ given by

$$f(x) = \begin{cases} x & x \in [0, 1) \\ x - 1 & x \in [2, 3] \end{cases}$$

The graph of $f(x)$ is shown to the right. This map looks discontinuous alright, but is it really discontinuous?

Consider $c = 2$. Let $\epsilon > 0$. Let us focus on left continuity at $c$, since that is the point that disturbs us. Let us choose $\delta = 1 + \epsilon > 0$. Now for any $y \in [0, 1) \cup [2, 3]$, since we are focusing on left continuity, we know that $f(y) = y$. Now if $\abs{c - y} = 2 - y < \delta$, we have that $1 - y < \epsilon$. Then, observe that

$$\abs{f(2) - f(y)} = 1 - y < \epsilon.$$

By definition, $f$ is indeed continuous at $c = 2$.

The keen and trained eye would immediately recognize where things have went wrong, but the innocent will most likely exclaim, “Hold up! WTF!?”

I shall now present the definition the Weierstrass’ definition of continuity.

## Definition of Weierstrass Continuity

Weierstrass’ flavour of continuity is one that those who have taken a serious course in Calculus should be familiar with.

Let $f : X \to Y$ be a function between two (metric) spaces (or just sets) $X$ and $Y$, and let $x \in X$. We (Weierstrass) say(s) that $f$ is continuous on $x$ if

$$\forall \epsilon > 0 \enspace \exists \delta > 0 \enspace \forall y \in X$$
$$|x - y| < \delta \implies |f(x) - f(y)| < \epsilon$$

## Problem Analysis

The problem lies in that misleading graph: the domain of $f(x)$ is a disjoint union, not a single “connected set” Connectedness is important in analysis. It is actually non-trivial to define what it means for a set to be connected. Here is a relevant article about connectedness. . Therefore, the usual Cartesian plane is not a good representation of the graph. In particular, we would have to truncate the area between 1 and 2, and one would then immediately notice (or at least agree) that $f$ is indeed continuous.

Of course, the choice of $\delta$ is particularly awkward: it is not applicable to the right side of $c = 2$, since $y = 3$, in particular, would not work for $\epsilon = 0.2 > 0$. This is a consequence of assuming the usual sense of distance between points, or metric, on the real numbers $\mathbb{R}$ (and its subsets). To make things work nicely again, i.e. to be able to choose a nicer $\delta$ that works on both sides, we would either have to define a new metric, or we can “identify” everything between 1 and 2 as simply 2. In other words, we are looking at a world where $0 + 1 = 2$.

Note that there is no problem in embedding the graph $f$ onto the usual Cartesian plane. It still helps us visualize how the function looks like on the usual world in our normal sense of distances.

## Conclusion

Weierstrass’ definition of continuity transcends our usual sense of distance and even rectified itself, to some extent, in this particular example. I now have newfound appreciation and respect for the thought that has gone into this way of thinking about continuity.

- Japorized -