2

Graph of a monotone real function defined on an interval is a nowhere dense set in $\mathbb{R}^{2}$.

I know that when $f$ is continuous, $G(f)$ is a closed set. Also, its interior is empty since any $\varepsilon$-ball around a point $(x_0, f(x_0))$ would have to contain a horizontal strip, which is not in $G(f)$. This implies that the graph of $f$ is nowhere dense. However, I'm not sure how to show the same result to a monotone function. I suspect we have to use Baire's Category Theorem somewhere, but I was unable to do so.

Any hints?

huh
  • 387
  • The graph of a monotone function is nowhere dense in a very strong way: For each point on the graph there exists a square neighborhood centered at that point and a pair of opposite "quadrants" of that square neighborhood that contain no points of the graph (quadrants II & IV if the function is non-decreasing; quadrants I & III if the function is non-increasing). Now look at the last equivalence for "nowhere dense" in this MSE answer. – Dave L. Renfro Feb 01 '24 at 03:47
  • @DaveL.Renfro I understand the idea, but I'm not sure whether I would know how to formalize it. I was thinking perhaps using the fact that a monotone function defined on an interval has countably many discontinuities. So we write $G(f)$ as a countable union of graphs of continuous functions (we do this by restricting $f$ to each continuous piece). In that way, $G(f)$ would be a meager set and, thus, nowhere dense by Baire's Theorem, since $\mathbb{R} \times \mathbb{R}$ is complete. Does this make sense? – huh Feb 01 '24 at 05:13
  • 1
    It is not obvious that $G(f)$ can be partitioned into countably many graphs of continuous functions. Let ${ q_n : n \in \mathbb{N} }$ be an enumeration of the rationals and consider $$f(x) = \sum_{q_n \leqslant x} \frac{1}{2^n}.$$ This is an increasing function (and even a cdf!), but the discontinuities are at the rational points and form a dense subset of $\mathbb{R}$. How would you decompose it? – Adayah Feb 01 '24 at 08:37
  • We don't really need to write $G(f)$ as a countable union exactly. It would be sufficient to show that it is contained in some countable union of graphs, which we can do by adding the discontinuity points, since lateral limits always exist. So what we are doing is taking $f$ restricted to the interval where it is continuous and extending it continuously to the closed interval. – huh Feb 01 '24 at 08:50
  • 1
    You can't restrict $f$ to the interval where it is continuous because there is no such interval - the discontinuities are dense in $\mathbb{R}$. Using a bit more advanced tools it is possible to show that $G(f)$ can not be covered with countably many graphs of continuous functions. By the way, it doesn't help to have $G(f)$ contained in, rather than equal to, the countable union of graphs of continuous functions, because if the former were possible, the latter would too by taking the intersection with $G(f)$. – Adayah Feb 01 '24 at 09:09
  • Oh, I see it now. That really doesn't work. I thought we could have countably many intervals where $f$ is continuous, but I overlooked the fact that there are dense sets of discontinuities. Thank you for your help! – huh Feb 01 '24 at 09:23

1 Answers1

2

Nothing as sublime as the Baire category theorem is needed here. Take the following characterization of a nowhere dense set:

A subset $N \subseteq \mathbb{R}^2$ is nowhere dense iff for every non-empty open subset $U \subseteq \mathbb{R}^2$ there is a non-empty open $V \subseteq U$ disjoint from $N$.

Using it, we prove that the graph of an increasing function $f : I \to \mathbb{R}$ is nowhere dense. Take any non-empty open $U \subseteq \mathbb{R}^2$. If $U \cap G(f) = \varnothing$, we are done. Otherwise pick $(p, q) \in U \cap G(f)$. Then

$$V := U \cap \big( (p, \infty) \times (-\infty, q) \big)$$

is a non-empty open set contained in $U$ and disjoint from $G(f)$, as required.

Adayah
  • 10,468
  • Awesome! I proved that characterization on a different exercise, but I didn't think of using it. I was thinking about Baire's Theorem, so I ended up going nowhere for a while. Would you mind taking a look at the idea I left under Dave's comment above just to see if it makes any sense? – huh Feb 01 '24 at 08:32