Graph of a monotone function is nowhere dense.

Question

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?

Answer 1

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.