The indicator of $\mathbb{Q}$ is continuous at no point of $\mathbb{R}$

Refuted claim: every function $\mathbb{R} \to \mathbb{R}$ is continuous at at least one point (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point).

The witness is the Dirichlet function, the indicator of the rationals: writing $\mathbb{Q}$ for the canonical copy of the rationals inside $\mathbb{R}$ (The rationals embed densely in the reals),

$\mathbf{1}_{\mathbb{Q}} : \mathbb{R} \to \mathbb{R}, \qquad \mathbf{1}_{\mathbb{Q}}(x) := \begin{cases} 1 & \text{if } x \in \mathbb{Q},\\ 0 & \text{if } x \notin \mathbb{Q}. \end{cases}$

It is continuous at no point of $\mathbb{R}$. The mechanism is that both $\mathbb{Q}$ and its complement are dense (Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$, and every nonempty open subset of $\mathbb{R}$ is uncountable), so every neighbourhood of every real contains a point of each, and the two values differ by $1$.

The argument is choice free. Density is used in the form "every neighbourhood of every point meets the set", which is The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points applied to a closure equal to $\mathbb{R}$; no sequence is built, so neither A point lies in the closure of $A \subseteq \mathbb{R}$ iff some sequence in $A$ converges to it, so a subset of $\mathbb{R}$ is closed iff it is sequentially closed nor $f$ is continuous at $c \in A$ if and only if $f(x_k) \to f(c)$ for every sequence in $A$ converging to $c$, the converse direction costing countable choice is invoked, and the countable choice those two spend is not spent here.

• Why $\varepsilon_0 = 1/2$ and not $1$. Any $\varepsilon_0 \le 1$ works, since the discrepancy produced is exactly $1$. Taking $1/2$ leaves the inequality strict and makes it visible that the failure is not a boundary effect.

• Restricting the domain repairs it completely. The restriction of $\mathbf{1}_{\mathbb{Q}}$ to $\mathbb{Q}$ is constantly $1$ and the restriction to the irrationals is constantly $0$; both are continuous. This is the standard warning that continuity is a property of the pair (function, domain), recorded in Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point: continuity passes to subsets of the domain, never up from them.

• A near miss worth naming. Multiplying by $x$ repairs continuity at exactly one point: $x \mapsto x\,\mathbf{1}_{\mathbb{Q}}(x)$ is continuous at $0$ and nowhere else, which is $x \cdot 1_{\mathbb{Q}}(x)$ is continuous at $0$ and at no other point. The same argument as above, applied to any function taking two distinct values densely, shows nowhere-continuity; in particular the function equal to $1$ on $\mathbb{Q}$ and $-1$ elsewhere is nowhere continuous while its absolute value is constant.