The indicator of $\mathbb{Q}$ has a limit at no point of $\mathbb{R}$

Write $\mathbb{Q}$ for the canonical copy of the rationals inside $\mathbb{R}$ (The rationals embed densely in the reals), $X := \mathbb{R} \setminus \mathbb{Q}$ for the irrationals, and let

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

Refuted claim: there is a point $c \in \mathbb{R}$ at which $\mathbf{1}_{\mathbb{Q}}$ has a limit (The $\varepsilon$-$\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$).

The refutation fixes an arbitrary real $c$ and produces two sequences tending to $c$, one of rationals and one of irrationals, both avoiding $c$; the image sequences are constantly $1$ and constantly $0$, and A function has no limit at $c$ as soon as two sequences in $A \setminus \{c\}$ tending to $c$ give different limits of the values applies. Since $c$ was arbitrary, the function has a limit nowhere.

Where the choice principle enters, and where it does not. Producing the two sequences is a use of 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, whose left-to-right direction spends countable choice, and that cost is inherited here and recorded by that item. The criterion applied afterwards is the choice-free one (A function has no limit at $c$ as soon as two sequences in $A \setminus \{c\}$ tending to $c$ give different limits of the values).

• Why the sets are punctured before the sequences are drawn. A function has no limit at $c$ as soon as two sequences in $A \setminus \{c\}$ tending to $c$ give different limits of the values requires all terms to lie in $A \setminus \{c\}$, since a sequence allowed to take the value $c$ would report on $\mathbf{1}_{\mathbb{Q}}(c)$, which the limit ignores (The $\varepsilon$-$\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$). Step 1.2 therefore verifies density of $\mathbb{Q} \setminus \{c\}$ and of $X \setminus \{c\}$ directly, by placing the auxiliary neighbourhood strictly to the right of $c$.

• The two densities are not proved the same way. $\mathbb{Q}$ is dense because it is built to approximate; $X$ is dense because a countable set cannot exhaust an interval. Both are claims 1 and 2 of Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$, and every nonempty open subset of $\mathbb{R}$ is uncountable, and this item uses them only through the neighbourhood formulation of [L1].

• The failure is as total as possible. Not merely does the limit fail at some points: it fails at every point of $\mathbb{R}$, while the function is bounded throughout, taking only the values $0$ and $1$. Multiplying by $x$ leaves exactly one point where a limit survives, which is $x \cdot 1_{\mathbb{Q}}(x)$ has a limit at $0$ and at no other point.