TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)

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Every $G_\delta$ subset of $\mathbb{R}$ is the set of continuity points of some $f : \mathbb{R} \to \mathbb{R}$ , so the $G_\delta$ sets are exactly the continuity sets

Statement

Let $G \subseteq \mathbb{R}$ be a $G_\delta$ set ( $F_\sigma$ and $G_\delta$ subsets of $\mathbb{R}$ ). Then there is a function $f : \mathbb{R} \to \mathbb{R}$ whose set of continuity points (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) is exactly $G$ .

Together with For $f : A \to \mathbb{R}$ the set of points of $A$ at which $f$ is discontinuous is the intersection with $A$ of an $F_\sigma$ subset of $\mathbb{R}$ , and the set of points at which $f$ is continuous is the intersection with $A$ of a $G_\delta$ subset; for $A = \mathbb{R}$ the two sets are $F_\sigma$ and $G_\delta$ outright, which says that the continuity set of every $f : \mathbb{R} \to \mathbb{R}$ is a $G_\delta$ set, this identifies the two classes:

$\{\, \text{continuity sets of functions } \mathbb{R} \to \mathbb{R} \,\} \;=\; \{\, G_\delta \text{ subsets of } \mathbb{R} \,\} .$

The construction. Write $G = \bigcap_{n \in \mathbb{N}} V_n$ with each $V_n$ open and put $W_n := V_0 \cap \dots \cap V_n$ , so that the $W_n$ are open and decreasing with $\bigcap_n W_n = G$ . For $x \notin G$ let $n(x)$ be the least $n$ with $x \notin W_n$ , and set

$f(x) := 0 \ \text{ for } x \in G, \qquad f(x) := \frac{1}{\iota(n(x)+1)} \ \text{ for } x \notin G,\ x \in \mathbb{Q}, \qquad f(x) := -\frac{1}{\iota(n(x)+1)} \ \text{ for } x \notin G,\ x \notin \mathbb{Q}.$

The sign carries the whole of the discontinuity: near a point outside $G$ there are points of the opposite rationality, where $f$ has the opposite sign or is $0$ , and the values cannot come close.

Facts & Assumptions

Given: A $G_\delta$ set $G = \bigcap_{n \in \mathbb{N}} V_n \subseteq \mathbb{R}$ with each $V_n$ open.

[L1]

A finite intersection of open subsets of $\mathbb{R}$ is open (Arbitrary unions and finite intersections of open subsets of $\mathbb{R}$ are open, and dually for closed sets, claim 2); a set $U$ is open exactly when every point of it has a neighbourhood $N_\rho(x) \subseteq U$ (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

[L2]

Every nonempty subset of $\mathbb{N}$ has a least element (The well-ordering principle).

[L3]

Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$ , so every neighbourhood of every real contains a rational and an irrational (Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$ , and every nonempty open subset of $\mathbb{R}$ is uncountable, 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, The rationals embed densely in the reals).

[L4]

For every real $\eta > 0$ there is a natural $m \ge 1$ with $1/\iota(m) < \eta$ , and $\iota$ is positive and strictly increasing on the naturals $\ge 1$ , so $j < k$ gives $1/\iota(k+1) < 1/\iota(j+1)$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , The canonical natural $\iota(n) = n \cdot 1_F$ of a field, Canonical naturals are positive and strictly increasing).

[L5]

The continuity set of any $h : \mathbb{R} \to \mathbb{R}$ is a $G_\delta$ subset of $\mathbb{R}$ (For $f : A \to \mathbb{R}$ the set of points of $A$ at which $f$ is discontinuous is the intersection with $A$ of an $F_\sigma$ subset of $\mathbb{R}$ , and the set of points at which $f$ is continuous is the intersection with $A$ of a $G_\delta$ subset; for $A = \mathbb{R}$ the two sets are $F_\sigma$ and $G_\delta$ outright, case $A = \mathbb{R}$ , $F_\sigma$ and $G_\delta$ subsets of $\mathbb{R}$ ).

Proof

technique · constructive

1.1

Put $W_n := \bigcap_{j \le n} V_j$ for $n \in \mathbb{N}$ . Each $W_n$ is open, being a finite intersection of open sets; $W_{n+1} \subseteq W_n$ ; and $\bigcap_{n} W_n = \bigcap_{n} V_n = G$ , since a point lies in every $W_n$ exactly when it lies in every $V_j$ .

L1construct

2.1

For $x \notin G$ the set $\{\, n \in \mathbb{N} : x \notin W_n \,\}$ is nonempty, so $n(x) := \min\{\, n : x \notin W_n \,\}$ is defined; and $x \in W_j$ for every $j < n(x)$ , by minimality.

step 1.1L2construct

2.2

$f$ is continuous at every $x \in G$ . Let $\varepsilon > 0$ be real and take a natural $m \ge 1$ with $1/\iota(m) < \varepsilon$ . Since $x \in G \subseteq W_{m}$ and $W_{m}$ is open, there is a real $\rho > 0$ with $N_\rho(x) \subseteq W_{m}$ .

step 1.1L1L4

3.1

Define $f : \mathbb{R} \to \mathbb{R}$ by $f(x) := 0$ for $x \in G$ , $f(x) := 1/\iota(n(x)+1)$ for $x \notin G$ with $x$ rational, and $f(x) := -1/\iota(n(x)+1)$ for $x \notin G$ with $x$ irrational. Then $f(x) = 0$ exactly for $x \in G$ , since $1/\iota(n+1) > 0$ for every $n \in \mathbb{N}$ ; moreover $f(x) > 0$ at a rational outside $G$ and $f(x) < 0$ at an irrational outside $G$ .

step 2.1L4construct

4.1

With $m$ and $\rho$ as in step 2.2, let $y \in N_\rho(x)$ . If $y \in G$ then $f(y) = 0$ . If $y \notin G$ then $y \in W_{m}$ , so $n(y) \ne m$ and indeed $n(y) > m$ , because $y \in W_{m}$ forces $y \in W_j$ for every $j \le m$ ; hence $|f(y)| = 1/\iota(n(y)+1) < 1/\iota(m) < \varepsilon$ , using $n(y) + 1 > m$ . In both cases $|f(y) - f(x)| = |f(y)| < \varepsilon$ , since $f(x) = 0$ .

step 1.1step 2.1step 3.1step 2.2L4

4.2

$f$ is discontinuous at every $x \notin G$ . Put $\varepsilon := 1/\iota(n(x)+1) > 0$ , so that $|f(x)| = \varepsilon$ , and let $\delta > 0$ be real. If $x$ is rational then $f(x) = \varepsilon > 0$ ; the neighbourhood $N_\delta(x)$ contains an irrational $y$ , and $f(y) \le 0$ , whether $y \in G$ or not. If $x$ is irrational then $f(x) = -\varepsilon < 0$ ; the neighbourhood $N_\delta(x)$ contains a rational $y$ , and $f(y) \ge 0$ .

step 2.1step 3.1L3

5.1

In either case of step 4.2 the point $y$ satisfies $|f(y) - f(x)| \ge \varepsilon$ , since $f(x)$ and $f(y)$ have opposite weak signs and $|f(x)| = \varepsilon$ . So no $\delta$ witnesses the continuity condition at $x$ for this $\varepsilon$ , and $f$ is discontinuous at $x$ .

step 4.2

6.1

By steps 4.1 and 5.1 the set of continuity points of the function $f$ constructed in step 3.1 is exactly $G$ , which proves the theorem. Combined with the fact that every continuity set is $G_\delta$ , the two classes coincide.

step 3.1step 4.1step 5.1L5discharge-construct∎

Remarks

Why the $V_n$ are replaced by the decreasing $W_n$ . The index $n(x)$ is useful only because $y \in W_m$ implies $y \in W_j$ for all $j \le m$ , which is what makes $n(y) > m$ in step 4.1. For an arbitrary sequence $(V_n)$ that implication fails, and $n(x)$ would carry no information about how deep $x$ sits in the intersection. Passing to the finite intersections costs nothing, since they are still open and still intersect to $G$ .
Two extreme cases. For $G = \mathbb{R}$ the construction gives $f = 0$ , continuous everywhere. For $G = \varnothing$ , obtained as the intersection of the sequence constantly $\varnothing$ , every $x$ lies outside $W_0 = \varnothing$ , so $n(x) = 0$ and $f$ takes the value $1$ at every rational and $-1$ at every irrational; it is nowhere continuous, as the Dirichlet function is (The Dirichlet function is continuous at no point of $\mathbb{R}$ , and Thomae's function is continuous at every irrational and at no rational, so its set of continuity points is exactly the set of irrationals and its oscillation at $c$ equals $t(c)$ ).
The construction does not guarantee monotonicity, and the theorem does not claim it. The function built above always takes values in $[-1,1]$ , so it is bounded; no further behaviour beyond its continuity set is asserted.

Depends on

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Sources

Gdelta set (Wikipedia) (standard reference, not scraped)
Classification of discontinuities (Wikipedia) (standard reference, not scraped)

Every GδG_\deltaGδ​ subset of R\mathbb{R}R is the set of continuity points of some f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R, so the GδG_\deltaGδ​ sets are exactly the continuity sets