TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Every $G_{δ}$ subset of $R$ is the set of continuity points of some $f : R \to R$ , so the $G_{δ}$ sets are exactly the continuity sets

Statement

Let $G \subseteq R$ be a $G_{δ}$ set ( $F_{σ}$ and $G_{δ}$ subsets of $R$ ). Then there is a function $f : R \to R$ whose set of continuity points (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ 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 $f : A \to R$ the set of points of $A$ at which $f$ is discontinuous is the intersection with $A$ of an $F_{σ}$ subset of $R$ , and the set of points at which $f$ is continuous is the intersection with $A$ of a $G_{δ}$ subset; for $A = R$ the two sets are $F_{σ}$ and $G_{δ}$ outright, which says that the continuity set of every $f : R \to R$ is a $G_{δ}$ set, this identifies the two classes:

${continuity sets of functions R \to R} = {G_{δ} subsets of R} .$

The construction. Write $G = ⋂_{n \in N} V_{n}$ 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 $⋂_{n} W_{n} = G$ . For $x \notin G$ let $n (x)$ be the least $n$ with $x \notin W_{n}$ , and set

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

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 $G_{δ}$ set $G = ⋂_{n \in N} V_{n} \subseteq R$ with each $V_{n}$ open.

[L1]

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

[L2]

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

[L3]

Both $Q$ and $R ∖ Q$ are dense in $R$ , so every neighbourhood of every real contains a rational and an irrational (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ 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 $η > 0$ there is a natural $m \geq 1$ with $1 / ι (m) < η$ , and $ι$ is positive and strictly increasing on the naturals $\geq 1$ , so $j < k$ gives $1 / ι (k + 1) < 1 / ι (j + 1)$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing).

[L5]

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

Proof

technique · constructive

1.1

Put $W_{n} : = ⋂_{j \leq n} V_{j}$ for $n \in N$ . Each $W_{n}$ is open, being a finite intersection of open sets; $W_{n + 1} \subseteq W_{n}$ ; and $⋂_{n} W_{n} = ⋂_{n} V_{n} = G$ , 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 ${n \in N : x \notin W_{n}}$ 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 $ε > 0$ be real and take a natural $m \geq 1$ with $1 / ι (m) < ε$ . Since $x \in G \subseteq W_{m}$ and $W_{m}$ is open, there is a real $ρ > 0$ with $N_{ρ} (x) \subseteq W_{m}$ .

step 1.1L1L4

3.1

Define $f : R \to R$ by $f (x) : = 0$ for $x \in G$ , $f (x) : = 1 / ι (n (x) + 1)$ for $x \notin G$ with $x$ rational, and $f (x) : = - 1 / ι (n (x) + 1)$ for $x \notin G$ with $x$ irrational. Then $f (x) = 0$ exactly for $x \in G$ , since $1 / ι (n + 1) > 0$ for every $n \in N$ ; 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 $ρ$ as in step 2.2, let $y \in N_{ρ} (x)$ . If $y \in G$ then $f (y) = 0$ . If $y \notin G$ then $y \in W_{m}$ , so $n (y) \neq m$ and indeed $n (y) > m$ , because $y \in W_{m}$ forces $y \in W_{j}$ for every $j \leq m$ ; hence $∣ f (y) ∣ = 1 / ι (n (y) + 1) < 1 / ι (m) < ε$ , using $n (y) + 1 > m$ . In both cases $∣ f (y) - f (x) ∣ = ∣ f (y) ∣ < ε$ , since $f (x) = 0$ .

step 1.1step 2.1step 3.1step 2.2L4

4.2

$f$ is discontinuous at every $x \notin G$ . Put $ε : = 1 / ι (n (x) + 1) > 0$ , so that $∣ f (x) ∣ = ε$ , and let $δ > 0$ be real. If $x$ is rational then $f (x) = ε > 0$ ; the neighbourhood $N_{δ} (x)$ contains an irrational $y$ , and $f (y) \leq 0$ , whether $y \in G$ or not. If $x$ is irrational then $f (x) = - ε < 0$ ; the neighbourhood $N_{δ} (x)$ contains a rational $y$ , and $f (y) \geq 0$ .

step 2.1step 3.1L3

5.1

In either case of step 4.2 the point $y$ satisfies $∣ f (y) - f (x) ∣ \geq ε$ , since $f (x)$ and $f (y)$ have opposite weak signs and $∣ f (x) ∣ = ε$ . So no $δ$ witnesses the continuity condition at $x$ for this $ε$ , 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 $G_{δ}$ , 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 $j \leq m$ , 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 $G = R$ the construction gives $f = 0$ , continuous everywhere. For $G = \emptyset$ , obtained as the intersection of the sequence constantly $\emptyset$ , every $x$ lies outside $W_{0} = \emptyset$ , 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 $R$ , 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.

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Sources

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

Every GδGδ​ subset of RR is the set of continuity points of some f:R→Rf:R→R, so the GδGδ​ sets are exactly the continuity sets