TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$f : A \to R$ is continuous on $A$ if and only if the preimage of every open subset of $R$ is the intersection with $A$ of an open subset of $R$ , and dually for closed sets

Statement

Let $A \subseteq R$ and $f : A \to R$ . Call a set $S \subseteq A$ relatively open in $A$ when $S = U \cap A$ for some open $U \subseteq R$ , and relatively closed in $A$ when $S = G \cap A$ for some closed $G \subseteq R$ (Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen). For $V \subseteq R$ write $f^{- 1} (V) : = {x \in A : f (x) \in V}$ . Then the following are equivalent.

$f$ is continuous on $A$ (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).
$f^{- 1} (V)$ is relatively open in $A$ for every open $V \subseteq R$ .
$f^{- 1} (F)$ is relatively closed in $A$ for every closed $F \subseteq R$ .

"Relatively open" is defined here inline, and on purpose. At this point in the reading order this library has no subspace-topology item for $R$ , and the metric one (Isometry, isometric embedding, and the subspace metric on a subset) may not be reached before Dictionary: for $A \subseteq R$ with the metric $d (x, y) = ∣ x - y ∣$ , continuity and uniform continuity of $f : A \to R$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $R$ is compact in the open-cover sense of $R$ exactly when it is a compact metric subspace has said that the two vocabularies agree, which is later on this page. The phrase above is therefore an abbreviation for the displayed condition and nothing more.

The preimage is taken inside $A$ . $f^{- 1} (V)$ is a subset of $A$ , never of $R$ , so claim 2 does not say that preimages of open sets are open. They are open only when $A$ is itself open: then $U \cap A$ is an intersection of two open sets, hence open (Arbitrary unions and finite intersections of open subsets of $R$ are open, and dually for closed sets). For $A = [0, 1]$ and $f$ the identity, $f^{- 1} ((- 1, 1 / 2)) = [0, 1 / 2)$ is not open, and it is the trace on $A$ of the open set $(- 1, 1 / 2)$ .

No choice principle is used. The open set witnessing claim 2 is not selected point by point; it is constructed as a single union over a family cut out by a property, which is the device the proof below makes explicit.

Facts & Assumptions

Given: A set $A \subseteq R$ and a function $f : A \to R$ ; for $V \subseteq R$ , $f^{- 1} (V) = {x \in A : f (x) \in V}$ .

[L1]

Continuity of $f$ at $c \in A$ : for every real $ε > 0$ there is a real $δ > 0$ with $∣ f (x) - f (c) ∣ < ε$ for every $x \in A$ satisfying $∣ x - c ∣ < δ$ ; equivalently $f (A \cap N_{δ} (c)) \subseteq N_{ε} (f (c))$ (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, The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L2]

Open sets of $R$ : $U$ is open when every $x \in U$ has some $N_{ε} (x) \subseteq U$ ; every neighbourhood $N_{ε} (x)$ is itself open; a set is closed exactly when its complement is open (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 $R$ ).

[L3]

An arbitrary union of open subsets of $R$ is open (Arbitrary unions and finite intersections of open subsets of $R$ are open, and dually for closed sets).

[L4]

Set algebra: for $V \subseteq R$ one has $f^{- 1} (R ∖ V) = A ∖ f^{- 1} (V)$ ; and for $U \subseteq R$ , $A ∖ (U \cap A) = (R ∖ U) \cap A$ .

Proof

technique · direct

1.1

From 1 to 2: the canonical witness. Assume $f$ is continuous on $A$ and let $V \subseteq R$ be open. Define $U : = ⋃ {N_{δ} (x) : x \in f^{- 1} (V), δ \in R, δ > 0, f (A \cap N_{δ} (x)) \subseteq V} .$ The family being united is cut out by a property of the pair $(x, δ)$ , so it is a set and nothing is selected from it. Each of its members is open by [L2], so $U$ is open by [L3].

L2L3

1.2

$f^{- 1} (V) \subseteq U \cap A$ . Let $x \in f^{- 1} (V)$ , so $x \in A$ and $f (x) \in V$ . Since $V$ is open, [L2] gives a real $ε > 0$ with $N_{ε} (f (x)) \subseteq V$ , and continuity at $x$ gives, by [L1], a real $δ > 0$ with $f (A \cap N_{δ} (x)) \subseteq N_{ε} (f (x)) \subseteq V$ . So this pair $(x, δ)$ contributes $N_{δ} (x)$ to the union, and $x \in N_{δ} (x)$ by [L2]. Hence $x \in U$ , and $x \in A$ .

L1L2

1.3

From 2 to 1. Assume claim 2, let $c \in A$ and let a real $ε > 0$ be given. The set $V : = N_{ε} (f (c))$ is open by [L2], so $f^{- 1} (V) = U \cap A$ for some open $U \subseteq R$ . Since $∣ f (c) - f (c) ∣ = 0 < ε$ we have $c \in f^{- 1} (V)$ , hence $c \in U$ , and [L2] gives a real $δ > 0$ with $N_{δ} (c) \subseteq U$ . Every $x \in A$ with $∣ x - c ∣ < δ$ then lies in $U \cap A = f^{- 1} (V)$ , so $f (x) \in N_{ε} (f (c))$ , that is $∣ f (x) - f (c) ∣ < ε$ . As $c$ and $ε$ were arbitrary, $f$ is continuous on $A$ .

L1L2

2.1

$U \cap A \subseteq f^{- 1} (V)$ . Let $y \in U \cap A$ . Then $y \in N_{δ} (x)$ for some pair $(x, δ)$ occurring in the union, so $y \in A \cap N_{δ} (x)$ and therefore $f (y) \in V$ by the defining property of that pair. Hence $y \in f^{- 1} (V)$ .

step 1.1

3.1

Claim 2 holds. By steps 1.2 and 2.1, $f^{- 1} (V) = U \cap A$ with $U$ open, so $f^{- 1} (V)$ is relatively open in $A$ ; and $V$ was an arbitrary open subset of $R$ .

step 1.1step 1.2step 2.1

4.1

2 and 3 are equivalent. Let $F \subseteq R$ be closed and put $V : = R ∖ F$ , which is open by [L2]. If claim 2 holds then $f^{- 1} (V) = U \cap A$ with $U$ open, and by [L4] $f^{- 1} (F) = A ∖ f^{- 1} (V) = A ∖ (U \cap A) = (R ∖ U) \cap A,$ with $R ∖ U$ closed by [L2]; so $f^{- 1} (F)$ is relatively closed. The converse runs the same computation in the other direction, starting from an open $V$ , putting $F : = R ∖ V$ and using $f^{- 1} (V) = A ∖ f^{- 1} (F)$ .

step 3.1L2L4

5.1

Statements 1, 2 and 3 are therefore equivalent, and the passage from 1 to 2 selected nothing.

step 3.1step 1.3step 4.1∎

Remarks

Why the union, and not a choice of $δ$ at each point. The textbook proof says "for each $x \in f^{- 1} (V)$ pick $δ_{x}$ ", which is a choice function on a subset of $R$ that may be uncountable. Nothing in this library licenses that. Uniting all admissible balls avoids the issue completely: the family is defined by a property, membership of $x$ in $U$ needs only the existence of one admissible $δ$ for that single $x$ , and the reverse inclusion needs only the defining property of whichever pair happens to catch $y$ .
The dual form is not "preimages of closed sets are closed". As with claim 2, the preimage lives in $A$ , and it is relatively closed. The image direction fails outright: a continuous function may carry a closed set to a set that is not closed, which is FALSE: the image of a closed subset of $R$ under a continuous real function is closed.
This is the statement that survives when $R$ is replaced by a metric space or a topological space. The metric version at this point in the reading order is The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement together with Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form, and the agreement of the two vocabularies for $A \subseteq R$ is Dictionary: for $A \subseteq R$ with the metric $d (x, y) = ∣ x - y ∣$ , continuity and uniform continuity of $f : A \to R$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $R$ is compact in the open-cover sense of $R$ exactly when it is a compact metric subspace, later on this page.

Depends on

Used by

Dependency tree · two levels

15 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Continuous function (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 4 (Thm 4.8) (standard reference, not scraped)
J. Lebl, Basic Analysis I, §3.2 (standard reference, not scraped)
D. Ernst, Continuous Real Functions (standard reference, not scraped)
MIT 18.100B lecture notes (standard reference, not scraped)

f:A→Rf:A→R is continuous on AA if and only if the preimage of every open subset of RR is the intersection with AA of an open subset of RR, and dually for closed sets