LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

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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.

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If $(U_{r})_{r \in D}$ are open with $\overline{U_{r}} \subseteq U_{s}$ whenever $r < s$ and $U_{1} = X$ , then $x \mapsto \inf {r \in D : x \in U_{r}}$ is a continuous map $X \to [0, 1]$ , and no choice principle is used

Statement

Let $(X, T)$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) and let $D$ be the dyadic rationals of $[0, 1]$ (The dyadic rationals of $[0, 1]$ , their finite levels $D_{n}$ , and their density in $[0, 1]$ ). Let $(U_{r})_{r \in D}$ be a family of open subsets of $X$ such that

$\overline{U_{r}} \subseteq U_{s} whenever r < s in D, and U_{1} = X .$

Then

$f (x) : = \inf ({r \in D : x \in U_{r}} \cup {1})$

defines a map $f : X \to [0, 1]$ , and $f$ is continuous.

No choice principle is used in passing from the family $(U_{r})_{r \in D}$ to $f$ . Every existential instantiation in the proof below is a single choice from a single nonempty set of reals, never a simultaneous selection over an infinite index; where the family $(U_{r})_{r \in D}$ itself is later built by a choice-consuming recursion, that cost is incurred in producing the family, not in this lemma.

Facts & Assumptions

Given: A topological space $(X, T)$ , the dyadic rationals $D$ of $[0, 1]$ , and a family $(U_{r})_{r \in D}$ of open subsets of $X$ with $\overline{U_{r}} \subseteq U_{s}$ whenever $r < s$ in $D$ , and $U_{1} = X$ .

[A1]

Shrinking hypothesis: for $r < s$ in $D$ , $\overline{U_{r}} \subseteq U_{s}$ .

[A2]

$U_{1} = X$ .

[L1]

$D \subseteq [0, 1]$ , and $D$ is dense in $[0, 1]$ : for every $x \in [0, 1]$ and every real $ε > 0$ there is $r \in D$ with $∣ x - r ∣ < ε$ (The dyadic rationals of $[0, 1]$ , their finite levels $D_{n}$ , and their density in $[0, 1]$ ).

[L2]

Infimum: a nonempty $S \subseteq R$ bounded below has $\inf S \in R$ (Every nonempty set bounded below has an infimum), which is a lower bound of $S$ and is $\geq$ every other lower bound of $S$ (Greatest lower bound (infimum)). Consequently, for a real $a$ : (i) if some $s \in S$ has $s < a$ then $\inf S \leq s < a$ ; (ii) if $\inf S < a$ then some $s \in S$ has $s < a$ , since otherwise $a$ would be a lower bound of $S$ forcing $a \leq \inf S$ ; (iii) if $r < \inf S$ then $r < s$ for every $s \in S$ , since $\inf S$ is itself a lower bound of $S$ .

[L3]

The traces on $[0, 1]$ of the order rays, $[0, a) : = (- \infty, a) \cap [0, 1]$ and $(a, 1] : = (a, \infty) \cap [0, 1]$ for $a \in R$ , form a subbasis for the subspace topology of $[0, 1]$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace). Indeed each ray $(- \infty, a)$ , $(a, \infty)$ is a union of bounded open intervals of $R$ , hence open in the usual topology (The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), so the topology the rays generate is contained in the usual topology of $R$ ; and every bounded open interval $(a, b)$ is the intersection $(a, \infty) \cap (- \infty, b)$ of two rays, so by A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis the finite intersections of the rays already form a basis containing every bounded open interval, hence the rays generate at least the usual topology. The two inclusions make the rays a subbasis for the usual topology of $R$ (Basis and subbasis for a topology, and the topology generated by a family of sets), and tracing a subbasis onto a subspace gives a subbasis for the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

[L4]

Checking preimages of a fixed subbasis suffices for continuity (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$ , clause (d) $\Leftrightarrow$ (a)).

[L5]

$A \subseteq \overline{A}$ for every $A \subseteq X$ , and $X ∖ \overline{A}$ is open (Interior, closure, boundary, exterior, derived set and isolated point in a topological space, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

Proof

technique · direct

1.1

For $x \in X$ put $S_{x} : = {r \in D : x \in U_{r}} \cup {1}$ ; then $S_{x}$ is a nonempty subset of $[0, 1]$ , since $1 \in S_{x}$ and $D \subseteq [0, 1]$ by [L1], so $S_{x}$ is bounded below by $0$ and above by $1$ .

givenL1

2.1

By step 1.1 and [L2], $\inf S_{x}$ exists in $R$ for every $x \in X$ , lies in $[0, 1]$ since $0$ is a lower bound of $S_{x}$ and $\inf S_{x} \leq 1$ as $1 \in S_{x}$ ; define $f : X \to [0, 1]$ by $f (x) : = \inf S_{x}$ .

step 1.1L2construct

3.1

For every $x \in X$ and real $a$ with $0 < a \leq 1$ : if there is $r \in D$ with $r < a$ and $x \in U_{r}$ , then $r \in S_{x}$ , so $f (x) \leq r < a$ by L2.

step 2.1L2

3.2

For every $x \in X$ and real $a$ with $0 < a \leq 1$ : if $f (x) < a$ , then by L2 some $s \in S_{x}$ has $s < a \leq 1$ , so $s \neq 1$ , hence $s \in D$ and $x \in U_{s}$ , with $s < a$ .

step 2.1L2

3.3

For real $a \leq 0$ : ${x : f (x) < a} = \emptyset$ , since $f (x) \geq 0$ always by step 2.1; for real $a > 1$ : ${x : f (x) < a} = X$ , since $f (x) \leq 1 < a$ always by step 2.1; both open.

step 2.1

3.4

For every $x \in X$ and real $a$ with $0 \leq a < 1$ : if $f (x) > a$ , put $x_{0} : = (a + f (x)) / 2 \in (a, f (x)) \subseteq [0, 1]$ and $δ : = (f (x) - a) / 2 > 0$ ; by [L1] fix $r_{1} \in D$ with $∣ x_{0} - r_{1} ∣ < δ$ , so $r_{1} \in (a, f (x))$ .

step 2.1L1choose

3.5

For every $x \in X$ , real $a$ with $0 \leq a < 1$ , and $r \in D$ with $r > a$ : if $x \notin \overline{U_{r}}$ , then $r$ is a lower bound of $S_{x}$ . Indeed, for $s = 1 \in S_{x}$ : $r \leq 1 = s$ , since $r \in D \subseteq [0, 1]$ by [L1]; for $s \in D$ with $x \in U_{s}$ : if $s < r$ then [A1] gives $\overline{U_{s}} \subseteq U_{r}$ , so $x \in U_{s} \subseteq \overline{U_{s}} \subseteq U_{r} \subseteq \overline{U_{r}}$ by [L5], contradicting $x \notin \overline{U_{r}}$ , so $s \geq r$ .

step 2.1A1L1L5

3.6

For real $a < 0$ : ${x : f (x) > a} = X$ , since $f (x) \geq 0 > a$ always by step 2.1; for real $a \geq 1$ : ${x : f (x) > a} = \emptyset$ , since $f (x) \leq 1 \leq a$ always.

step 2.1

4.1

For real $a$ with $0 < a \leq 1$ : ${x \in X : f (x) < a} = ⋃_{r \in D, r < a} U_{r}$ , by steps 3.1 and 3.2 giving the two inclusions; a union of open sets, hence open.

step 3.1step 3.2

4.2

Continuing under the hypothesis of step 3.4: since $a < r_{1}$ , by [L1] fix $r_{2} \in D$ with $∣ (a + r_{1}) / 2 - r_{2} ∣ < (r_{1} - a) / 2$ , so $r_{2} \in (a, r_{1})$ .

step 3.4L1choose

4.3

Continuing under the hypothesis of step 3.5: since $r$ is a lower bound of $S_{x}$ by step 3.5, [L2] gives $r \leq \inf S_{x} = f (x)$ ; combined with $r > a$ , $f (x) > a$ .

step 3.5step 2.1L2

5.1

Continuing, with $r_{1}, r_{2}$ as in step 4.2: since $r_{1} < f (x) = \inf S_{x}$ , L2 gives $r_{1} < s$ for every $s \in S_{x}$ ; in particular $r_{1} \neq 1$ , since $r_{1} < f (x) \leq 1$ , so $r_{1} \notin S_{x}$ forces $x \notin U_{r_{1}}$ , as otherwise $r_{1}$ itself would lie in $S_{x}$ .

step 3.4step 2.1L2

6.1

Continuing: since $r_{2} < r_{1}$ in $D$ , [A1] gives $\overline{U_{r_{2}}} \subseteq U_{r_{1}}$ ; if $x \in \overline{U_{r_{2}}}$ then $x \in U_{r_{1}}$ , contradicting step 5.1; so $x \notin \overline{U_{r_{2}}}$ , and $r_{2} > a$ .

step 4.2step 5.1A1

7.1

For real $a$ with $0 \leq a < 1$ : ${x \in X : f (x) > a} = ⋃_{r \in D, r > a} (X ∖ \overline{U_{r}})$ . A point of the left side has, by steps 3.4 and 6.1, some $r = r_{2} \in D$ with $r > a$ and $x \in X ∖ \overline{U_{r}}$ ; a point $x$ of the right side lies in $X ∖ \overline{U_{r}}$ for some such $r$ , hence $x \notin \overline{U_{r}}$ , giving $f (x) > a$ by step 4.3. Each $X ∖ \overline{U_{r}}$ is open by [L5], so the union is open.

step 6.1step 4.3L5

8.1

By [L3], the sets $[0, a)$ and $(a, 1]$ , $a \in R$ , form a subbasis for the subspace topology of $[0, 1]$ ; and $f^{- 1} ([0, a)) = {x : f (x) < a}$ , $f^{- 1} ((a, 1]) = {x : f (x) > a}$ are open in $X$ for every real $a$ , by steps 4.1, 3.3, 7.1 and 3.6.

step 4.1step 3.3step 7.1step 3.6L3

9.1

By [L4], since the preimage of every member of that subbasis is open, $f$ is continuous as a map $X \to [0, 1]$ ; together with step 2.1 this proves the statement.

step 8.1step 2.1L4∎

Remarks

Why the $\cup {1}$ in the definition of $f$ . It is what makes $S_{x}$ manifestly nonempty and bounded above by $1$ without first invoking $U_{1} = X$ ; under that hypothesis $1 \in D$ already forces $1 \in S_{x}$ on its own (since every $x \in X = U_{1}$ ), so the union is not strictly necessary here, but it keeps well-definedness visible from the definition of $S_{x}$ alone, which matters when this lemma is quoted with a family for which the reader has not yet checked $U_{1} = X$ line by line.
Where density of $D$ is spent, and only there. The forward half of the " $f (x) > a$ " characterisation (steps 3.4, 4.2, 5.1 and 6.1) is the only place two dyadic points strictly between $a$ and $f (x)$ are extracted; the " $f (x) < a$ " half needs no density at all, only the defining property of an infimum. This asymmetry mirrors the asymmetry of the hypothesis: the shrinking clause $\overline{U_{r}} \subseteq U_{s}$ supplies a closed set inside an open one, and closing the resulting gap is what the second dyadic point is for.
The subbasis fact (Fact [L3]) has no home elsewhere in this library at this point in the reading order: no earlier item states that the order rays generate the usual topology of $R$ , so it is derived here from the basis criterion rather than cited as a single fact.

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Sources

Urysohn's lemma (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §33 (standard reference, not scraped)
Bernard Badzioch, MTH 427 Topology I, Notes 10 (standard reference, not scraped)

If (Ur)r∈D(Ur​)r∈D​ are open with Ur‾⊆UsUr​​⊆Us​ whenever r<sr<s and U1=XU1​=X, then x↦inf⁡{r∈D:x∈Ur}x↦inf{r∈D:x∈Ur​} is a continuous map X→[0,1]X→[0,1], and no choice principle is used