LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

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Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
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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 $(X, \mathcal{T})$ 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

$\overline{U_r} \subseteq U_s \quad \text{whenever } r < s \text{ in } D, \qquad \text{and} \qquad U_1 = X.$

Then

$f(x) \;:=\; \inf\big(\{\, r \in D : x \in U_r \,\} \cup \{1\}\big)$

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 $(X,\mathcal{T})$ , 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 $\varepsilon > 0$ there is $r \in D$ with $|x-r| < \varepsilon$ (The dyadic rationals of $[0,1]$ , their finite levels $D_n$ , and their density in $[0,1]$ ).

[L2]

Infimum: a nonempty $S \subseteq \mathbb{R}$ bounded below has $\inf S \in \mathbb{R}$ (Every nonempty set bounded below has an infimum), which is a lower bound of $S$ and is $\ge$ 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 $\inf S \le s < a$ ; (ii) if $\inf S < a$ then some $s \in S$ has $s < a$ , since otherwise $a$ would be a lower bound of $S$ forcing $a \le \inf S$ ; (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 $a \in \mathbb{R}$ , 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 $\mathbb{R}$ , hence open in the usual topology (The absolute value makes $\mathbb{R}$ 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 $\mathbb{R}$ ; 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 $\mathbb{R}$ (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 $X \setminus \overline{A}$ 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 $\mathbb{R}$ for every $x \in X$ , lies in $[0,1]$ since $0$ is a lower bound of $S_x$ and $\inf S_x \le 1$ 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 $0 < a \le 1$ : if there is $r \in D$ with $r < a$ and $x \in U_r$ , then $r \in S_x$ , so $f(x) \le r < a$ by L2.

step 2.1L2

3.2

For every $x \in X$ and real $a$ with $0 < a \le 1$ : if $f(x) < a$ , then by L2 some $s \in S_x$ has $s < a \le 1$ , so $s \ne 1$ , hence $s \in D$ and $x \in U_s$ , with $s < a$ .

step 2.1L2

3.3

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

step 2.1

3.4

For every $x \in X$ and real $a$ with $0 \le a < 1$ : if $f(x) > a$ , put $x_0 := (a+f(x))/2 \in (a,f(x)) \subseteq [0,1]$ and $\delta := (f(x)-a)/2 > 0$ ; by [L1] fix $r_1 \in D$ with $|x_0 - r_1| < \delta$ , so $r_1 \in (a,f(x))$ .

step 2.1L1choose

3.5

For every $x \in X$ , real $a$ with $0 \le a < 1$ , and $r \in D$ with $r > a$ : if $x \notin \overline{U_r}$ , then $r$ is a lower bound of $S_x$ . Indeed, for $s = 1 \in S_x$ : $r \le 1 = s$ , 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 $s \ge r$ .

step 2.1A1L1L5

3.6

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

step 2.1

4.1

For real $a$ with $0 < a \le 1$ : $\{\, x \in X : f(x) < a \,\} = \bigcup_{r \in D,\, r<a} U_r$ , 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 $r \le \inf S_x = f(x)$ ; 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 $r_1 \ne 1$ , since $r_1 < f(x) \le 1$ , 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 $0 \le a < 1$ : $\{\, x \in X : f(x) > a \,\} = \bigcup_{r \in D,\, r>a} \big(X \setminus \overline{U_r}\big)$ . A point of the left side has, by steps 3.4 and 6.1, some $r = r_2 \in D$ with $r > a$ and $x \in X \setminus \overline{U_r}$ ; a point $x$ of the right side lies in $X \setminus \overline{U_r}$ for some such $r$ , hence $x \notin \overline{U_r}$ , giving $f(x) > a$ by step 4.3. Each $X \setminus \overline{U_r}$ is open by [L5], so the union is open.

step 6.1step 4.3L5

8.1

By [L3], the sets $[0,a)$ and $(a,1]$ , $a \in \mathbb{R}$ , 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 $\mathbb{R}$ , 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)