LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck 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.
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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Finite pointwise minima of continuous maps to $[0,1]$ are continuous

Statement

If $f_0,\ldots,f_{n-1}:X\to[0,1]$ are continuous, then $x\mapsto\min_{i<n}f_i(x)$ is continuous; for $n=0$ this minimum is the constant-one map.

Facts & Assumptions

Given: A space $X$ and a finite family of continuous maps $f_i:X\to[0,1]$ .

[F1]

The product of two continuous maps is continuous into the product, and a map is continuous exactly when inverse images of open sets are open (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, 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 $f(\overline{A}) \subseteq \overline{f(A)}$ ).

[F2]

In the usual topology of $\mathbb R$ , a set is open exactly when each of its points lies in a bounded open interval contained in it; bases trace to subspaces, so the sets $(a,b)\cap[0,1]$ form a basis for $[0,1]$ (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, clause 3; Basis and subbasis for a topology, and the topology generated by a family of sets; 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).

Proof

technique · induction

1.1

The map $m:[0,1]^2\to[0,1]$ , $m(s,t)=\min(s,t)$ is continuous: for an interval $(a,b)$ , $m^{-1}((a,b))=\{s>a,t>a\}\cap(\{s<b\}\cup\{t<b\})$ , which is open in the product.

1.2

The empty minimum is constant one, hence continuous, and the one-term minimum is $f_0$ .

givenbase

2.1

Assume the minimum $g_n=\min_{i<n}f_i$ is continuous. Then $(g_n,f_n):X\to[0,1]^2$ is continuous and $g_{n+1}=m\circ(g_n,f_n)$ is continuous.

F1step 1.1ih

3.1

Induction gives the claim for every finite family.

step 1.2step 2.1discharge-induction∎

Depends on

Used by

Arbitrary products of completely regular spaces are completely regular Lemma

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 64 results over 11 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. R. Munkres, Topology, 2nd ed., §18 (standard reference, not scraped)
Boundedness Properties in Functionlattices (Canadian Journal of Mathematics) (standard reference, not scraped)

Finite pointwise minima of continuous maps to [0,1][0,1][0,1] are continuous