PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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.

Under choice, a continuous surjection does not increase density or Lindelöf degree

Statement

Assume the Axiom of Choice. If $f : X \to Y$ is continuous and onto, then $d (Y) \leq d (X)$ and $L (Y) \leq L (X)$ .

Facts & Assumptions

Given: The Axiom of Choice and a continuous surjection $f : X \to Y$ (The Axiom of Choice).

[L1]

A continuous map has open preimages of open sets (Continuity of a map of topological spaces at a point and globally, 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 $f (\overline{A}) \subseteq \overline{f (A)}$ ).

[L2]

The least dense-set cardinality $d (Z)$ and the least cardinal bounding subcovers $L (Z)$ exist for every topological space $Z$ (Under choice, $d (X)$ is a well-defined cardinal, Under choice, $L (X)$ is a well-defined cardinal).

Proof

technique · direct

1.1

If $D \subseteq X$ is dense, then $f [D]$ is dense in $Y$ : a nonempty open $V \subseteq Y$ has nonempty open preimage by surjectivity and [L1], so that preimage meets $D$ and $V$ meets $f [D]$ .

1.2

For an open cover $U$ of $Y$ , the family ${f^{- 1} [U] : U \in U}$ is an open cover of $X$ ; a subfamily indexed by at most $L (X)$ members covers $X$ , and the corresponding members of $U$ cover $Y$ by surjectivity.

L1L2

2.1

Taking $D$ with $∣ D ∣ = d (X)$ , step 1.1 gives a dense subset of $Y$ of cardinality at most $d (X)$ ; hence $d (Y) \leq d (X)$ by [L2].

step 1.1L2

3.1

Step 1.2 gives $L (Y) \leq L (X)$ by [L2], and together with step 2.1 this proves both inequalities.

step 2.1step 1.2L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

21 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

D. H. Fremlin, Measure Theory, Chapter 5A (standard reference, not scraped)