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We show that the submonoid $F^{\\circ}(A^{\\ast}):= F(A^{\\ast})\\setminus \\{\\emptyset\\}$ is free. The MacNeille completion $N(A^{\\ast})$ of $A^{\\ast}$ is a submonoid of $F(A^{\\ast})$. As a corollary, we obtain that the monoid $N^{\\circ}(A^{\\ast}):=N(A^{\\ast})\\setminus \\{\\emptyset\\}$ is free. We give an interpretation of the freeness of $F^{\\circ}(A^{\\ast})$ in the category"},"verification_status":{"content_addressed":true,"pith_receipt":true,"author_attested":false,"weak_author_claims":0,"strong_author_claims":0,"externally_anchored":false,"storage_verified":false,"citation_signatures":0,"replication_records":0,"graph_snapshot":true,"references_resolved":false,"formal_links_present":false},"canonical_record":{"source":{"id":"1705.09750","kind":"arxiv","version":1},"metadata":{"license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","primary_cat":"math.CO","submitted_at":"2017-05-27T01:18:39Z","cross_cats_sorted":[],"title_canon_sha256":"4b2c2f088d919a9ffed7075c53a6416b5bfe3bf73a6650b86e5a0342b1a7c20a","abstract_canon_sha256":"e3b99a83742299c49883a57b3c1ba7a4ff631ead0ad39ae6857d858137bd8623"},"schema_version":"1.0"},"receipt":{"kind":"pith_receipt","key_id":"pith-v1-2026-05","algorithm":"ed25519","signed_at":"2026-05-18T00:43:34.575817Z","signature_b64":"sBqm+tBlBLCxw5GBtgUIyiIN2TvrmS97R9spVDn3ZgBK6X6eqjQioozswwwRZ7Uy6LsL/0Rq8QKjQbeFcRXyAg==","signed_message":"canonical_sha256_bytes","builder_version":"pith-number-builder-2026-05-17-v1","receipt_version":"0.3","canonical_sha256":"1473c9f667a9ed5d35401e14c3ba63f2c37279c322381ffa072a50580e58eed1","last_reissued_at":"2026-05-18T00:43:34.575385Z","signature_status":"signed_v1","first_computed_at":"2026-05-18T00:43:34.575385Z","public_key_fingerprint":"8d4b5ee74e4693bcd1df2446408b0d54"},"graph_snapshot":{"paper":{"title":"Free monoids and generalized metric spaces","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":[],"primary_cat":"math.CO","authors_text":"Ivo Rosenberg, Maurice Pouzet, Mustapha Kabil","submitted_at":"2017-05-27T01:18:39Z","abstract_excerpt":"Let $A$ be an ordered alphabet, $A^{\\ast}$ be the free monoid over $A$ ordered by the Higman ordering, and let $F(A^{\\ast})$ be the set of final segments of $A^{\\ast}$. With the operation of concatenation, this set is a monoid. We show that the submonoid $F^{\\circ}(A^{\\ast}):= F(A^{\\ast})\\setminus \\{\\emptyset\\}$ is free. The MacNeille completion $N(A^{\\ast})$ of $A^{\\ast}$ is a submonoid of $F(A^{\\ast})$. As a corollary, we obtain that the monoid $N^{\\circ}(A^{\\ast}):=N(A^{\\ast})\\setminus \\{\\emptyset\\}$ is free. 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