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| Format: | Preprint |
| Published: |
2025
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| Online Access: | https://arxiv.org/abs/2509.20726 |
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| _version_ | 1866912620180668416 |
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| author | Grätzer, G. |
| author_facet | Grätzer, G. |
| contents | Let $A$ be a finite ordered set. Define the ordered set $A^A$ as the set of all maps from $A$ to $A$, ordered pointwise. Let ${}^{A} A$ be the dual of $A^A$.
We prove results in the spirit of Parts~I--III, but now using both $A^A$ and ${}^{A}A$. For example, if \[ \Bigl({}^{{}^{ {}^{ {}^{A}A}A}A}A\Bigr)^{A^{A^{A}}} \] is isomorphic to \[ \Bigl({}^{ {}^{ {}^{ {}^{B}B}B}B}B\Bigr)^{B^{B^{B}}} \] for finite ordered sets $A$ and $B$, then $A$ is isomorphic to $B$. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2509_20726 |
| institution | arXiv |
| publishDate | 2025 |
| record_format | arxiv |
| spellingShingle | Notes on the ordered set $A^A$. Part IV. The dual ${}^{A}\!A$ of $A^A$ for finite ordered sets Grätzer, G. Rings and Algebras 06 Let $A$ be a finite ordered set. Define the ordered set $A^A$ as the set of all maps from $A$ to $A$, ordered pointwise. Let ${}^{A} A$ be the dual of $A^A$. We prove results in the spirit of Parts~I--III, but now using both $A^A$ and ${}^{A}A$. For example, if \[ \Bigl({}^{{}^{ {}^{ {}^{A}A}A}A}A\Bigr)^{A^{A^{A}}} \] is isomorphic to \[ \Bigl({}^{ {}^{ {}^{ {}^{B}B}B}B}B\Bigr)^{B^{B^{B}}} \] for finite ordered sets $A$ and $B$, then $A$ is isomorphic to $B$. |
| title | Notes on the ordered set $A^A$. Part IV. The dual ${}^{A}\!A$ of $A^A$ for finite ordered sets |
| topic | Rings and Algebras 06 |
| url | https://arxiv.org/abs/2509.20726 |