I "discovered" a few minutes ago that every poset can be embedded into a meet-semilattice.

Let $\mathfrak{A}$ be a poset. Then it is embedded into the meet-semilattice generated by sets $\{ x \in \mathfrak{A} \mid x \le a \}$ where $a$ ranges through $\mathfrak{A}$.

I'm sure I am not the first person who discovered this. Which book could you suggest to read about such things?

  • 3
    This is a special case of the Yoneda embedding, which you can read about in any category theory textbook. – Zhen Lin Aug 11 '12 at 18:13
  • 3
    @ZhenLin why don't you post this as an answer. – user2468 Aug 11 '12 at 18:17
  • 2
    Section 3.4 of Stanley's Enumerative Combinatorics Vol 1 might be helpful. It focusses on finite posets, but gives some nicer results. I believe your observation is implicit in that section. – Jack Schmidt Aug 11 '12 at 18:22
  • 2
    I don't understand the downvote. – anon Aug 11 '12 at 18:24
  • 4
    @JD It isn't an answer – just a remark. The question deserves an answer phrased in purely order-theoretic terms. – Zhen Lin Aug 11 '12 at 18:25

This result is mentioned for example as theorem 1.1 in chapter 1 of J.B. Nation's "Revised Notes on Lattice Theory". See also theorem 2.2 in chapter 2. One advantage of Nation's text is that it is freely available.

Simply searching for every poset embedded in Google Books returns some reasonably looking references. (Of course, you can try some other similar search queries.)

For example Theorem 1.11 in the book Steven Roman: Lattices and Ordered Sets uses precisely the embedding you suggested to show that every poset $P$ can be order embedded in a powerset $\mathscr P(P)$.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.