# Did Dedekind prove this lemma about posets (or an equivalent)?

Here's an easy lemma:

Any poset $(S, \preceq)$ is order-isomorphic to a subset of the powerset $\mathcal{P}(S)$ ordered by set-inclusion.

I seem to recall having seen this attributed to Dedekind.

Am I right that Dedekind proves this little result? And if so, where does he do it?

• I'm not quite sure about the relevance of this question to the history of mathematics. See this post I made – Danu Nov 17 '14 at 21:32

I think that the source is Richard Dedekind's Was sind und was sollen die Zahlen ? (Vieweg: Braunschweig, 1888).

I quote from the English translation : THE NATURE AND MEANING OF NUMBERS, (The Open Court Publishing Co., 1901) :

98. Definition [page 37]. If $$n$$ is any number, then will we denote by $$Z_n$$ the system [set] of all numbers that are not greater than $$n$$ [...].

106. Theorem [page 38]. If $$m < n$$, then is $$Z_m$$ proper part of $$Z_n$$ and conversely.

I've found it through :

• Gregory Moore, Zermelo's Axiom of Choice : Its Origin Development and Influence (1982), page 26.

We can easily generalise it to a poset $$M$$ whatever, defining the mapping :

$$\pi : a \to M_a$$, for any $$a \in M.$$

where $$M_a = \{ x \in M : x \le a \}$$

• I suspect you are right that (106) in his truly great paper is the nearest Dedekind gets. But is that really enough to warrant attributing Dedekind the generalization to [what we now call] any poset? Perhaps not ...? – Peter Smith Nov 18 '14 at 20:08
• @PeterSmith - I agree; with insight, it is very easy to "generalize" it --- now that we have one hundred years of development of set theory and abstract algebra (due also to Dedekind). – Mauro ALLEGRANZA Nov 18 '14 at 20:20