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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?

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  • 1
    $\begingroup$ I'm not quite sure about the relevance of this question to the history of mathematics. See this post I made $\endgroup$ – Danu Nov 17 '14 at 21:32
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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 $M_a = \{ x \in M : x \le a \}$

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  • $\begingroup$ 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 ...? $\endgroup$ – Peter Smith Nov 18 '14 at 20:08
  • $\begingroup$ @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). $\endgroup$ – Mauro ALLEGRANZA Nov 18 '14 at 20:20

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