The property is that for $f:D\to D$ is a function on some partially order set: $f(x)\geq x$ for every $x\in D$. That is $f≥I$ where $I$ is the identity function on $D$. Of course we can replace $\geq $ by $\leq$. In some settings we know that projections ($p^2=p$) satisfy the property for $\leq$. Thanks in advance.
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$\begingroup$ Thank you very much Todd and everyone for informations you provided. $\endgroup$– hafid1990Aug 25, 2016 at 9:22
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4 Answers
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The property $(\forall x)\,x\leq f(x)$ was called "inflationary" in Peter Freyd's paper "Aspects of topoi". Years later, Yuri Gurevich was working on a paper in theoretical computer science and asked me what I called this property. I told him "inflationary" but mentioned that this was not, as far as I knew, standard terminology. Yuri liked the name and used it, and it seems to have caught on among people working in computer science and finite model theory, so that, for example, "inflationary fixed point" is now standard terminology.
answered Aug 23, 2016 at 20:43
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$\begingroup$ This is the term I've heard too. For example, in the context of the Bourbaki-Witt fixed point theorem, or the (unfortunately) lesser known fixed point theorem of Pataraia. $\endgroup$– Todd Trimble ♦Aug 23, 2016 at 21:18
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$\begingroup$ Thank you very much for you answers. I think if for the other case (when we inverse the order) the function will be called deflationary? $\endgroup$ Aug 23, 2016 at 22:07
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$\begingroup$ Hafid, that certainly sounds reasonable, although I've not heard that used specifically. $\endgroup$– Todd Trimble ♦Aug 23, 2016 at 22:13
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$\begingroup$ I like and use the "inflationary" terminology, but one problem with it is that it suggests $x<f(x)$ rather than $x\leq f(x)$, since otherwise there is no inflation happening. But I don't think we would prefer "nondeflationary"? $\endgroup$ Aug 24, 2016 at 13:24
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There is a related usage in set theory in the context of Fodor's lemma, where we often consider functions $f$ on the ordinals with the property that $f(\alpha)<\alpha$. These are called the regressive functions, and sometimes they are called pressing-down functions; I suppose we could also now call them strictly deflationary. Fodor's lemma asserts that every regressive function on an uncountable regular cardinal is constant on a stationary set.
answered Aug 24, 2016 at 12:57
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I think if we use the term "progressing" it will be reasonable to use "regressing" for the contrary. I have found the term progressing in the following reference: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed., 2010). But, I would be gratful if one can provide me some references in which there is a use of the term "inflationary". Many thanks everyone.
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$\begingroup$ Besides the Peter Freyd reference already given to you in Andreas's answer, see Lattices and Ordered Sets by Steven Roman, p. 263. Not to mention en.wikipedia.org/wiki/Bourbaki%E2%80%93Witt_theorem Now that you have the term, I think googling will turn up lots more. By the way, I think "progressive" is more usual, not "progressing", and similarly "regressive". $\endgroup$– Todd Trimble ♦Aug 24, 2016 at 18:45
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The property $∀x:x≤f(x)$ is also called „extensive“, especially with respect to closure systems or lattices.
In general, a closure system is defined as the image of a closure operator, which is an idempotent and extensional endomorphism of an ordered set.
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$\begingroup$ I think "extensive" is more common. Sometimes also "expansive". $\endgroup$ Aug 24, 2016 at 21:41
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$\begingroup$ You are right. Extensive fits better than extensional. $\endgroup$ Aug 26, 2016 at 12:17