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Let $X$ be an a set and let $\mathcal{C}$ be a collection of subsets of $X$ satisfying the following property:

If $A$ and $A^\prime$ are subsets of $X$ with $A \in \mathcal{C}$ and $A^\prime \subseteq A$, then $A^\prime \in \mathcal{C}$.

I have heard this described variously as "$\mathcal{A}$ is down-closed" and "$\mathcal{A}$ is a down ideal", but neither of these phrases seem very prevalent on the internet. Is there a more common name for this property?

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    Somewhat confusing to use the same letter (albeit in a different font) for the set and its members. Do you mean $A' \in {\cal A}$?2011-12-13
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    This reminds me of dedekend cuts.2011-12-13
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    How about "downward closed"?2011-12-13
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    @RobertIsrael My explanation was incomplete because I didn't define an ambient set. Thank you for pointing that out.2011-12-13
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    @AsafKaragila Isn't closure under unions also required for an ideal?2011-12-13
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    @QuinnCulver: I withdraw my previous comment :-)2011-12-13

4 Answers 4

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As this page indicates, you're missing a condition required for the given collection to be an 'ideal'. As this page indicates, the terms 'downward closed', 'down set', 'lower set', et al. are appropriate in this case.

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The term "downset" is sometimes used. See e.g. Anderson, "Combinatorics of Finite Sets".

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    Google books [link](http://books.google.com/books?id=RjDd4RaqrIwC&q=downset&pg=PA87&redir_esc=y#v=snippet&q=downset).2011-12-14
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In the case where $\mathcal{C}$ consists of finite sets only, $\mathcal{C}$ is an abstract simplicial complex.

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    According to the wikipedia article, this is in the case that $\mathcal C$ consists of finite sets only, right?2011-12-14
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    Good point. I've edited accordingly.2011-12-14
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In "The Probabilistic Method", Alon and Spencer call such a collection "monotone decreasing".

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    Sorry for answering my own question. I found this reference after posting and thought it should be included.2011-12-13
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    As are the terms I listed, 'monotone decreasing' is a general order-theoretic term. Note that 'decreasing' was listed in the link I provided.2011-12-13