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Imagine that we have a collection of sets $\cal C$, let $S\notin \cal C$ and $A \in \cal C$.

In algebra, topology, etc... we often see statements like " ----- is the smallest set containing -----", but I'm always not sure how to formalize these kinds of statements. How can I write that $A$ is the smallest set in $\cal C$ containing the set $S$? How can we measure the "length" of these sets in order to say that one is smaller than the other?

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  • $\begingroup$ For sets of discrete elements $|S|$ is the cardinality or "size" of a set—the number of its elements. $\endgroup$
    – David G. Stork
    Nov 9, 2020 at 16:10
  • $\begingroup$ But, for example, If I'm talking about sets in $\Bbb R$, how can I say that some set is the smallest set containing another? All intervals in $\Bbb R$ have the same cardinality @DavidG.Stork $\endgroup$
    – Eduardo Magalhães
    Nov 9, 2020 at 16:11

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In this use we order sets by inclusion, so you would say that $A$ is in $\mathcal C$, contains $S$, and all other sets in $\mathcal C$ that contain $S$ also contain $A$.

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  • $\begingroup$ So, $S \subset B \Rightarrow A \subset B$? $\endgroup$
    – Eduardo Magalhães
    Nov 9, 2020 at 16:12
  • $\begingroup$ Yes, for all sets $B \subset \mathcal C$ $\endgroup$
    – Ross Millikan
    Nov 9, 2020 at 16:19
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The statement usually means the property $$A\in\mathcal C\land A\supseteq S\land (\forall B\in \mathcal C,(A\subseteq B\lor S\nsubseteq B))$$ id est that $A\in\mathcal C$ is a set containing $S$ and, for all other sets $B\in\mathcal C$ which contain $S$, $A\subseteq B$.

Personally, I would not use the word "smallest" when I'm referring to cardinality. I would say "of least cardinality" (or "least size", or "least measure", or "least greatest lower bund of the sine of an odd integer contained in it"...).

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  • $\begingroup$ So is this the same as saying that if $\mathcal {P} = \{U \in \mathcal C : S \subseteq U\}$, then $\bigcap_{U \in \mathcal P} U = A$? $\endgroup$
    – Eduardo Magalhães
    Nov 9, 2020 at 16:16
  • $\begingroup$ @EduardoMagalhães There is one if and only if $\bigcap\mathcal P\in\mathcal C$, in which case $A=\bigcap \mathcal P$. $\endgroup$
    – user239203
    Nov 9, 2020 at 16:18

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