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I am not sure what to call this but in the preliminaries for chapter 2 on sets in Alexander Schrijver's Combinatorial Optimization book he states the following:

A set $U$ is said to separate $s$ and $t$ if $s \neq t$ and $|U \cap \{s,t\}| = 1$

The part I am confused about is what operation would result in $|U \cap \{s,t\}| = 1$. I would understand if it was $\emptyset$ or $\{s,t\}$ or a subset of $\{s,t\}$, but I don't understand how it would equate to $1$.

Thanks for the guidance.

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    $\begingroup$ $|A|$ is the cardinality of the set $A$, i.e. the number of its elements. Thus $|A|=1$ means that set $A$ ans only one element. $\endgroup$ – Mauro ALLEGRANZA Apr 19 at 12:53
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    $\begingroup$ The vertical bars mean "number of elements", so there's exactly one of $s$ or $t$ in the intersection. $\endgroup$ – dbx Apr 19 at 12:53
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    $\begingroup$ This says simply that $U$ contains just one of $s$ or $t$, not both. So it "separates" that one from the rest of the underlying set. $\endgroup$ – Ethan Bolker Apr 19 at 12:54
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    $\begingroup$ More specifically, $|U \cap \{ s,t \} |=1$ means that the intersection of sets $U$ and $\{ s,t \}$ has only one element, i.e. that either $s \in U$ or $t \in U$, but not both. $\endgroup$ – Mauro ALLEGRANZA Apr 19 at 12:55
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    $\begingroup$ The condition could also be written as $s\in U\leftrightarrow t\notin U$ $\endgroup$ – Hagen von Eitzen Apr 19 at 12:57
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$|\cdot|$ means cardinality, so the condition is that exactly one of $s,t$ is an element of $U$

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