# Size of Set Equal to 1? $|U \cap \{s,t\}| = 1$

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.

• $|A|$ 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. – Mauro ALLEGRANZA Apr 19 at 12:53
• The vertical bars mean "number of elements", so there's exactly one of $s$ or $t$ in the intersection. – dbx Apr 19 at 12:53
• 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. – Ethan Bolker Apr 19 at 12:54
• 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. – Mauro ALLEGRANZA Apr 19 at 12:55
• The condition could also be written as $s\in U\leftrightarrow t\notin U$ – Hagen von Eitzen Apr 19 at 12:57

$$|\cdot|$$ means cardinality, so the condition is that exactly one of $$s,t$$ is an element of $$U$$