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I am currently reading an introduction to domain theory. In the discussion of bases of dcpos it is stated that every basis contains the compact elements of said dcpo.

Both these notions depend on the notion of en element $x$ of a dcpo $(D, \sqsubseteq)$ approximating another element $y$, denoted $x \ll y$, meaning for ever directed subset $A$ such that $y \in \; \downarrow \! \! \left(\bigsqcup \, A\right)$ one has $x \in \; \downarrow \!\! A$.

A compact element, now, means an element approximates itself. Denoting the set of elements approximating some element $x$ by $\downdownarrows \!\! x$ one defines a basis as a subset $B$ of $D$ such that every $B_x = \; \downdownarrows \!\! x \cap B$ is directed and with supremum $x$.

The proof given in the text states that that for some compact element $c \in D$:

$\ldots$ since $c = \bigsqcup B_c$ there must be an element $b \in B_c$ above $c$.

Perhaps I am missing something obvious, but I cannot see why this is the case.

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up vote 2 down vote accepted

You have $c = \bigsqcup B_c$, so $c \ll c \in \downarrow \bigsqcup B_c$, and by definition of approximates, $c \in \downarrow B_c$, and hence there must be some $b \in B_c$ such that $c \le b$.

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Doh! You are quite right, thank you for clearing it up! –  Tilo Wiklund Aug 16 '11 at 20:50
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