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First two axioms of directed sets readily follow for $(B,\leq)$ by the virtue of being a subset of $(A,\leq)$ but I don't see how the third one follows.

Directed set $(A,\leq)$ is a set with the order relation $\leq$, where the order relation $\leq$ is reflexive, transitive and that every pair of elements has an upper bound in $(A,\leq)$.

By "$B$ is cofinal in $A$.", I mean, that each element of $A$ is bounded above by some element of $B$.

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  • $\begingroup$ What assumptions do you have on $A$? I mean, you could take a set $A$ that is not directed, and $B=A$ would be cofinal in $A$ and not directed, so you must have some hypotheses on $A$. What are they? $\endgroup$ – Arturo Magidin Mar 20 at 2:17
  • $\begingroup$ (Also, not everyone defines things identically; I don’t know what the “axioms of directed sets” you have, or how you order them. Unless you expect us to read your mind (and the government gets really annoyed when I do that without a warrant), it’s impossible to know what you are asking.) $\endgroup$ – Arturo Magidin Mar 20 at 2:18
  • $\begingroup$ ok wait I'll add more details, sorry about this $\endgroup$ – hh11 apples Mar 20 at 2:19
  • $\begingroup$ If you are going to say “reflexive and transitive” explicitly, surely you are missing anti-symmetry. $\endgroup$ – Arturo Magidin Mar 20 at 2:25
  • $\begingroup$ Anti-symmetry is not an axiom that I am using $\endgroup$ – hh11 apples Mar 20 at 2:27
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Since $\leq$ is a partial order on $A$, it is a partial order, by restriction on any subset; this is a property of partial orders, directed or not.

Let $b_1,b_2\in B$. Since $A$ is directed, there exists $a\in A$ such that $b_1\leq a$ and $b_2\leq a$. Since $B$ is cofinal in $A$, there exists $b\in B$ such that $a\leq b$. Thus, there exists $b\in B$ such that $b_1\leq b$ and $b_2\leq b$. Hence, $B$ is directed.

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  • $\begingroup$ Oh god, thanks for your help, now that I see it, its not difficult. $\endgroup$ – hh11 apples Mar 20 at 2:29

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