# How does one show that if $B \subset A$ (where $A$ is a directed set) is cofinal in $(A, \leq)$ then $(B, \leq)$ is a directed set?

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 $$B$$.

• 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? – Arturo Magidin Mar 20 at 2:17
• (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.) – Arturo Magidin Mar 20 at 2:18
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.