Tail of Directed Set is Directed

Theorem

Let $\struct {\Lambda, \preceq}$ be a directed set.

Let $\lambda_0 \in \Lambda$.

Let $\Lambda_0 = \set {\lambda \in \Lambda : \lambda \succeq \lambda_0}$.

Let $\preceq_0$ be the restriction of $\preceq$ to $\Lambda_0$.


Then $\struct {\Lambda_0, \preceq_0}$ is directed.


Proof

Let $\mu_1, \mu_2 \in \Lambda_0$.

We can find $\lambda \in \Lambda$ such that $\mu_1 \preceq \lambda$ and $\mu_2 \preceq \lambda$.

Since $\lambda_0 \preceq \mu_1$ and $\lambda_0 \preceq \mu_2$, we have that $\lambda_0 \preceq \lambda$ by transitivity.

Hence $\lambda \in \Lambda_0$, with $\mu_1 \preceq \lambda$ and $\mu_2 \preceq \lambda$.

Hence $\struct {\Lambda_0, \preceq_0}$ is directed.

$\blacksquare$

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