Refinement of a Refinement is Refinement of Cover

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Theorem

Let $S$ be a set.

Let $\UU = \set {U_\alpha}$, $\VV = \set {V_\beta}$ and $\WW = \set {W_\gamma}$ be covers of $S$.

Let $\VV$ be a refinement of $\UU$.

Let $\WW$ be a refinement of $\VV$.

Then:

$\WW$ is a refinement of $\UU$

Proof

Let $W \in \WW$.

By definition of refinement:

$\exists V \in \VV : W \subseteq V$

Similarly:

$\exists U \in \UU : V \subseteq U$

From Subset Relation is Transitive:

$W \subseteq U$

Since $W$ was arbitrary:

$\forall W \in \WW : \exists U \in \UU : W \subseteq U$

It follows that $\WW$ is a refinement of $\UU$ by definition.

$\blacksquare$