Subcover is Refinement of Cover/Corollary

Corollary to Subcover is Refinement of Cover

Let $T = \struct {X, \tau}$ be a topological space.

Let $\UU$ be an open cover for $S$.

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

Then $\VV$ is an open refinement of $\UU$.

Proof

As all the elements of $\UU$ are open, all the elements of $\VV$ are likewise open.

Hence the result from definition of open refinement.

$\blacksquare$