Subcover is Refinement of Cover/Corollary

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Corollary to Subcover is Refinement of Cover

Let $T = \left({X, \tau}\right)$ be a topological space.


Let $\mathcal U$ be an open cover for $S$.

Let $\mathcal V$ be a subcover of $\mathcal U$.


Then $\mathcal V$ is an open refinement of $\mathcal U$.


Proof

As all the elements of $\mathcal U$ are open, all the elements of $\mathcal V$ are likewise open.

Hence the result from definition of open refinement.

$\blacksquare$