Subcover is Refinement of Cover

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Theorem

Let $S$ be a set.

Let $\mathcal C$ be a cover for $S$.

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


Then $\mathcal V$ is a refinement of $\mathcal C$.


Corollary

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

From the definition of subcover:

$\mathcal V \subseteq \mathcal C$

That is, every element of $\mathcal V$ is an element of $\mathcal C$.

From the definition of subset, every element of $\mathcal V$ is the subset of some element of $\mathcal C$.

This is precisely the definition of refinement.

$\blacksquare$