# Equivalent Conditions for Cover by Collection of Subsets

## Theorem

Let $X$ be a set.

Then the following conditions are equivalent for a subset $\mathcal C \subseteq \mathcal P \left({X}\right)$ of the power set of $X$:

$\left({1}\right): \quad$ $\mathcal C$ is a cover for $X$.
$\left({2}\right): \quad$ $\displaystyle X = \bigcup \mathcal C$.
$\left({3}\right): \quad$ $\displaystyle \exists \mathcal S \subseteq \mathcal C: X = \bigcup \mathcal S$.

## Proof

### $(1)$ implies $(2)$

By definition, $\mathcal C$ covers $X$ if and only if $X \subseteq \displaystyle \bigcup \mathcal C$.

By Union of Subsets is Subset, we have that:

$\displaystyle \bigcup \mathcal C \subseteq X$

since $\mathcal C \subseteq \mathcal P \left({X}\right)$.

By definition of set equality, it follows that $X = \displaystyle \bigcup \mathcal C$.

$\Box$

### $(2)$ implies $(3)$

Since $\mathcal C \subseteq \mathcal C$, we can take $\mathcal S = \mathcal C$.

Hence $(3)$ is immediate from $(2)$.

$\Box$

### $(3)$ implies $(1)$

By assumption, $X = \displaystyle \bigcup \mathcal S$ for some $\mathcal S \subseteq \mathcal C$.

By Union is Increasing, we have that:

$\displaystyle \bigcup \mathcal S \subseteq \bigcup \mathcal C$

Hence, $X \subseteq \displaystyle \bigcup \mathcal C$, i.e. $\mathcal C$ is a cover for $X$.

$\blacksquare$
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