Closure of Union contains Union of Closures

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Theorem

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


Let $\mathbb H$ be a set of subsets of $S$.

That is, let $\mathbb H \subseteq \powerset S$ where $\powerset S$ denotes the power set of $S$.


Then the union of the closures of the elements of $\mathbb H$ is a subset of the closure of the union of $\mathbb H$:

$\displaystyle \bigcup_{H \mathop \in \mathbb H} \map \cl H \subseteq \map \cl {\bigcup_{H \mathop \in \mathbb H} H}$


Proof

Let $\displaystyle K = \bigcup_{H \mathop \in \mathbb H} \map \cl H$ and $\displaystyle L = \bigcup_{H \mathop \in \mathbb H} H$.

We have:

$\forall H \in \mathbb H: H \subseteq L$

so from Topological Closure of Subset is Subset of Topological Closure:

$\map \cl H \subseteq \map \cl L$

It follows from Union is Smallest Superset: General Result that:

$K \subseteq \map \cl L$

$\blacksquare$


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