Closure of Infinite Union may not equal Union of Closures/Proof 1

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Theorem

Let $T$ be a topological space.

Let $I$ be an infinite indexing set.

Let $\family {H_i}_{i \mathop \in I}$ be an indexed family of subsets of a set $S$.


Let $\displaystyle H = \bigcup_{i \mathop \in I} H_i$ be the union of $\family {H_i}_{i \mathop \in I}$.


Then it is not always the case that:

$\displaystyle \bigcup_{i \mathop \in I} \map \cl {H_i} = \map \cl {\bigcup_{i \mathop \in I} H_i}$

where $\map \cl {H_i}$ denotes the closure of $H_i$.


Proof

Proof by Counterexample:

Consider the real number line $\struct {\R, \tau_d}$ under the usual (Euclidean) topology $\tau_d$.

Let:

$H_n \subseteq \R: H_n = \closedint {\dfrac 1 n} 1$ for $n \ge 2$

where $\closedint {\dfrac 1 n} 1$ denotes the closed real interval from $\dfrac 1 n$ to $1$.

From Closed Real Interval is Closed Set, $\closedint {\dfrac 1 n} 1$ is a closed set of $\struct {\R, \tau_d}$.

Then from Set is Closed iff Equals Topological Closure:

$\map \cl {H_n} = H_n$

Also:

$\displaystyle \bigcup_{n \mathop \ge 2} \map \cl {H_n} = \bigcup_{n \mathop \ge 2} H_n = \hointl 0 1$

However:

$\displaystyle \map \cl {\bigcup_{n \mathop \ge 2} H_n} = \closedint 0 1$

So:

$\displaystyle \bigcup_{n \mathop \ge 2} \map \cl {H_n} \ne \map \cl {\bigcup_{n \mathop \ge 2} H_n}$


Sources