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 $\ds H = \bigcup_{i \mathop \in I} H_i$ be the union of $\family {H_i}_{i \mathop \in I}$.
Then it is not always the case that:
- $\ds \bigcup_{i \mathop \in I} \map \cl {H_i} = \map \cl {\bigcup_{i \mathop \in I} H_i}$
where $\map \cl {H_i}$ denotes the closure of $H_i$.
Proof
Consider the real number line $\struct {\R, \tau_d}$ with 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:
- $\ds \bigcup_{n \mathop \ge 2} \map \cl {H_n} = \bigcup_{n \mathop \ge 2} H_n = \hointl 0 1$
However:
- $\ds \map \cl {\bigcup_{n \mathop \ge 2} H_n} = \closedint 0 1$
So:
- $\ds \bigcup_{n \mathop \ge 2} \map \cl {H_n} \ne \map \cl {\bigcup_{n \mathop \ge 2} H_n}$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Example $3.7.19$