2022-07-17

317: Intersection or Finite Union of Closed Sets Is Closed

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A description/proof of that intersection or finite union of closed sets is closed

Topics


About: topological space

The table of contents of this article


Starting Context



Target Context


  • The reader will have a description and a proof of the proposition that the intersection of any possibly uncountable number of closed sets or the union of any finite number of closed sets is closed.

Orientation


There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.


Main Body


1: Description


For any topological space, T, the intersection of any possibly uncountable number of closed sets, \(\cap_\alpha C_\alpha\) where \(C_\alpha \subseteq T\) closed where \({\alpha}\) is any possibly uncountable indices set, or the union of any finite number of closed sets, \(\cup_i C_i\) where \(C_i \subseteq T\) closed where \({i}\) is any finite indices set, is closed.


2: Proof


The complement of the intersection, \(T \setminus \cap_\alpha C_\alpha\), equals \(T \setminus \cap_\alpha (T \setminus U_\alpha)\) where \(U_\alpha := T \setminus C_\alpha\) open. But \(\cap_\alpha (T \setminus U_\alpha) = T \setminus \cup_\alpha U_\alpha\) by the proposition for any set, the intersection of the compliments of any possibly uncountable number of subsets is the complement of the union of the subsets. As \(\cup_\alpha U_\alpha\) is open, \(T \setminus \cup_\alpha U_\alpha\) is closed, so, \(T \setminus \cap_\alpha C_\alpha\) is open.

The complement of the union, \(T \setminus \cup_i C_i\), equals \(T \setminus \cup_i (T \setminus U_i)\) where \(U_i := T \setminus C_i\) open. But \(\cup_i (T \setminus U_i) = T \setminus \cap_i U_i\) by the proposition that for any set, the union of the complements of any subsets is the complement of the intersection of the subsets. As \(\cap_i U_i\) is open, \(T \setminus \cap_i U_i\) is closed, so, \(T \setminus \cup_i C_i\) is open.


References


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