description/proof of that for topological space, closure of intersection of subsets is contained in intersection of closures of subsets
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of closure of subset of topological space.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space, the closure of the intersection of any subsets is contained in the intersection of the closures of the subsets.
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: Structured Description
Here is the rules of Structured Description.
Entities:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_j \subseteq T \vert j \in J\}\):
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Statements:
\(\overline{\cap_{j \in J} S_j} \subseteq \cap_{j \in J} \overline{S_j}\)
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2: Note
Equality does not necessarily hold even if \(J\) is finite.
For example, let \(T = \mathbb{R}\) with the Euclidean topology and \(\{S_1 = (-1, 0), S_2 = (0, 1)\}\), then, \(\overline{\cap_{j \in J} S_j} = \overline{(-1, 0) \cap (0, 1)} = \overline{\emptyset} = \emptyset \subset \cap_{j \in J} \overline{S_j} = \overline{(-1, 0)} \cap \overline{(0, 1)} = [-1, 0] \cap [0, 1] = \{0\}\).
3: Proof
Whole Strategy: Step 1: see that \(\cap_{j \in J} S_j \subseteq \cap_{j \in J} \overline{S_j}\) and \(\cap_{j \in J} \overline{S_j}\) is closed.
Step 1:
As \(S_j \subseteq \overline{S_j}\), \(\cap_{j \in J} S_j \subseteq \cap_{j \in J} \overline{S_j}\).
\(\cap_{j \in J} \overline{S_j}\) is closed as an intersection of closed subsets.
So, \(\cap_{j \in J} \overline{S_j}\) is a closed subset that contains \(\cap_{j \in J} S_j\).
As \(\overline{\cap_{j \in J} S_j}\) is the intersection of all the closed subsets that contains \(\cap_{j \in J} S_j\), \(\overline{\cap_{j \in J} S_j} \subseteq \cap_{j \in J} \overline{S_j}\).
