description/proof of that for topological space, subset is closed iff closure of subset is subset
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 for any topological space, any subset is closed if and only if the closure of the subset is the subset.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(S\): \(\subseteq T\)
//
Statements:
\(S \in \{\text{ the closed subsets }\}\)
\(\iff\)
\(\overline{S} = S\)
//
2: Proof
Whole Strategy: Step 1: suppose that \(S\) is closed; Step 2: see that \(\overline{S} = S\); Step 3: suppose that \(\overline{S} = S\); Step 4: see that \(S\) is closed.
Step 1:
Let us suppose that \(S\) is closed.
Step 2:
\(\overline{S} = \cap \{C \in \{\text{ the closed subsets of } T\} \vert S \subseteq C\}\), by the definition of closure of subset.
\(\overline{S} \subseteq S\), because \(S \in \{C \in \{\text{ the closed subsets of } T\} \vert S \subseteq C\}\).
\(S \subseteq \overline{S}\), because for each \(s \in S\), \(s \in S \subseteq C\) for each \(C \in \{C \in \{\text{ the closed subsets of } T\} \vert S \subseteq C\}\), so, \(s \in \cap \{C \in \{\text{ the closed subsets of } T\} \vert S \subseteq C\} = \overline{S}\).
So, \(\overline{S} = S\).
Step 3:
Let us suppose that \(\overline{S} = S\).
Step 4:
\(\overline{S} = \cap \{C \in \{\text{ the closed subsets of } T\} \vert S \subseteq C\}\), by the definition of closure of subset.
\(\overline{S}\) is closed, by 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.
So, \(S = \overline{S}\) is closed.
