description/proof of that for topological space and subsets, interior of intersection of subsets is interior of intersection of interiors of subsets, and is contained in intersection of interiors of subsets
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 and any subsets, the interior of the intersection of the subsets is the interior of the intersection of the interiors of the subsets, and is contained in the intersection of the interiors 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:
\(Int (\cap_{j \in J} S_j) = Int (\cap_{j \in J} Int (S_j)) \subseteq \cap_{j \in J} Int (S_j)\)
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2: Proof
Whole Strategy: Step 1: see that \(Int (\cap_{j \in J} S_j) \subseteq Int (\cap_{j \in J} Int (S_j))\); Step 2: see that \(Int (\cap_{j \in J} Int (S_j)) \subseteq Int (\cap_{j \in J} S_j)\); Step 3: see that \(Int (\cap_{j \in J} S_j) \subseteq \cap_{j \in J} Int (S_j)\); Step 4: conclude the proposition.
Step 1:
\(Int (\cap_{j \in J} S_j) \subseteq \cap_{j \in J} S_j\).
So, for each \(j \in J\), \(Int (\cap_{j \in J} S_j) \subseteq \cap_{j \in J} S_j \subseteq S_j\).
So, for each \(j \in J\), \(Int (\cap_{j \in J} S_j) \subseteq Int (S_j)\), because \(Int (\cap_{j \in J} S_j)\) is an open subset contained in \(S_j\) while \(Int (S_j)\) is the union of the open subsets contained in \(S_j\).
So, \(Int (\cap_{j \in J} S_j) \subseteq \cap_{j \in J} Int (S_j)\).
So, \(Int (\cap_{j \in J} S_j) \subseteq Int (\cap_{j \in J} Int (S_j))\), as before.
Step 2:
\(Int (\cap_{j \in J} Int (S_j)) \subseteq \cap_{j \in J} Int (S_j) \subseteq \cap_{j \in J} S_j\).
So, \(Int (\cap_{j \in J} Int (S_j)) \subseteq Int (\cap_{j \in J} S_j)\).
Step 3:
\(Int (\cap_{j \in J} S_j) = \{t \in T \vert \exists U_t \subseteq T \in \{\text{ the open neighborhoods of } t\} (U_t \subseteq \cap_{j \in J} S_j)\}\), by the proposition that for any topological space and any subset, the interior of the subset is the set of the points of the space that have some open neighborhoods contained in the subset.
But \(U_t \subseteq \cap_{j \in J} S_j\) equals that for each \(j \in J\), \(U_t \subseteq S_j\).
So, for each \(t \in Int (\cap_{j \in J} S_j)\), for each \(j \in J\), \(t \in \{t \in T \vert \exists U_t \subseteq T \in \{\text{ the open neighborhoods of } t\} (U_t \subseteq S_j)\} = Int (S_j)\).
So, \(t \in \cap_{j \in J} Int (S_j)\).
So, \(Int (\cap_{j \in J} S_j) \subseteq \cap_{j \in J} Int (S_j)\).
Step 4:
So, \(Int (\cap_{j \in J} S_j) = Int (\cap_{j \in J} Int (S_j)) \subseteq \cap_{j \in J} Int (S_j)\).
