description/proof of that for topological space, finite union of nowhere dense subsets is nowhere dense
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, the union of any finite number of nowhere dense subsets is nowhere dense.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(J\): \(\in \{\text{ the finite index sets }\}\)
\(\{S_j \in \{\text{ the nowhere dense subsets of } T\} \vert j \in J\}\):
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Statements:
\(\cup_{j \in J} S_j \in \{\text{ the nowhere dense subsets of } T\}\)
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2: Proof
Whole Strategy: prove it inductively with respect to \(\vert J \vert\); Step 1: see that it holds when \(\vert J \vert = 1, 2\); Step 2: suppose that it holds when \(1 \le \vert J \vert \le n' - 1\), and see that it holds when \(\vert J \vert = n'\); Step 3: conclude the proposition.
Step 1:
Let \(J = \{1, ..., n\}\) without loss of generality, just for convenience of expressions: \(\vert J \vert = n\).
When \(\vert J \vert = 1\), it holds, because \(\cup_{j \in J} S_j = S_1\) is nowhere dense.
Let us suppose that \(\vert J \vert = 2\).
\(Int (\overline{\cup_{j \in J} S_j}) = Int (\cup_{j \in J} \overline{S_j})\), by the proposition that the closure of the union of any finite number of subsets is the union of the closures of the subsets.
Let us suppose that there was a nonempty open subset, \(U \subseteq T\), such that \(U \subseteq \cup_{j \in J} \overline{S_j}\).
Let \(U_1 := (T \setminus \overline{S_1}) \cap U \subseteq T\), which would be an open subset.
\(U_1 \subseteq \overline{S_2}\), because for each \(p \in U_1\), \(p \notin \overline{S_1}\) and \(p \in U \subseteq \cup_{j \in J} \overline{S_j}\), which would mean that \(p \in \overline{S_2}\).
\(U_1 \neq \emptyset\), because otherwise, \(U \subseteq T \setminus (T \setminus \overline{S_1}) = \overline{S_1}\), which would mean that \(S_1\) was not nowhere dense, a contradiction.
But that would mean that \(S_2\) was not nowhere dense, a contradiction.
So, there is no such \(U\).
That means that \(Int (\overline{\cup_{j \in J} S_j}) = Int (\cup_{j \in J} \overline{S_j}) = \emptyset\).
So, \(\cup_{j \in J} S_j\) is nowhere dense.
Step 2:
Let us suppose that the proposition holds when \(1 \le \vert J \vert \le n' - 1\) where \(n' \le 3\).
Let us suppose that \(n = n'\).
\(\cup_{j \in J} S_j = (\cup_{j \in J \setminus \{n'\}} S_j) \cup S_{n'}\).
By the induction hypothesis, \(\cup_{j \in J \setminus \{n'\}} S_j\) is nowhere dense, and so, \((\cup_{j \in J \setminus \{n'\}} S_j) \cup S_{n'}\) is nowhere dense by Step 1.
Step 3:
So, by the induction principle, for any \(\vert J \vert \in \mathbb{N} \setminus \{0\}\), \(\cup_{j \in J} S_j\) is nowhere dense.
