description/proof of that for nonempty topological space, complement of open dense subset is not 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 nonempty topological space, the complement of any open dense subset is not 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:
\(T\): \(\in \{\text{ the nonempty topological spaces }\}\)
\(U\): \(\in \{\text{ the open dense subsets of } T\}\)
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Statements:
\(T \setminus U \in \{\text{ the non-dense subsets of } T\}\)
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2: Note
When \(T = \emptyset\), \(U = \emptyset\), which is indeed open dense, and \(T \setminus S = \emptyset\), also which is dense.
\(U\) needs to be open: for \(T = \mathbb{R}\), as the Euclidean topological space, and \(S = \mathbb{Q}\), \(S\) is non-open dense (for each \(r \in T \setminus S\), which is any irrational number, each open ball around \(r\) contains a rational number), but \(T \setminus S\) is dense (for each \(r \in S\), which is any rational number, each open ball around \(r\) contains a irrational number).
3: Proof
Whole Strategy: Step 1: suppose that \(T \setminus U\) was dense, and find a contradiction.
Step 1:
Let us suppose that \(T \setminus U\) was dense.
\(\overline{T \setminus U} = T\), by the definition of dense subset of topological space.
\(T \setminus U\) would be nowhere-dense, by the proposition that for any topological space and its any open dense subset, the complement of the subset is nowhere dense, which would mean that \(Int (\overline{T \setminus U}) = \emptyset\).
But that would be \(Int (T) = T = \emptyset\), a contradiction against \(T\)'s being nonempty.
So, \(T \setminus U\) is not dense.
