description/proof of that for topological space, subspace, and subset of subspace, boundary of subset w.r.t. subspace is contained in boundary of subset w.r.t. superspace
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological subspace.
- The reader knows a definition of boundary of subset of topological space.
- The reader admits the proposition that for any topological space and any subset of any subspace, the closure of the subset on the subspace is contained in the closure of the subset on the base space.
- The reader admits the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space, any subspace, and any subset of the subspace, the boundary of the subset with respect to the subspace is contained in the boundary of the subset with respect to the superspace.
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:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(T\): \(\in \{\text{ the topological subspaces of } T'\}\)
\(S\): \(\subseteq T\)
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Statements:
\(\dot{S}_T \subseteq \dot{S}_{T'}\)
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2: Note
The equality does not necessarily hold.
For example, let \(T' = \mathbb{R}\) with the Euclidean topology, \(T = (-1, 1)\), and \(S = (-1, 1)\), then, \(\dot{S}_T = \emptyset \subset [-1, 1] = \dot{S}_{T'}\).
3: Proof
Whole Strategy: Step 1: see that \(\overline{S}_T \subseteq \overline{S}_{T'}\); Step 2: \(\overline{T \setminus S}_T \subseteq \overline{T' \setminus S}_{T'}\); Step 3: conclude the proposition.
Step 1:
\(\overline{S}_T \subseteq \overline{S}_{T'}\), by the proposition that for any topological space and any subset of any subspace, the closure of the subset on the subspace is contained in the closure of the subset on the base space.
Step 2:
We profusely use the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset hereafter.
Let us see that \(\overline{T \setminus S}_T \subseteq \overline{T' \setminus S}_{T'}\).
Let \(p \in \overline{T \setminus S}_T\) be any.
\(p \in T \setminus S\) or \(p\) is an accumulation point of \(T \setminus S\) on \(T\).
When \(p \in T \setminus S\), \(p \in T' \setminus S\), and so, \(p \in \overline{T' \setminus S}_{T'}\).
When \(p\) is an accumulation point of \(T \setminus S\) on \(T\), let \(U'_p \subseteq T'\) be any open neighborhood of \(p\) on \(T'\), then, \(U'_p \cap T\) is an open neighborhood of \(p\) on \(T\) by the definition of subspace topology, so, \(U'_p \cap T \cap (T \setminus S) \neq \emptyset\), so, \(\emptyset \subset U'_p \cap T \cap (T \setminus S) \subseteq U'_p \cap (T' \setminus S) \neq \emptyset\), which means that \(p\) is an accumulation point of \(T' \setminus S\) on \(T'\), which means that \(p \in \overline{T' \setminus S}_{T'}\).
So, \(\overline{T \setminus S}_T \subseteq \overline{T' \setminus S}_{T'}\).
Step 3:
As \(\dot{S}_T = \overline{S}_T \cap \overline{T \setminus S}_T\) and \(\dot{S}_{T'} = \overline{S}_{T'} \cap \overline{T' \setminus S}_{T'}\), by Step 1 and Step 2, \(\dot{S}_T = \overline{S}_T \cap \overline{T \setminus S}_T \subseteq \overline{S}_{T'} \cap \overline{T' \setminus S}_{T'} = \dot{S}_{T'}\).
