description/proof of that for continuous injection between topological spaces, image of boundary of subset is contained in boundary of image of subset
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of continuous, topological spaces map.
- The reader knows a definition of injection.
- The reader knows a definition of boundary of subset of topological space.
- The reader admits the proposition that for any continuous map between any topological spaces, the image of the closure of any subset of the domain is contained in the closure of the image of the subset.
Target Context
- The reader will have a description and a proof of the proposition that for any continuous injection between any topological spaces, the image of the boundary of any subset is contained in the boundary of the image of 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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the continuous injections }\}\)
\(S_1\): \(\subseteq T_1\)
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Statements:
\(f (Bou (S_1)) \subseteq Bou (f (S_1))\), where \(Bou (\bullet)\) denotes the boundary
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2: Note
When \(f\) is not injective, \(f (Bou (S_1)) \subseteq Bou (f (S_1))\) does not necessarily hold.
For example, let \(T_1 = T_2 = \mathbb{R}\), \(S_1 = (\pi / 4, \pi)\), and \(f = sin\), then, \(Bou (S_1) = \{\pi / 4, \pi\}\), \(f (S_1) = (0, 1]\), \(Bou (f (S_1)) = \{0, 1\}\), and \(f (\pi / 4) \notin Bou (f (S_1))\).
3: Proof
Whole Strategy: Step 1: see that \(f (\overline{S_1}) \subseteq \overline{f (S_1)}\) and \(f (\overline{T_1 \setminus S_1}) \subseteq \overline{T_2 \setminus f (S_1)}\); Step 2: see that for each \(p \in Bou (S_1)\), \(f (p) \in Bou (f (S_1))\).
Step 1:
\(Bou (S_1) = \overline{S_1} \cap \overline{T_1 \setminus S_1}\) and \(Bou (f (S_1)) = \overline{f (S_1)} \cap \overline{T_2 \setminus f (S_1)}\) by definition.
\(f (\overline{S_1}) \subseteq \overline{f (S_1)}\), by the proposition that for any continuous map between any topological spaces, the image of the closure of any subset of the domain is contained in the closure of the image of the subset.
\(f (\overline{T_1 \setminus S_1}) \subseteq \overline{f (T_1 \setminus S_1)}\), likewise.
But \(f (T_1 \setminus S_1) \subseteq T_2 \setminus f (S_1)\), because for each \(p \in T_1 \setminus S_1\), \(f (p) \notin f (S_1)\), because \(f\) is injective, so, \(f (p) \in T_2 \setminus f (S_1)\).
So, \(\overline{f (T_1 \setminus S_1)} \subseteq \overline{T_2 \setminus f (S_1)}\).
So, \(f (\overline{T_1 \setminus S_1}) \subseteq \overline{f (T_1 \setminus S_1)} \subseteq \overline{T_2 \setminus f (S_1)}\).
Step 2:
Let \(p \in Bou (S_1)\) be any.
As \(p \in \overline{S_1}\), \(f (p) \in \overline{f (S_1)}\), by Step 1.
As \(p \in \overline{T_1 \setminus S_1}\), \(f (p) \in \overline{T_2 \setminus f (S_1)}\), by Step 1.
So, \(f (p) \in \overline{f (S_1)} \cap \overline{T_2 \setminus f (S_1)} = Bou (f (S_1))\).
