description/proof of that for continuous map between topological spaces, boundary of preimage of subset is contained in but not necessarily equal to preimage of boundary 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 boundary of subset of topological space.
- The reader admits the proposition that for any topological space and any subset, the space is the disjoint union of the interior of the subset, the boundary of the subset, and the complement of the closure of the subset.
- The reader admits the proposition that for any map, the preimage of any subset minus any subset is the preimage of the 1st subset minus the preimage of the 2nd subset.
- The reader admits the proposition that for any continuous map between any topological spaces, the closure of the map preimage of any subset of the codomain is contained in but not necessarily equal to the preimage of the closure of the subset.
- The reader admits the proposition that for any continuous map between any topological spaces, the preimage of the interior of any subset of the codomain is contained in but not necessarily equal to the interior of the preimage of the subset.
Target Context
- The reader will have a description and a proof of the proposition that for any continuous map between any topological spaces, the boundary of the preimage of any subset of the codomain is contained in but not necessarily equal to the preimage of the boundary 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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(S_2\): \(\subseteq T_2\)
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Statements:
\(Bou (f^{-1} (S_2)) \subseteq f^{-1} (Bou (S_2))\)
\(\land\)
Not necessarily "\(Bou (f^{-1} (S_2)) = f^{-1} (Bou (S_2))\)"
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2: Proof
Whole Strategy: Step 1: see that \(Bou (f^{-1} (S_2)) = \overline{f^{-1} (S_2)} \setminus Int (f^{-1} (S_2))\) and \(Bou (S_2) = \overline{S_2} \setminus Int (S_2)\); Step 2: apply the proposition that for any map, the preimage of any subset minus any subset is the preimage of the 1st subset minus the preimage of the 2nd subset, the proposition that for any continuous map between any topological spaces, the closure of the map preimage of any subset of the codomain is contained in but not necessarily equal to the preimage of the closure of the subset, and the proposition that for any continuous map between any topological spaces, the preimage of the interior of any subset of the codomain is contained in but not necessarily equal to the interior of the preimage of the subset; Step 3: see an example that "\(Bou (f^{-1} (S_2)) = f^{-1} (Bou (S_2))\)" does not hold.
Step 1:
\(Bou (f^{-1} (S_2)) = \overline{f^{-1} (S_2)} \setminus Int (f^{-1} (S_2))\), by the proposition that for any topological space and any subset, the space is the disjoint union of the interior of the subset, the boundary of the subset, and the complement of the closure of the subset.
\(Bou (S_2) = \overline{S_2} \setminus Int (S_2)\), likewise.
Step 2:
\(f^{-1} (Bou (S_2)) = f^{-1} (\overline{S_2} \setminus Int (S_2)) = f^{-1} (\overline{S_2}) \setminus f^{-1} (Int (S_2))\), by the proposition that for any map, the preimage of any subset minus any subset is the preimage of the 1st subset minus the preimage of the 2nd subset.
\(\overline{f^{-1} (S_2)} \subseteq f^{-1} (\overline{S_2})\), by the proposition that for any continuous map between any topological spaces, the closure of the map preimage of any subset of the codomain is contained in but not necessarily equal to the preimage of the closure of the subset.
\(f^{-1} (Int (S_2)) \subseteq Int (f^{-1} (S_2))\), by the proposition that for any continuous map between any topological spaces, the preimage of the interior of any subset of the codomain is contained in but not necessarily equal to the interior of the preimage of the subset.
So, \(\overline{f^{-1} (S_2)} \setminus Int (f^{-1} (S_2)) \subseteq f^{-1} (\overline{S_2}) \setminus f^{-1} (Int (S_2))\).
But the left hand side is \(Bou (f^{-1} (S_2))\), by Step 1, while the right hand side is \(f^{-1} (Bou (S_2))\), as has been seen above.
So, \(Bou (f^{-1} (S_2)) \subseteq f^{-1} (Bou (S_2))\).
Step 3:
Let us see an example that "\(Bou (f^{-1} (S_2)) = f^{-1} (Bou (S_2))\)" does not hold.
Let \(T_1 := \mathbb{R}\) and \(T_2 := \mathbb{R}\) as the Euclidean topological spaces, \(f: T_1 \to T_2, t_1 \mapsto {t_1}^2\), and \(S_2 = [0, 1]\).
\(f\) is well known to be continuous.
\(f^{-1} (S_2) = [-1, 1]\) and \(Bou (f^{-1} (S_2)) = \{-1, 1\}\).
But \(Bou (S_2) = \{0, 1\}\) and \(f^{-1} (Bou (S_2)) = \{-1, 0, 1\}\).
So, \(Bou (f^{-1} (S_2)) = \{-1, 1\} \neq \{-1, 0, 1\} = f^{-1} (Bou (S_2))\).
