description/proof of that for continuous map between topological spaces, preimage of interior of subset is contained in but not necessarily equal to interior of preimage 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 interior of subset of topological space.
- The reader admits the local criterion for openness.
Target Context
- The reader will have a description and a proof of 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.
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:
\(f^{-1} (Int (S_2)) \subseteq Int (f^{-1} (S_2))\)
\(\land\)
Not necessarily "\(f^{-1} (Int (S_2)) = Int (f^{-1} (S_2))\)"
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2: Proof
Whole Strategy: Step 1: for each \(t_1 \in f^{-1} (Int (S_2))\), see that \(t_1 \in Int (f^{-1} (S_2))\); Step 2: see an example that "\(f^{-1} (Int (S_2)) = Int (f^{-1} (S_2))\)" does not hold.
Step 1:
Let \(t_1 \in f^{-1} (Int (S_2))\) be any.
\(f (t_1) \in Int (S_2)\).
As \(Int (S_2) \subseteq T_2\) is open, there is an open neighborhood of \(f (t_1)\), \(U_{f (t_1)} \subseteq T_2\), such that \(U_{f (t_1)} \subseteq Int (S_2)\), by the local criterion for openness.
There is an open neighborhood of \(t_1\), \(U_{t_1} \subseteq T_1\), such that \(f (U_{t_1}) \subseteq U_{f (t_1)}\), because \(f\) is continuous.
\(f (U_{t_1}) \subseteq U_{f (t_1)} \subseteq Int (S_2) \subseteq S_2\).
So, \(U_{t_1} \subseteq f^{-1} (S_2)\).
That means that \(U_{t_1} \subseteq Int (f^{-1} (S_2))\), because \(U_{t_1}\) is open and the interior is the union of all the open subsets contained in \(f^{-1} (S_2)\) while \(U_{t_1}\) is one of such open subsets.
So, \(t_1 \in Int (f^{-1} (S_2))\).
So, \(f^{-1} (Int (S_2)) \subseteq Int (f^{-1} (S_2))\).
Step 2:
Let us see an example that "\(f^{-1} (Int (S_2)) = Int (f^{-1} (S_2))\)" 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.
\(Int (S_2) = (0, 1)\) and \(f^{-1} (Int (S_2)) = (-1, 0) \cup (0, 1)\).
But \(f^{-1} (S_2) = [-1, 1]\) and \(Int (f^{-1} (S_2)) = (-1, 1)\).
So, \(f^{-1} (Int (S_2)) = (-1, 0) \cup (0, 1) \neq (-1, 1) = Int (f^{-1} (S_2))\).
