description/proof of that for continuous map from upper-closed interval into topological space and subset of codomain, if subset of domain that is mapped into complement of subset is not empty and image of upper end is in subset, image of supremum of subset of domain is in boundary of subset
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean topological space.
- The reader knows a definition of continuous, topological spaces map.
- The reader knows a definition of supremum of subset of partially-ordered set.
- The reader knows a definition of boundary of subset of topological space.
- The reader admits the proposition that for any upper-closed interval and any nonempty subset, the supremum of the subset exists and equals the supremum of the subset on the real numbers set.
- 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 continuous map from any upper-closed interval into any topological space and any subset of the codomain, if the subset of the domain that is mapped into the complement of the subset is not empty and the image of the upper end is in the subset, the image of the supremum of the subset of the domain is in 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:
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\) with the canonical linear ordering
\(J_1\): \(\in \{\text{ the upper-closed intervals of } \mathbb{R}\}\), with the upper end, \(r_2\), with the subspace topology
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(S_2\): \(\subseteq T_2\)
\(f\): \(: J_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(S\): \(= \{r \in J_1 \vert f (r) \in T_2 \setminus S_2\}\)
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Statements:
\(S \neq \emptyset \land f (r_2) \in S_2\)
\(\implies\)
\(f (Sup (S)) \in Bou (S_2)\)
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2: Proof
Whole Strategy: Step 1: see that \(t := Sup (S)\) exists; Step 2: take any open neighborhood of \(f (t)\), \(U_{f (t)}\), and a \(B_{t, \delta}\) such that \(f (B_{t, \delta} \cap J_1) \subseteq U_{f (t)}\); Step 3: see that \(f (B_{t, \delta} \cap J_1) \cap (T_2 \setminus S_2) \neq \emptyset\); Step 4: see that \(f (B_{t, \delta} \cap J_1) \cap S_2 \neq \emptyset\); Step 5: conclude the proposition.
Step 1:
\(Sup (S)\) exists, by the proposition that for any upper-closed interval and any nonempty subset, the supremum of the subset exists and equals the supremum of the subset on the real numbers set.
Let \(t := Sup (S)\).
Step 2:
Let \(U_{f (t)} \subseteq T_2\) be any open neighborhood of \(f (t)\).
As \(f\) is continuous, there is an open neighborhood of \(t\), \(U_t \subseteq J_1\), such that \(f (U_t) \subseteq U_{f (t)}\).
\(U_t = U'_t \cap J_1\), where \(U'_t \subseteq \mathbb{R}\) is an open neighborhood of \(t\) on \(\mathbb{R}\), by the definition of subspace topology.
There is a \(B_{t, \delta} \subseteq U'_t\), by the definition of Euclidean topology.
\(B_{t, \delta} \cap J_1 \subseteq U_t\), and \(f (B_{t, \delta} \cap J_1) \subseteq U_{f (t)}\).
Step 3:
Let us see that \(f (t) \in \overline{T_2 \setminus S_2}\).
When \(f (t) \in T_2 \setminus S_2\), \(f (B_{t, \delta} \cap J_1) \cap (T_2 \setminus S_2) \neq \emptyset\).
When \(f (t) \in S_2\), \(t\) is not the lower end of \(J_1\) even if \(J_1\) is lower-closed, because otherwise, \(S = \emptyset\), because \((t, r_2]\) would not contain any \(S\) point (otherwise, \(t\) would not be any upper bound) and \(t \notin S\), so, a \((t - \delta', t]\) where \(0 \lt \delta' \lt \delta\) is contained in \(J_1\), but not the whole of \((t - \delta', t]\) can be mapped into \(S_2\), because if so, \(t\) would not be any minimum of the upper bounds, so, \(f (B_{t, \delta} \cap J_1) \cap (T_2 \setminus S_2) \neq \emptyset\).
So, anyway, \(f (B_{t, \delta} \cap J_1) \cap (T_2 \setminus S_2) \neq \emptyset\), so, \(U_{f (t)} \cap (T_2 \setminus S_2) \neq \emptyset\).
That means that \(f (t) \in \overline{T_2 \setminus S_2}\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
Step 4:
When \(t = r_2\), as \(f (r_2) \in S_2\), \(f (B_{t, \delta} \cap J_1) \cap S_2 \neq \emptyset\).
When \(t \lt r_2\), there is a \(\delta' \in \mathbb{R}\) such that \(0 \lt \delta' \lt \delta\) and \((t, t + \delta')\) is contained in \(J_1\), and the whole \((t, t + \delta')\) is mapped into \(S_2\), because otherwise, \(t\) would not be any upper bound of \(S\), so, \(f (B_{t, \delta} \cap J_1) \cap S_2 \neq \emptyset\).
So, anyway, \(f (B_{t, \delta} \cap J_1) \cap S_2 \neq \emptyset\), so, \(U_{f (t)} \cap S_2 \neq \emptyset\).
That means that \(f (t) \in \overline{S_2}\), by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.
Step 5:
So, \(f (t) \in \overline{S_2} \cap \overline{T_2 \setminus S_2} = Bou (S_2)\).
