description/proof of that locally compact Hausdorff topological space is completely regular
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of locally compact topological space.
- The reader knows a definition of Hausdorff topological space.
- The reader knows a definition of completely regular topological space.
- The reader admits the proposition that for any locally compact Hausdorff topological space, in any neighborhood around any point, there is an open neighborhood of the point whose (the open neighborhood's) closure is compact and contained in the former neighborhood.
- The reader admits the proposition that for any locally compact Hausdorff topological space and any 1st open subset whose closure is compact and contained in any 2nd open subset, there is an open subset that contains the closure of the 1st subset whose (the 3rd subset's) closure is compact and contained in the 2nd subset.
- The reader admits the proposition that for any \(2\) distinct non-negative real numbers and any natural number larger than \(1\), there are some 2nd and 3rd natural numbers such that the 3rd number divided by the number to the power of the 2nd number is exactly between the real numbers.
- The reader admits the proposition that for the \(1\)-dimensional Euclidean topological space, the set of the upper bounded open intervals and the lower bounded open intervals is a subbasis.
- The reader admits the proposition that in order to check the continuousness of any map between any topological spaces, the preimages of only any basis or any subbasis are enough to be checked.
Target Context
- The reader will have a description and a proof of the proposition that any locally compact Hausdorff topological space is completely regular.
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:
\(T\): \(\in \{\text{ the locally compact Hausdorff topological spaces }\}\)
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Statements:
\(T \in \{\text{ the completely regular topological spaces }\}\)
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2: Note
Proof resembles that of the proposition that for any normal topological space, any closed subset, and any open subset that contains the closed subset, there is a continuous map into any closed Interval that maps the closed subset to any boundary of the closed interval and the complement of the open subset to the other boundary (Urysohn's lemma): the only difference is how to construct \(\{U_j \vert j \in J\}\).
In fact, \(f (U_0) = \{0\}\) and \(f (C) = \{1\}\) where \(U_0\) is an open neighborhood of \(t\), and as in Urysohn's lemma, a continuous map, \(f: T \to [r_1, r_2]\) such that \(f (C) = \{r_1\} \land f (U_0) = \{r_2\}\) or \(f (C) = \{r_2\} \land f (U_0) = \{r_1\}\) can be taken.
3: Proof
Whole Strategy: Step 1: take \(J := \{m / 2^n \vert n \in \mathbb{N}, m \in \mathbb{N} \text{ such that } 0 \le m \le 2^n\}\) and for each \(t \notin C\), take \(\{U_j \vert j \in J\}\) such that \(t \in U_0\) and \(U_1 = T \setminus C\), and for each \(j_1, j_2 \in J\) such that \(j_1 \lt j_2\), \(\overline{U_{j_1}} \subseteq U_{j_2}\); Step 2: take \(f: T \to [0, 1], t \mapsto 1 \text{ when } t \notin U_1; \mapsto Inf (\{j \in J \vert t \in U_j\}) \text{ when } t \in U_1\) and see that \(f\) is a continuous map that satisfies the conditions for \(T\) to be completely regular.
Step 1:
Let us take \(J := \{m / 2^n \vert n \in \mathbb{N}, m \in \mathbb{N} \text{ such that } 0 \le m \le 2^n\}\).
Let \(t \in T\) be any.
Let \(C \subseteq T\) be any any closed subset such that \(t \notin C\).
Let \(U_1 := T \setminus C\), open.
\(t \in U_1\), so, \(U_1\) is an open neighborhood of \(t\).
There is an open neighborhood of \(t\), \(U_0 \subseteq T\), such that \(\overline{U_0} \subseteq T\) is compact and \(\overline{U_0} \subseteq U_1\), by the proposition that for any locally compact Hausdorff topological space, in any neighborhood around any point, there is an open neighborhood of the point whose (the open neighborhood's) closure is compact and contained in the former neighborhood.
Now, we have for \(n = 0\), for \(J_n := \{m / 2^n \vert m \in \mathbb{N} \text{ such that } 0 \le m \le 2^n\}\), \(\{U_j \vert j \in J_n\}\) such that for each \(j_1, j_2 \in J_n\) such that \(j_1 \lt j_2\), \(\overline{U_{j_1}} \subseteq U_{j_2}\).
There is an open \(U_{1 / 2^1} \subseteq T\) such that \(\overline{U_0} \subseteq U_{1 / 2^1} \subseteq \overline{U_{1 / 2^1}} \subseteq U_1\), where \(\overline{U_{1 / 2^1}} \subseteq T\) is compact, by the proposition that for any locally compact Hausdorff topological space and any 1st open subset whose closure is compact and contained in any 2nd open subset, there is an open subset that contains the closure of the 1st subset whose (the 3rd subset's) closure is compact and contained in the 2nd subset.
Now, we have for \(n = 1\), for \(J_n := \{m / 2^n \vert m \in \mathbb{N} \text{ such that } 0 \le m \le 2^n\}\), \(\{U_j \vert j \in J_n\}\) such that for each \(j_1, j_2 \in J_n\) such that \(j_1 \lt j_2\), \(\overline{U_{j_1}} \subseteq U_{j_2}\) and \(\overline{U_j} \subseteq T\) is compact for each \(j \in J_n \setminus \{1\}\).
Let us suppose that we have for \(n = n' - 1\), for \(J_n := \{m / 2^n \vert m \in \mathbb{N} \text{ such that } 0 \le m \le 2^n\}\), \(\{U_j \vert j \in J_n\}\) such that for each \(j_1, j_2 \in J_n\) such that \(j_1 \lt j_2\), \(\overline{U_{j_1}} \subseteq U_{j_2}\) and \(\overline{U_j} \subseteq T\) is compact for each \(j \in J_n \setminus \{1\}\).
For \(n = n'\), for \(J_n := \{m / 2^n \vert m \in \mathbb{N} \text{ such that } 0 \le m \le 2^n\}\), we take \(\{U_j \vert j \in J_n\}\) such that for each \((2 p + 1) / 2^n \in J_n\) where \(p \in \mathbb{N}\) such that \(0 \le p \le 2^{n - 1} - 1\), \(\overline{U_{(2 p) / 2^n}} \subseteq U_{(2 p + 1) / 2^n} \subseteq \overline{U_{(2 p + 1) / 2^n}} \subseteq U_{(2 p + 2) / 2^n}\) where \(\overline{U_{(2 p + 1) / 2^n}} \subseteq T\) is compact, which is possible, by the proposition that for any locally compact Hausdorff topological space and any 1st open subset whose closure is compact and contained in any 2nd open subset, there is an open subset that contains the closure of the 1st subset whose (the 3rd subset's) closure is compact and contained in the 2nd subset: each \(U_{(2 p) / 2^n} = U_{p / 2^{n - 1}}\) and \(U_{2^n / 2^n} = U_1\) have been already determined.
Now, we have for \(n = n'\), for \(J_n := \{m / 2^n \vert m \in \mathbb{N} \text{ such that } 0 \le m \le 2^n\}\), \(\{U_j \vert j \in J_n\}\) such that for each \(j_1, j_2 \in J_n\) such that \(j_1 \lt j_2\), \(\overline{U_{j_1}} \subseteq U_{j_2}\) and \(\overline{U_j} \subseteq T\) is compact for each \(j \in J_n \setminus \{1\}\), because when \(j_1, j_2 \in J_{n - 1}\), it holds; when \(j_1 = (2 p + 1) / 2^n \in J_n \setminus J_{n - 1}\), \((2 p + 2) / 2^n \le j_2\), and while \(\overline{U_{j_1}} \subseteq U_{(2 p + 2) / 2^n}\), when \(j_2 \in J_{n - 1}\), \(U_{(2 p + 2) / 2^n} \subseteq U_{j_2}\) holds and otherwise, \(j_2 = (2 p' + 1) / 2^n\), and \(\overline{U_{(2 p') / 2^n}} \subseteq U_{j_2}\), but \(U_{(2 p + 2) / 2^n} \subseteq U_{(2 p') / 2^n}\), because \(p \lt p'\), so, \(p + 1 \le p'\), and \(2 (p + 1) \le 2 p'\); when \(j_2 = (2 p + 1) / 2^n \in J_n \setminus J_{n - 1}\), \(j_1 \le (2 p) / 2^n\), and while \(\overline{U_{(2 p) / 2^n}} \subseteq U_{j_2}\), when \(j_1 \in J_{n - 1}\), \(\overline{U_{j_1}} \subseteq \overline{U_{(2 p) / 2^n}}\) holds and otherwise, \(j_1 = (2 p' + 1) / 2 ^n\), and \(\overline{U_{j_1}} \subseteq U_{(2 p' + 2) / 2 ^n}\), but \(U_{(2 p' + 2) / 2 ^n} \subseteq U_{(2 p) / 2^n}\), because \(p' \lt p\), so, \(p' + 1 \lt p\), and \(2 (p' + 1) \lt 2 p\).
So, inductively, we have \(\{U_j \vert j \in J\}\).
For each \(j_1, j_2 \in J\) such that \(j_1 \lt j_2\), \(\overline{U_{j_1}} \subseteq U_{j_2}\), because while \(j_1 = m / 2^n\) and \(j_2 = m' / 2^{n'}\), when \(n = n'\), we already know that it holds, because \(j_1, j_2 \in J_n\); when \(n \lt n'\), \(j_1 = (2^{n ' - n} m) / 2^{n'}\), so, \(j_1, j_2 \in J_{n'}\), so, it holds; when \(n' \lt n\), \(j_2 = (2^{n - n'} m') / 2^n\), so, \(j_1, j_2 \in J_n\), so, it holds.
Step 2:
Let us define \(f: T \to [0, 1], t' \mapsto 1 \text{ when } t' \notin U_1; \mapsto Inf (\{j \in J \vert t' \in U_j\}) \text{ when } t' \in U_1\).
The definition is valid, because \(\{j \in J \vert t \in U_j\}\) is not empty when \(t' \in U_1\), because \(1 \in \{j \in J \vert t' \in U_j\}\), and \(\{j \in J \vert t' \in U_j\}\) is lower bounded by \(0\).
Let us see that \(f\) is continuous.
Let \((- \infty, r) \subseteq \mathbb{R}\) be any.
When \(r \le 0\), \(f^{-1} ((- \infty, r)) = \emptyset\), because \(f\) is into \([0, 1]\).
When \(0 \lt r\), \(f^{-1} ((- \infty, r)) = \cup \{U_j \vert j \lt r\}\), because for each \(t' \in f^{-1} ((- \infty, r))\), \(f (t') \lt r\), there is a \(j \in J\) such that \(f (t') \lt j \lt r\), by the proposition that for any \(2\) distinct non-negative real numbers and any natural number larger than \(1\), there are some 2nd and 3rd natural numbers such that the 3rd number divided by the number to the power of the 2nd number is exactly between the real numbers, then, \(t' \in U_j\), because if \(t' \notin U_j\), \(t' \notin U_{j^`}\) for each \(j^` \in J\) such that \(j^` \lt j\), a contradiction against that \(f (t')\) was the infimum, so, \(t' \in \cup \{U_j \vert j \lt r\}\); for each \(t' \in \cup \{U_j \vert j \lt r\}\), \(t' \in U_j\) for a \(j \in J\) such that \(j \lt r\), then, \(f (t') \le j \lt r\), so, \(t' \in f^{-1} ((- \infty, r))\).
So, \(f^{-1} ((- \infty, r)) \subseteq T\) is open.
Let \((r, \infty) \subseteq \mathbb{R}\) be any.
When \(r \lt 0\), \(f^{-1} ((r, \infty)) = T\), because \(f\) is into \([0, 1]\).
When \(0 \le r\), \(f^{-1} ((r, \infty)) = \cup \{T \setminus U_j \vert r \lt j\}\), because for each \(t' \in f^{-1} ((r, \infty))\), \(r \lt f (t')\), there is a \(j \in J\) such that \(r \lt j \lt f (t')\), by the proposition that for any \(2\) distinct non-negative real numbers and any natural number larger than \(1\), there are some 2nd and 3rd natural numbers such that the 3rd number divided by the number to the power of the 2nd number is exactly between the real numbers, then, \(t' \notin U_j\), because if \(t' \in U_j\), \(f (t') \le j\), a contradiction, so, \(t' \in T \setminus U_j\), so, \(t' \in \cup \{T \setminus U_j \vert r \lt j\}\); for each \(t' \in \cup \{T \setminus U_j \vert r \lt j\}\), \(t' \in T \setminus U_j\) for a \(j \in J\) such that \(r \lt j\), \(t' \notin U_j\), so, \(j \le f (t')\), because if \(f (t') \lt j\), \(t' \in U_j\), a contradiction, so, \(r \lt j \le f (t')\), so, \(t' \in f^{-1} ((r, \infty))\).
But \(\cup \{T \setminus U_j \vert r \lt j\} = \cup \{T \setminus \overline{U_j} \vert r \lt j\}\), because for each \(t' \in \cup \{T \setminus U_j \vert r \lt j\}\), \(t' \in T \setminus U_j\) for a \(j\) such that \(r \lt j\), there is a \(j'\) such that \(r \lt j' \lt j\), by the proposition that for any \(2\) distinct non-negative real numbers and any natural number larger than \(1\), there are some 2nd and 3rd natural numbers such that the 3rd number divided by the number to the power of the 2nd number is exactly between the real numbers, \(\overline{U_{j'}} \subseteq U_j\), so, \(t' \in T \setminus \overline{U_{j'}}\), so, \(t' \in \cup \{T \setminus \overline{U_j} \vert r \lt j\}\); for each \(t' \in \cup \{T \setminus \overline{U_j} \vert r \lt j\}\), \(t' \in T \setminus \overline{U_j}\) for a \(j\) such that \(r \lt j\), so, \(t' \notin \overline{U_j}\), so, \(t' \notin U_j\), so, \(t' \in T \setminus U_j\), so, \(t' \in \cup \{T \setminus U_j \vert r \lt j\}\).
So, \(f^{-1} ((r, \infty)) \subseteq T\) is open.
By the proposition that for the \(1\)-dimensional Euclidean topological space, the set of the upper bounded open intervals and the lower bounded open intervals is a subbasis and the proposition that in order to check the continuousness of any map between any topological spaces, the preimages of only any basis or any subbasis are enough to be checked, \(f\) is continuous.
\(f (t) = 0\), because \(t \in U_0\).
\(f (C) = \{1\}\), because for each \(c \in C = T \setminus U_1\), \(c \notin U_1\).
So, \(f\) is a continuous map that satisfies the conditions for \(T\) to be completely regular.
So, \(T\) is completely regular.
