description/proof of that for normal topological space, closed subset, and open subset that contains closed subset, there is continuous map into closed Interval that maps closed subset to boundary and complement of open subset to other boundary (Urysohn's lemma)
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of normal topological space.
- The reader knows a definition of closed subset of topological space.
- The reader knows a definition of Euclidean topological space.
- The reader knows a definition of topological subspace.
- The reader knows a definition of continuous, topological spaces map.
- The reader admits the proposition that any topological space is normal if and only if for each open subset, for each closed subset contained in the open subset, there is an open subset such that it contains the closed subset and its closure is contained in the open 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.
- The reader admits the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
Target Context
- The reader will have a description and a proof 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).
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:
\(T\): \(\in \{\text{ the normal topological spaces }\}\)
\(C\): \(\in \{\text{ the closed subsets of } T\}\)
\(U\): \(\in \{\text{ the opens subsets of } T\}\), such that \(C \subseteq U\)
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\)
\([r_1, r_2]\): \(\subseteq \mathbb{R}\), with the subspace topology
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Statements:
\(\exists f: T \to [r_1, r_2] \in \{\text{ the continuous maps }\} (f (C) = \{r_1\} \land f (T \setminus U) = \{r_2\})\)
\(\land\)
\(\exists f: T \to [r_1, r_2] \in \{\text{ the continuous maps }\} (f (C) = \{r_2\} \land f (T \setminus U) = \{r_1\})\)
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2: 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 take \(\{U_j \vert j \in J\}\) such that 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 continuous; Step 3: take a continuous \(g: [0, 1] \to [r_1, r_2]\) such that \(g (0) = r_1\) and \(g (1) = r_2\) or \(g (0) = r_2\) and \(g (1) = r_1\); Step 4: conclude the proposition.
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 \(U_1 := U\).
There is an open \(U_0 \subseteq T\) such that \(C \subseteq U_0 \subseteq \overline{U_0} \subseteq U_1\), by the proposition that any topological space is normal if and only if for each open subset, for each closed subset contained in the open subset, there is an open subset such that it contains the closed subset and its closure is contained in the open subset.
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 \(\overline{U_0} \subseteq U_{1 / 2^1} \subseteq \overline{U_{1 / 2^1}} \subseteq U_1\), as before.
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 \(j_1 \lt j_2\), \(\overline{U_{j_1}} \subseteq U_{j_2}\).
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 \(j_1 \lt j_2\), \(\overline{U_{j_1}} \subseteq U_{j_2}\).
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 \(0 \le p \lt 2^{n - 1} - 1\), \(\overline{U_{(2 p) / 2^n}} \subseteq U_{(2 p + 1) / 2^n} \subseteq \overline{U_{(2 p + 1) / 2^n}} \subseteq U_{(2 p + 2) / 2^n}\), which is possible, because \(\overline{U_{(2 p) / 2^n}} = \overline{U_{p / 2^{n - 1}}} \subseteq U_{(p + 1) / 2^{n - 1}} = U_{(2 p + 2) / 2^n}\): 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 \(j_1 \lt j_2\), \(\overline{U_{j_1}} \subseteq U_{j_2}\), 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 \(U_{(2 p + 2) / 2^n} \subseteq U_{(2 p') / 2^n}\); 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 \(U_{(2 p' + 2) / 2 ^n} \subseteq U_{(2 p) / 2^n}\).
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, \(t \in f'^{-1} ((r, \infty))\).
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.
\(f' (C) = \{0\}\), because for each \(c \in C\), \(c \in U_0\).
\(f' (T \setminus U) = \{1\}\), because for each \(t \in T \setminus U\), \(t \notin U = U_1\).
Step 3:
Let us take \(g: [0, 1] \to [r_1, r_2], t \mapsto (r_2 - r_1) (t + r_1 / (r_2 - r_1))\).
That is indeed into \([r_1, r_2]\), because \(g (0) = (r_2 - r_1) r_1 / (r_2 - r_1) = r_1\) and \(g (1) = (r_2 - r_1) (1 + r_1 / (r_2 - r_1)) = (r_2 - r_1) ((r_2 - r_1 + r_1) / (r_2 - r_1)) = (r_2 - r_1) r_2 / (r_2 - r_1) = r_2\) while \(g\) is increasing.
\(g\) is continuous, because it is linear.
\(g \circ f': T \to [r_1, r_2]\) is continuous, by the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
\(g \circ f' (C) = \{r_1\}\), because \(f' (C) = \{0\}\) and \(g (0) = r_1\).
\(g \circ f' (T \setminus U) = \{r_2\}\), because \(f' (T \setminus U) = \{1\}\) and \(g (1) = r_2\).
Let us take \(g: [0, 1] \to [r_1, r_2], t \mapsto - (r_2 - r_1) (t - r_2 / (r_2 - r_1))\).
That is indeed into \([r_1, r_2]\), because \(g (0) = - (r_2 - r_1) (- r_2 / (r_2 - r_1)) = r_2\) and \(g (1) = - (r_2 - r_1) (1 - r_2 / (r_2 - r_1)) = - (r_2 - r_1) (r_2 - r_1 - r_2) / (r_2 - r_1) = - (r_2 - r_1) (- r_1) / (r_2 - r_1) = r_1\) while \(g\) is decreasing.
\(g\) is continuous, because it is linear.
\(g \circ f': T \to [r_1, r_2]\) is continuous, by the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
\(g \circ f' (C) = \{r_2\}\), because \(f' (C) = \{0\}\) and \(g (0) = r_2\).
\(g \circ f' (T \setminus U) = \{r_1\}\), because \(f' (T \setminus U) = \{1\}\) and \(g (1) = r_1\).
Step 4:
So, taking \(g \circ f'\) as \(f\), there is an \(f\) that satisfies the conditions for this proposition.
