description/proof of that for continuous map from closed subset of normal topological space into interval, there is continuous extension of map into same interval (Tietze extension theorem)
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 topological subspace.
- The reader knows a definition of continuous, topological spaces map.
- The reader admits the proposition that any constant map between any topological spaces is continuous.
- The reader admits the proposition that any expansion of any continuous map on the codomain is continuous.
- The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
- The reader admits the proposition that for any continuous map from any closed subset of any topological space and any closed subset and any open subset containing the closed subset of the codomain, there is an open subset of the domain space that contains the preimage of the closed subset such that the complement of the open subset contains the preimage of the complement of the open subset.
- The reader admits the proposition that for any uniformly convergent sequence of continuous maps from any topological space into any metric space with the induced topology, the convergence is continuous.
- The reader admits 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 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 continuous map from any closed subset of any normal topological space into any interval, there is a continuous extension of the map into the same interval (Tietze extension theorem).
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:
\(T\): \(\in \{\text{ the normal topological spaces }\}\)
\(C\): \(\in \{\text{ the closed subsets of } T\}\), with the subspace topology
\(I\): \(\in \{\text{ the intervals of } \mathbb{R}\}\), with \(I = (r_1, r_2), (r_1, r_2], [r_1, r_2), \text{ or } [r_1, r_2]\), where \(r_1\) or \(r_2\) may be \(- \infty\) or \(\infty\)
\(f\): \(: C \to I\), \(\in \{\text{ the continuous maps }\}\)
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Statements:
\(\exists g: T \to I \in \{\text{ the continuous maps }\} (g \vert_C = f)\)
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2: Note
\(I\) is valid only for these cases: for \((r_1, r_2), (r_1, r_2], \text{ and } [r_1, r_2)\), \(r_1 \lt r_2\); for \([r_1, r_2]\), \(r_1 \le r_2\).
Any closed boundary does not have \(\infty\) or \(- \infty\).
The lower boundary of \(I\) can (but does not have to) be taken to be \(Inf (\{f (c) \vert c \in C\})\) and the upper boundary of \(I\) can (but does not have to) be taken to be \(Sup (\{f (c) \vert c \in C\})\); when so, when the infimum or supremum is not the minimum or maximum, \(I\) can (but does not have to) be taken lower open or upper open, otherwise, \(I\) needs to be taken lower closed or upper closed.
This proposition is claiming that when \(I = (r_1, r_2)\) for example, \(g\) can be taken to be into \((r_1, r_2)\) instead of into \([r_1, r_2]\): \(Inf (\{g (t) \vert t \in T\}) = Inf (\{f (c) \vert c \in C\})\) and \(Sup (\{g (t) \vert t \in T\}) = Sup (\{f (c) \vert c \in C\})\) is a weaker claim that guarantees only that \(g\) is into \([r_1, r_2]\).
3: Proof
Whole Strategy: Step 1: suppose that \(I\) is bounded; Step 2: define \(r := r_2 - r_1\), deal with the case that \(r = 0\), and suppose that \(0 \lt r\) thereafter; Step 3: define \(f_0 := f - r_1: C \to [0, r]\), \(g_1: T \to [0, 1 / 3 r]\) such that \(g_1 ({f_0}^{-1} ([0, 1 / 3 r])) = \{0\} \land g_1 ({f_0}^{-1} ([2 / 3 r, r])) = \{1 / 3 r\}\), and \(f_1 := f_0 - g_1 \vert_C: C \to [0, 2 / 3 r]\); Step 4: inductively define \(g_n: T \to [0, 1 / 3 (2 / 3)^{n - 1} r]\) such that \(g_n ({f_{n - 1}}^{-1} ([0, 1 / 3 (2 / 3)^{n - 1} r])) = \{0\} \land g_n ({f_{n - 1}}^{-1} ([2 / 3 (2 / 3)^{n - 1} r, (2 / 3)^{n - 1} r])) = \{1 / 3 (2 / 3)^{n - 1} r\}\) and \(f_n : C \to [0, (2 / 3)^n r] = f_{n - 1} - g_n \vert_C\); Step 5: see that \(g := \sum_{n \in \mathbb{N} \setminus \{0\}} g_n : T \to [0, r]\) converges uniformly and \(g \vert_C = f_0\); Step 6: deal with the case, \(I = [r_1, r_2]\), and deal with the other cases by taking, for example, \(g - 1 / 2 r\), \(C' \subseteq T := (g - 1 / 2 r)^{-1} (\{- r / 2, r / 2\})\), a continuous \(h: T \to [0, 1]\) such that \(h (C') = \{0\} \land h (C) = \{1\}\), and \((g - 1 / 2 r) h + 1 / 2 r + r_1: T \to I\); Step 7: suppose that \(f\) is not value-bounded; Step 8: take \(tan: (- \pi / 2, \pi / 2) \to (- \infty, \infty)\) and value-bounded \(tan^{-1} \circ f\), and take a continuous \(g\) such that \(g \vert_C = tan^{-1} \circ f\) and see that \(tan \circ g: T \to I\) is a one that satisfies the requirements.
Step 1:
Let us suppose that \(I\) is bounded, which means that \(r_1 \neq - \infty\) and \(r_2 \neq \infty\).
Step 2:
Let us define \(r := r_2 - r_1\).
Let us suppose that \(r = 0\).
That means that \(I = [r_1, r_1]\): refer to Note.
Then, \(g = r_1: T \to I\), constant, is enough: \(g\) is into \(I\), continuous, by the proposition that any constant map between any topological spaces is continuous, and \(g_C = f\).
Let us suppose that \(0 \lt r\) hereafter.
Step 3:
Let us define \(f_0 := f - r_1: C \to [0, r]\): even if \(I\) is not closed, it is valid, because enlarging the codomain is allowed.
\(f_0\) is continuous, because \(f: C \to \mathbb{R}\) is continuous, by the proposition that any expansion of any continuous map on the codomain is continuous, \(f - r_1: C \to \mathbb{R}\) is obviously continuous, and \(f - r_1: C \to [0, r]\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
\([0, 1 / 3 r] \subseteq [0, r]\) is closed and \([0, 2 / 3 r) \subseteq [0, r]\) is open such that \([0, 1 / 3 r] \subseteq [0, 2 / 3 r)\).
So, there is a continuous \(g_1: T \to [0, 1 / 3 r]\) such that \(g_1 ({f_0}^{-1} ([0, 1 / 3 r])) = \{0\} \land g_1 ({f_0}^{-1} ([2 / 3 r, r])) = \{1 / 3 r\}\), by the proposition that for any continuous map from any closed subset of any topological space and any closed subset and any open subset containing the closed subset of the codomain, there is an open subset of the domain space that contains the preimage of the closed subset such that the complement of the open subset contains the preimage of the complement of the open subset.
Let us define \(f_1 := f_0 - g_1 \vert_C: C \to [0, 2 / 3 r]\), which is valid, because while \(C = {f_0}^{-1} ([0, r]) = {f_0}^{-1} ([0, 1 / 3 r]) \cup {f_0}^{-1} ((1 / 3 r, 2 / 3 r)) \cup {f_0}^{-1} ([2 / 3 r, r])\), for each \(c \in {f_0}^{-1} ([0, 1 / 3 r])\), \((f_0 - g_1 \vert_C) (c) = f_0 (c) \in [0, 1 / 3 r]\); for each \(c \in {f_0}^{-1} ((1 / 3 r, 2 / 3 r))\), \(0 = 1 / 3 r - 1 / 3 r \lt (f_0 - g_1 \vert_C) (c) \lt 2 / 3 r - 0 = 2 / 3 r\); for each \(c \in {f_0}^{-1} ([2 / 3 r, r])\), \((f_0 - g_1 \vert_C) (c) = f_0 (c) - 1 / 3 r\) and \(1 / 3 r = 2 / 3 r - 1 / 3 r \le f_0 (c) - 1 / 3 r \le r - 1 / 3 r = 2 / 3 r\).
\(f_1\) is continuous, because \(g_1: T \to \mathbb{R}\) and \(f_0: C \to \mathbb{R}\) are continuous, by the proposition that any expansion of any continuous map on the codomain is continuous, \(g_1 \vert_C: C \to \mathbb{R}\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous, \(f_0 - g_1 \vert_C: C \to \mathbb{R}\) is obviously continuous, and \(f_0 - g_1 \vert_C: C \to [0, 2 / 3 r]\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
Step 4:
Let us inductively define for \(n \in \mathbb{N} \setminus \{0\}\), continuous \(g_n: T \to [0, 1 / 3 (2 / 3)^{n - 1} r]\) such that \(g_n ({f_{n - 1}}^{-1} ([0, 1 / 3 (2 / 3)^{n - 1} r])) = \{0\} \land g_n ({f_{n - 1}}^{-1} ([2 / 3 (2 / 3)^{n - 1} r, (2 / 3)^{n - 1} r])) = \{1 / 3 (2 / 3)^{n - 1} r\}\) and continuous \(f_n : C \to [0, (2 / 3)^n r] = f_{n - 1} - g_n \vert_C\), as follows.
Let us suppose that we have continuous \(f_{n' - 1}: C \to [0, (2 / 3)^{n' - 1} r]\).
\([0, 1 / 3 (2 / 3)^{n' - 1} r] \subseteq [0, (2 / 3)^{n' - 1} r]\) is closed and \([0, 2 / 3 (2 / 3)^{n' - 1} r) \subseteq [0, (2 / 3)^{n' - 1} r]\) is open such that \([0, 1 / 3 (2 / 3)^{n' - 1} r] \subseteq [0, 2 / 3 (2 / 3)^{n' - 1} r)\).
So, there is a continuous \(g_{n'}: T \to [0, 1 / 3 (2 / 3)^{n' - 1} r]\) such that \(g_{n'} ({f_{n' - 1}}^{-1} ([0, 1 / 3 (2 / 3)^{n' - 1} r])) = \{0\} \land g_{n'} ({f_0}^{-1} ([2 / 3 (2 / 3)^{n' - 1} r, (2 / 3)^{n' - 1} r])) = \{1 / 3 (2 / 3)^{n' - 1} r\}\), by Note for the proposition that for any continuous map from any closed subset of any topological space and any closed subset and any open subset containing the closed subset of the codomain, there is an open subset of the domain space that contains the preimage of the closed subset such that the complement of the open subset contains the preimage of the complement of the open subset.
Let us define \(f_{n'} := f_{n' - 1} - g_{n'} \vert_C: C \to [0, (2 / 3)^{n'} r]\), which is valid, because while \(C = {f_{n' - 1}}^{-1} ([0, (2 / 3)^{n' - 1} r]) = {f_{n' - 1}}^{-1} ([0, 1 / 3 (2 / 3)^{n' - 1} r]) \cup {f_{n' - 1}}^{-1} ((1 / 3 (2 / 3)^{n' - 1} r, 2 / 3 (2 / 3)^{n' - 1} r)) \cup {f_{n' - 1}}^{-1} ([2 / 3 (2 / 3)^{n' - 1} r, (2 / 3)^{n' - 1} r])\), for each \(c \in {f_{n' - 1}}^{-1} ([0, 1 / 3 (2 / 3)^{n' - 1} r])\), \((f_{n' - 1} - g_{n'} \vert_C) (c) = f_{n' - 1} (c) \in [0, 1 / 3 (2 / 3)^{n' - 1} r]\); for each \(c \in {f_{n' - 1}}^{-1} ((1 / 3 (2 / 3)^{n' - 1} r, 2 / 3 (2 / 3)^{n' - 1} r))\), \(0 = 1 / 3 (2 / 3)^{n' - 1} r - 1 / 3 (2 / 3)^{n' - 1} r \lt (f_{n' - 1} - g_{n'} \vert_C) (c) \lt 2 / 3 (2 / 3)^{n' - 1} r - 0 = 2 / 3 (2 / 3)^{n' - 1} r\); for each \(c \in {f_{n' - 1}}^{-1} ([2 / 3 (2 / 3)^{n' - 1} r, (2 / 3)^{n' - 1} r])\), \((f_{n' - 1} - g_1 \vert_C) (c) = f_{n' - 1} (c) - 1 / 3 (2 / 3)^{n' - 1} r\) and \(1 / 3 (2 / 3)^{n' - 1} r = 2 / 3 (2 / 3)^{n' - 1} r - 1 / 3 (2 / 3)^{n' - 1} r \le f_{n' - 1} (c) - 1 / 3 (2 / 3)^{n' - 1} r \le (2 / 3)^{n' - 1} r - 1 / 3 (2 / 3)^{n' - 1} r = 2 / 3 (2 / 3)^{n' - 1} r\).
\(f_{n'}\) is continuous, because \(g_{n'}: T \to \mathbb{R}\) and \(f_{n' - 1}: C \to \mathbb{R}\) are continuous, by the proposition that any expansion of any continuous map on the codomain is continuous, \(g_{n' - 1} \vert_C: C \to \mathbb{R}\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous, \(f_{n' - 1} - g_{n'} \vert_C: C \to \mathbb{R}\) is obviously continuous, and \(f_{n' - 1} - g_{n'} \vert_C: C \to [0, (2 / 3)^{n'} r]\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
Step 5:
Let us define \(g := \sum_{n \in \mathbb{N} \setminus \{0\}} g_n : T \to [0, r]\), which is valid, because for each \(t \in T\), \(0 = \sum_{n \in \mathbb{N} \setminus \{0\}} 0 \le \sum_{n \in \mathbb{N} \setminus \{0\}} g_n (t) \le \sum_{n \in \mathbb{N} \setminus \{0\}} 1 / 3 (2 / 3)^{n - 1} r = 1 / 3 r (1 + 2 / 3 + (2 / 3)^2 + ...) = 1 / 3 r 1 / (1 - 2 / 3) = r\).
\(g\) is uniformly convergent, because for each \(t \in T\), \(\vert (g - \sum_{n \in \{1, ..., n'\}} g_n) (t) \vert = \sum_{n \in \{n' + 1, ...\}} g_n (t) \le \sum_{n \in \{n' + 1, ...\}} 1 / 3 (2 / 3)^{n - 1} r = 1 / 3 r ((2 / 3)^{n'} + (2 / 3)^{n' + 1} + ...) = 1 / 3 r (2 / 3)^{n'} (1 + (2 / 3)^1 + ...) = 1 / 3 r (2 / 3)^{n'} 1 / (1 - 2 / 3) = r (2 / 3)^{n'}\), and for each \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is an \(N \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N \lt n\), \(r (2 / 3)^n \lt \epsilon\), independent of \(t\).
\(g\) is continuous, by the proposition that for any uniformly convergent sequence of continuous maps from any topological space into any metric space with the induced topology, the convergence is continuous.
\(f_n = f_{n - 1} - g_n \vert_C = f_{n - 2} - g_{n - 1} \vert_C - g_n \vert_C = ... = f_0 - g_1 \vert_C - ... - g_n \vert_C = f_0 - (g_1+ ... + g_n) \vert_C\), so, \(lim_{n \to \infty} f_n = lim_{n \to \infty} f_0 - (g_1+ ... + g_n) \vert_C = f_0 - g \vert_C\), but \(lim_{n \to \infty} f_n = 0\), because \(f_n\) is into \([0, (2 / 3)^n r]\).
So, \(0 = f_0 - g \vert_C\), so, \(g \vert_C = f_0\).
Step 6:
When \(I = [r_1, r_2]\), \(g + r_1: T \to [r_1, r_2] = I\) is continuous, and \((g + r_1)\vert_C = (f_0 + r_1) = f\), so, \(g + r_1\) is a one required for this proposition.
Otherwise, \(g + r_1\) is not enough, because \(I \subset [r_1, r_2]\), so, let us remedy that.
\(g - 1 / 2 r: T \to [-r / 2, r / 2]\) is continuous, obviously.
When \(I = (r_1, r_2)\), let us take \(C' \subseteq T := (g - 1 / 2 r)^{-1} (\{- r / 2, r / 2\})\), closed; when \(I = (r_1, r_2]\), let us take \(C' \subseteq T := (g - 1 / 2 r)^{-1} (\{- r / 2\})\), closed; when \(I = [r_1, r_2)\), let us take \(C' \subseteq T := (g - 1 / 2 r)^{-1} (\{r / 2\})\), closed.
\(C \cap C' = \emptyset\), because when \(I = (r_1, r_2)\), for each \(c \in C\), \(f (c) \notin \{r_1, r_2\}\) and \((g - 1 / 2 r) (c) = (f_0 - 1 / 2 r) (c) = (f - r_1 - 1 / 2 r) (c) = f (c) - r_1 - 1 / 2 r \in (- 1 / 2 r, 1 / 2 r)\), so, \((g - 1 / 2 r) (c) \notin \{- 1 / 2 r, 1 / 2 r\}\), so, \(c \notin C'\); when \(I = (r_1, r_2]\), for each \(c \in C\), \(f (c) \notin \{r_1\}\) and \((g - 1 / 2 r) (c) = (f_0 - 1 / 2 r) (c) = (f - r_1 - 1 / 2 r) (c) = f (c) - r_1 - 1 / 2 r \in (- 1 / 2 r, 1 / 2 r]\), so, \((g - 1 / 2 r) (c) \notin \{- 1 / 2 r\}\), so, \(c \notin C'\); when \(I = [r_1, r_2)\), for each \(c \in C\), \(f (c) \notin \{r_2\}\) and \((g - 1 / 2 r) (c) = (f_0 - 1 / 2 r) (c) = (f - r_1 - 1 / 2 r) (c) = f (c) - r_1 - 1 / 2 r \in [- 1 / 2 r, 1 / 2 r)\), so, \((g - 1 / 2 r) (c) \notin \{1 / 2 r\}\), so, \(c \notin C'\).
So, \(C' \subseteq T \setminus C\).
There is a continuous \(h: T \to [0, 1]\) such that \(h (C') = \{0\} \land h (T \setminus (T \setminus C)) = h (C) = \{1\}\), by 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).
When \(I = (r_1, r_2)\), \((g - 1 / 2 r) h: T \to (-r / 2, r / 2)\), because for each \(c' \in C'\), \((g - 1 / 2 r) h (c') = 0\); for each \(t \in T \setminus C'\), \((g - 1 / 2 r) (t) \in (-r / 2, r / 2)\); when \(I = (r_1, r_2]\), \((g - 1 / 2 r) h: T \to (-r / 2, r / 2]\), because for each \(c' \in C'\), \((g - 1 / 2 r) h (c') = 0\); for each \(t \in T \setminus C'\), \((g - 1 / 2 r) (t) \in (-r / 2, r / 2]\); when \(I = [r_1, r_2)\), \((g - 1 / 2 r) h: T \to [-r / 2, r / 2)\), because for each \(c' \in C'\), \((g - 1 / 2 r) h (c') = 0\); for each \(t \in T \setminus C'\), \((g - 1 / 2 r) (t) \in [-r / 2, r / 2)\).
\((g - 1 / 2 r) h\) is continuous, as the product of the continuous maps.
\(((g - 1 / 2 r) h) \vert_C = (g - 1 / 2 r) \vert_C\).
\((g - 1 / 2 r) h + 1 / 2 r + r_1: T \to I\), where \(I = (r_1, r_2), (r_1, r_2], \text{ or } [r_1, r_2)\), is continuous.
\(((g - 1 / 2 r) h + 1 / 2 r + r_1) \vert_C = ((g - 1 / 2 r) h) \vert_C + 1 / 2 r + r_1 = (g - 1 / 2 r) \vert_C + 1 / 2 r + r_1 = f_0 - 1 / 2 r + 1 / 2 r + r_1 = f_0 + r_1 = f\).
So, \((g - 1 / 2 r) h + 1 / 2 r + r_1\) is a one that satisfies the requirements for this proposition.
Step 7:
Let us suppose that \(f\) is not value-bounded, which means that \(r_1 = - \infty\) or \(r_2 = \infty\).
Step 8:
Let us take \(tan: (- \pi / 2, \pi / 2) \to (- \infty, \infty)\), which is an increasing homeomorphism, as is well known.
When \(I = (- \infty, \infty)\), \(tan^{-1} \circ f: C \to (- \pi / 2, \pi / 2)\) is value-bounded continuous; when \(I = (- \infty, r_2) \text{ or } (- \infty, r_2]\), \(tan^{-1} \circ f: C \to (- \pi / 2, tan^{-1} (r_2)) \text{ or } (- \pi / 2, tan^{-1} (r_2)]\) is value-bounded continuous; when \(I = (r_1, \infty) \text{ or } [r_1, \infty)\), \(tan^{-1} \circ f: C \to (tan^{-1} (r_1), \pi / 2) \text{ or } [tan^{-1} (r_1), \pi / 2)\) is value-bounded continuous.
By Step 6, there is a continuous \(g: T \to (- \pi / 2, \pi / 2), (- \pi / 2, tan^{-1} (r_2)), (- \pi / 2, tan^{-1} (r_2)], (tan^{-1} (r_1), \pi / 2), \text{ or } [tan^{-1} (r_1), \pi / 2)\) such that \(g \vert_C = tan^{-1} \circ f\).
\(tan \circ g: T \to I\), where \(I = (- \infty, \infty), (- \infty, r_2), (- \infty, r_2], \text{ or } (r_1, \infty), [r_1, \infty)\), 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.
\(tan \circ g \vert_C = tan \circ tan^{-1} \circ f = f\).
So, \(tan \circ g\) is a one that satisfies the requirements for this proposition.
