<The previous article in this series | The table of contents of this series |
description/proof of that for measure space and measurable subset, Lebesgue integral of measurable extended real function over space is integral over subset plus integral over complement of subset
Topics
About:
measure space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any measure space and for any measurable subset, the Lebesgue integral of any measurable extended real function over the space is the integral over the subset plus the integral over the complement 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:
\((M, A, \mu)\): \(\in \{\text{ the measure spaces }\}\)
\(\overline{\mathbb{R}}\): \(= \text{ the extended Euclidean topological space }\) with the Borel \(\sigma\)-algebra
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\) with the Borel \(\sigma\)-algebra
\(f\): \(: M \to \overline{\mathbb{R}}\), \(\in \{\text{ the measurable maps }\}\)
\(a\): \(\in A\)
//
Statements:
\(\int_M f d \mu = \int_a f d \mu + \int_{M \setminus a} f d \mu\)
//
2: Proof
Whole Strategy: Step 1: see what \(\int_M f d \mu\), \(\int_a f d \mu\), and \(\int_{M \setminus a} f d \mu\) mean; Step 2: see that for each \(g = f^+ \text{ or } f^-\), \(\int_a g d \mu + \int_{M \setminus a} g d \mu \le \int_M g d \mu\); Step 3: see that for each \(g = f^+ \text{ or } f^-\), \(\int_M g d \mu \le \int_a g d \mu + \int_{M \setminus a} f g \mu\); Step 4: conclude the proposition.
Step 1:
\(\int_M f d \mu = \int_M f^+ d \mu - \int_M f^- d \mu\), \(\int_a f d \mu = \int_M \chi_a f^+ d \mu - \int_M \chi_a f^- d \mu\), and \(\int_{M \setminus a} f d \mu = \int_M \chi_{M \setminus a} f^+ d \mu - \int_M \chi_{M \setminus a} f^- d \mu\), by definition.
Let \(g := f^+ \text{ or } f^-\).
\(\int_M g d \mu = Sup (\{\int_M h d \mu \vert h \in P^+ \text{ such that } h \le g\})\), where \(\int_M h d \mu\) means for \(h (M) = \{t_1, ..., t_n\}\) (which means that \(h = \sum_{j \in \{1, ..., n\}} t_j \chi_{h^{-1} (t_j)}\)), \(\sum_{j \in \{1, ..., n\}} t_j \mu (h^{-1} (t_j))\), by definition.
\(\{h^{-1} (t_j) \vert j \in \{1, ..., n\}\}\) are some disjoint measurable subsets of \(M\).
\(\int_a g d \mu = Sup (\{\int_M h d \mu \vert h \in P^+ \text{ such that } h \le \chi_a g\})\), where \(h = \sum_{j \in \{1, ..., n_a\}} r_j \chi_{h^{-1} (r_j)}\) and \(\int_M h d \mu = \sum_{j \in \{1, ..., n_a\}} r_j \mu (h^{-1} (r_j))\).
\(\{h^{-1} (r_j) \vert j \in \{1, ..., n_a\}\}\) are some disjoint measurable subsets of \(a\).
\(\int_{M \setminus a} g d \mu = Sup (\{\int_M h d \mu \vert h \in P^+ \text{ such that } h \le \chi_{M \setminus a} g\})\), where \(h = \sum_{j \in \{1, ..., n_{M \setminus a}\}} s_j \chi_{h^{-1} (s_j)}\) and \(\int_M h d \mu = \sum_{j \in \{1, ..., n_{M \setminus a}\}} s_j \mu (h^{-1} (s_j))\).
\(\{h^{-1} (s_j) \vert j \in \{1, ..., n_{M \setminus a}\}\}\) are some disjoint measurable subsets of \(M \setminus a\).
Step 2:
Let us see that \(\int_a g d \mu + \int_{M \setminus a} g d \mu \le \int_M g d \mu\).
Let \(h_a = \sum_{j \in \{1, ..., n_a\}} r_j \chi_{{h_a}^{-1} (r_j)} \in \{h \in P^+ \text{ such that } h \le \chi_a g\}\) and \(h_{M \setminus a} = \sum_{j \in \{1, ..., n_{M \setminus a}\}} s_j \chi_{{h_{M \setminus a}}^{-1} (s_j)} \in \{h \in P^+ \text{ such that } h \le \chi_{M \setminus a} g\}\}\) be any.
\(h_a + h_{M \setminus a} \in \{h \in P^+ \text{ such that } h \le g\}\), because it is a simple function with the finite values, \(\{r_1, ..., r_{n_a}, s_1, ..., s_{n_{M \setminus a}}\}\) (which may contain some duplications), where \(\{(h_a + h_{M \setminus a})^{-1} (r_1), ..., (h_a + h_{M \setminus a})^{-1} (r_{n_a}), (h_a + h_{M \setminus a})^{-1} (s_1), ..., (h_a + h_{M \setminus a})^{-1} (s_{n_{M \setminus a}})\}\) (which may contain some duplications) are some disjoint measurable subsets of \(M\): \((h_a + h_{M \setminus a})^{-1} (r_1) = {h_a}^{-1} (r_1) \cup {h_{M \setminus a}}^{-1} (r_1)\) where \({h_{M \setminus a}}^{-1} (r_1)\) is for when \(r_1 \in \{s_1, ..., s_{n_{M \setminus a}}\}\), and so on, and \(h_a + h_{M \setminus a} \le g\) holds over \(a\), because \(h_a + h_{M \setminus a} = h_a + 0 \le \chi_a g = g\), and it holds over \(M \setminus a\), because \(h_a + h_{M \setminus a} = 0 + h_{M \setminus a} \le \chi_{M \setminus a} g = g\), so it holds over \(M\).
So, \(\int_M h_a d \mu + \int_M h_{M \setminus a} d \mu = \int_M (h_a + h_{M \setminus a}) d \mu \le \int_M g d \mu\).
By the proposition that for the real numbers field with the canonical linear ordering and any finite set of subsets, if the sum of each set of representatives of the subsets is equal to or smaller than any number, the sum of the supremums of the subsets is equal to or smaller than the number, \(\int_a g d \mu + \int_{M \setminus a} g d \mu = Sup (\{\int_M h_a d \mu\}) + Sup (\{\int_M h_{M \setminus a} d \mu\}) \le \int_M g d \mu\).
Step 3:
Let us see that \(\int_M g d \mu \le \int_a g d \mu + \int_{M \setminus a} g d \mu\).
Let \(h = \sum_{j \in \{1, ..., n\}} t_j \chi_{h^{-1} (t_j)} \in \{h \in P^+ \text{ such that } h \le g\}\) be any.
Let us take \(\{r_1, ..., r_{n_a}\} = \{r_j \in \{t_1, ..., t_n\} \vert h^{-1} (t_1) \cap a \neq \emptyset\}\), which may be empty, and \(\{s_1, ..., s_{n_{M \setminus a}}\} = \{s_j \in \{t_1, ..., t_n\} \vert h^{-1} (t_1) \cap (M \setminus a) \neq \emptyset\}\), which may be empty.
Let us take \(h_a = \sum_{j \in \{1, ..., n_a\}} r_j \chi_{h^{-1} (r_j) \cap a}\), which is regarded to be \(0\) when \(\{1, ..., n_a\}\) is empty and \(h_{M \setminus a} = \sum_{j \in \{1, ..., n_{M \setminus a}\}} s_j \chi_{h^{-1} (s_j) \cap (M \setminus a)}\), which is regarded to be \(0\) when \(\{1, ..., n_{M \setminus a}\}\) is empty.
Then, \(h = h_a + h_{M \setminus a}\), because over \(a\), \(h = \sum_{j \in \{1, ..., n\}} t_j \chi_{h^{-1} (t_j)} = \sum_{j \in \{1, ..., n_a\}} r_j \chi_{h^{-1} (r_j) \cap a}\), because only the terms with \(h^{-1} (t_j) \cap a \neq \emptyset\) do not vanish, \(= h_a = h_a + h_{M \setminus a}\), and over \(M \setminus a\), \(h = \sum_{j \in \{1, ..., n\}} t_j \chi_{h^{-1} (t_j)} = \sum_{j \in \{1, ..., n_{M \setminus a}\}} s_j \chi_{h^{-1} (s_j) \cap (M \setminus a)}\), because only the terms with \(h^{-1} (t_j) \cap (M \setminus a) \neq \emptyset\) do not vanish, \(= h_{M \setminus a} = h_a + h_{M \setminus a}\).
\(h_a \in \{h \in P^+ \text{ such that } h \le \chi_a g\}\), because it is a simple function with the finite values, \(\{r_1, ..., r_{n_a}\}\), where \(\{{h_a}^{-1} (r_1), ..., {h_a}^{-1} (r_{n_a})\}\) are some disjoint measurable subsets of \(M\): \({h_a}^{-1} (r_1) = h^{-1} (r_1) \cap a\), and so on, and \(h_a \le \chi_a g\) holds over \(a\), because \(h_a = h \le g = \chi_a g\), and it holds over \(M \setminus a\), because \(h_a = 0 \le \chi_a g\), so it holds over \(M\).
\(h_{M \setminus a} \in \{h \in P^+ \text{ such that } h \le \chi_{M \setminus a} g\}\), because it is a simple function with the finite values, \(\{s_1, ..., s_{n_{M \setminus a}}\}\), where \(\{{h_{M \setminus a}}^{-1} (s_1), ..., {h_{M \setminus a}}^{-1} (s_{n_{M \setminus a}})\}\) are some disjoint measurable subsets of \(M\): \({h_{M \setminus a}}^{-1} (s_1) = h^{-1} (s_1) \cap (M \setminus a)\), and so on, and \(h_{M \setminus a} \le \chi_{M \setminus a} g\) holds over \(a\), because \(h_{M \setminus a} = 0 \le \chi_{M \setminus a} g\), and it holds over \(M \setminus a\), because \(h_{M \setminus a} = h \le g = \chi_{M \setminus a} g\), so it holds over \(M\).
\(\int_M h d \mu = \int_M (h_a + h_{M \setminus a}) d \mu = \int_M h_a d \mu + \int_M h_{M \setminus a} d \mu \le \int_a g d \mu + \int_{M \setminus a} g d \mu\).
That implies that \(\int_M g d \mu \le \int_a g d \mu + \int_{M \setminus a} g d \mu\).
Step 4:
So, \(\int_M g d \mu = \int_a g d \mu + \int_{M \setminus a} g d \mu\).
So, \(\int_M f d \mu = \int_M f^+ d \mu - \int_M f^- d \mu = \int_a f^+ d \mu + \int_{M \setminus a} f^+ d \mu - (\int_a f^- d \mu + \int_{M \setminus a} f^- d \mu) = \int_a f^+ d \mu - \int_a f^- d \mu + \int_{M \setminus a} f^+ d \mu - \int_{M \setminus a} f^- d \mu = \int_a (f^+ - f^-) d \mu + \int_{M \setminus a} (f^+ - f^-) d \mu = \int_a f d \mu + \int_{M \setminus a} f d \mu\).
References
<The previous article in this series | The table of contents of this series |
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that topological space is compact iff for each set of closed subsets whose intersection is empty, there is nonempty finite subset whose intersection is empty
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that any topological space is compact if and only if for each possibly uncountable set of closed subsets whose intersection is empty, there is a nonempty finite subset whose intersection is empty.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
//
Statements:
\(T \in \{\text{ the compact topological spaces }\}\)
\(\iff\)
\(\forall \{C_j \in \{\text{ the closed subsets of } T\} \vert j \in J\} \text{ where } J \text{ is any possibly uncountable index set } \text{ such that } \cap_{j \in J} C_j = \emptyset (\exists J^` \subseteq J \text{ such that } \vert J \vert \in \mathbb{N} \setminus \{0\} (\cap_{j \in J^`} C_j = \emptyset))\)
//
2: Proof
Whole Strategy: Step 1: suppose that there is a \(J^`\) for each such \(\{C_j \vert j \in J\}\); Step 2: see that for each open cover of \(T\), there is a finite subcover; Step 3: suppose that \(T\) is compact; Step 4: see that there is a \(J^`\) for each such \(\{C_j \vert j \in J\}\).
Step 1:
Let us suppose that for each such \(\{C_j \vert j \in J\}\), there is a \(J^`\).
Step 2:
Let \(\{U_j \vert j \in J\}\), where \(J\) is a possibly uncountable index set, be any open cover of \(T\).
Let us take \(\{T \setminus U_j \vert j \in J\}\), which is a set of some closed subsets.
\(\cap_{j \in J} (T \setminus U_j) = \emptyset\), because if there was a \(t \in \cap_{j \in J} (T \setminus U_j)\), \(t \in T \setminus U_j\) for each \(j \in J\), so, \(t \notin U_j\) for each \(j \in J\), so, \(t \notin \cup_{j \in J} U_j = T\), a contradiction.
By the supposition, there is a nonempty finite subset, \(J^` \subseteq J\), such that \(\cap_{j \in J^`} (T \setminus U_j) = \emptyset\).
\(\cup_{j \in J^`} U_j = T\), because if there was a \(t \in T \setminus \cup_{j \in J^`} U_j\), \(t \notin \cup_{j \in J^`} U_j\), so, \(t \notin U_j\) for each \(j \in J^`\), so, \(t \in T \setminus U_j\) for each \(j \in J^`\), so, \(t \in \cap_{j \in J^`} (T \setminus U_j) = \emptyset\), a contradiction.
So, \(\{U_j \vert j \in J^`\}\) is a finite subcover.
So, \(T\) is compact.
Step 3:
Let us suppose that \(T\) is compact.
Step 4:
\(\cup_{j \in J} (T \setminus C_j) = T\), because while \(\cup_{j \in J} (T \setminus C_j) \subseteq T\) is obvious, for each \(t \in T\), \(t \notin C_j\) for a \(j \in J\), because otherwise, \(t \in \cap_{j \in J} C_j = \emptyset\), a contradiction, so, \(t \in T \setminus C_j\) for a \(j \in J\), so, \(t \in \cup_{j \in J} (T \setminus C_j)\), so, \(T \subseteq \cup_{j \in J} (T \setminus C_j)\).
So, \(\{T \setminus C_j \vert j \in J\}\) is an open cover of \(T\).
There is a finite subcover, \(\{T \setminus C_j \vert j \in J^`\}\), because \(T\) is compact.
\(\cap_{j \in J^`} C_j = \emptyset\), because if there was a \(t \in \cap_{j \in J^`} C_j\), \(t \in C_j\) for each \(j \in J^`\), so, \(t \notin T \setminus C_j\) for each \(j \in J^`\), so, \(t \notin \cup_{j \in J^`} T \setminus C_j = T\), a contradiction.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for map from metric space minus point into metric space, iff for each sequence on domain that converges to point, its image converges to codomain point, map converges w.r.t. point to codomain point
Topics
About:
metric space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any map from any metric space minus any point into any metric space, if and only if for each sequence on the domain that converges to the point, its image converges to any same codomain point, the map converges with respect to the point to the codomain point.
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:
\(M_1\): \(\in \{\text{ the metric spaces }\}\), with the topology induced by the metric
\(M_2\): \(\in \{\text{ the metric spaces }\}\), with the topology induced by the metric
\(m_1\): \(\in M_1\), such that \(\{m_1\} \subseteq M_1 \notin \{\text{ the open subsets of } M_1\}\)
\(m_2\): \(\in M_2\)
\(f\): \(: M_1 \setminus \{m_1\} \to M_2\)
//
Statements:
\(\forall s: J \to M_1 \setminus \{m_1\} \in \{\text{ the sequences that converge to } m_1\} (f \circ s: J \to M_2 \in \{\text{ the sequences that converge to } m_2\})\)
\(\iff\)
\(lim_{m'_1 \to m_1} f (m'_1) = m_2\)
//
2: Note
The reason why the case that \(\{m_1\} \subseteq M_1\) is open is excluded is that for that case, each point of \(M_2\) will be a convergence of \(f\) with respect to \(m_1\), as is mentioned in Note for the definition of convergence of map from topological space minus point into topological space with respect to point, and so, \(lim_{m'_1 \to m_1} f (m'_1)\) will not make sense in general and also there may not be any sequence that converges to \(m_1\), so, the condition can become vacuous.
The mechanism of this proposition is essentially the same with that of the proposition that for any map between any metric spaces and any domain point, if and only if for each sequence on the domain that converges to the point, its image converges to the image of the point, the map is continuous at the point: the difference is that for this proposition, as the point is not on the domain, "continuous at the point" does not make sense.
3: Proof
Whole Strategy: Step 1: suppose that \(f \circ s\) converges to \(m_2\) for each \(s\); Step 2: for each \(\epsilon\), take \(\delta\) such that \(f (B_{m_1, \delta} \setminus \{m_1\}) \subseteq B_{m_2, \epsilon}\); Step 3: suppose that \(f\) converges to \(m_2\) with respect to \(m_1\); Step 4: see that \(f \circ s\) converges to \(m_2\) for each \(s\).
Step 1:
Let us suppose that for each \(s: J \to M_1\) that converges to \(m_1\), \(f \circ s: J \to M_2\) converges to \(m_2\).
Step 2:
Step 2 Strategy: for each \(\epsilon\), take \(\delta\) such that \(f (B_{m_1, \delta} \setminus \{m_1\}) \subseteq B_{m_2, \epsilon}\), as follows: Step 2-1: take the set of the infinite sequences that converge to \(m_1\), \(\{s_l \vert l \in L\}\), for each \(l \in L\), take \(N_l\) such that for each \(N_l \lt n\), \(f \circ s_l (n) \in B_{m_2, \epsilon}\), take \(\delta_l := Min (\{dist (m_1, s_l ({J_l}_n)) \vert n \le N_l \land f \circ s_l ({J_l}_n) \notin B_{m_2, \epsilon}\})\), and take \(\delta := Inf (\{\delta_l \vert l \in L^`\})\), where \(L^` \subseteq L\); Step 2-2: see that \(0 \lt \delta\); Step 2-3: see that \(f (B_{m_1, \delta} \setminus \{m_1\}) \subseteq B_{m_2, \epsilon}\); Step 2-4: see that the convergence is unique.
Step 2-1:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
Let \(\{s_l: J_l \to M_1 \setminus \{m_1\} \vert l \in L\}\) be the set of the infinite sequences that converge to \(m_1\), where \(L\) is a possibly uncountable index set and \(J_l \subseteq \mathbb{N}\).
\(L\) is not empty, because there is at least \(s_l: \mathbb{N} \to M_1 \setminus \{m_1\}\) such that \(s_l (n) \in B_{m_1, (1 / 2)^n} \setminus \{m_1\}\): as \(\{m_1\} \subseteq M_1\) is not open, there is such an \(s_l (n)\), because if \(B_{m_1, (1 / 2)^n} \setminus \{m_1\} = \emptyset\), \(B_{m_1, (1 / 2)^n} = \{m_1\}\), which would imply that \(\{m_1\}\) was open.
Let \(l \in L\) be any.
There is an \(N_l \subseteq \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N_l \lt n\), \(f \circ s_l ({J_l}_n) \in B_{m_2, \epsilon}\), because \(f \circ s_l\) converges to \(m_2\).
Let \(L^` := \{l \in L \vert \{n \in \mathbb{N} \vert n \le N_l \land f \circ s_l ({J_l}_n) \notin B_{m_2, \epsilon}\} \neq \emptyset\}\).
When \(L^` = \emptyset\), let \(\delta \in \mathbb{R}\) be any such that \(0 \lt \delta\). Then, \(f (B_{m_1, \delta} \setminus \{m_1\}) \subseteq B_{m_2, \epsilon}\), because if there was an \(m'_1 \in B_{m_1, \delta} \setminus \{m_1\}\) such that \(f (m'_1) \notin B_{m_2, \epsilon}\), there would be an \(s_l: \mathbb{N} \to M_1 \setminus \{m_1\}\) such that \(s_l (0) = m'_1\) and \(s_l (n) \in B_{m_1, (1 / 2)^n} \setminus \{m_1\}\) for each \(0 \lt n\), which would converge to \(m_1\) and \(l \in L^`\): \(1 \le N_l\) and \(s_l (0) \notin B_{m_2, \epsilon}\), a contradiction.
Let us suppose that \(L^` \neq \emptyset\), hereafter.
Let \(l \in L^`\) be any.
Let us take \(\delta_l := Min (\{dist (m_1, s_l ({J_l}_n)) \vert n \in \mathbb{N} \text{ such that } n \le N_l \land f \circ s_l ({J_l}_n) \notin B_{m_2, \epsilon}\})\), which exists, because \(\{n \in \mathbb{N} \vert n \le N_l \land f \circ s_l ({J_l}_n) \notin B_{m_2, \epsilon}\}\) is not empty.
Let us take \(\delta := Inf (\{\delta_l \vert l \in L^`\})\), which exists because \(\{\delta_l \vert l \in L^`\}\) is not empty and lower bounded by \(0\).
Step 2-2:
While \(0 \le \delta\), let us see that \(0 \lt \delta\).
Let us suppose that \(0 = \delta\).
Let us take \(s: \mathbb{N} \to M_1 \setminus \{m_1\}\) as follows.
There would be a \(\delta_{l_0} \lt (1 / 2)^0\), by the proposition that for any linearly-ordered set and any subset, any element of the set is the infimum of the subset if and only if the element is equal to or smaller than each element of the subset and for each element of the set larger than the element, there is an element of the subset smaller: \(\delta \lt (1 / 2)^0\).
That would means that \(f \circ s_{l_0} ({J_{l_0}}_n) \notin B_{m_2, \epsilon}\) and \(dist (s_{l_0} ({J_{l_0}}_n), m_1) = \delta{l_0}\) for an \(n \le N_{l_0}\), and let \(s (0) = s_{l_0} ({J_{l_0}}_n)\), so, \(dist (s (0), m_1) = \delta{l_0} \lt (1 / 2)^0\) and \(f \circ s (0) \notin B_{m_2, \epsilon}\).
Then, there would be a \(\delta_{l_1} \lt (1 / 2)^1\), as before.
That would mean that \(f \circ s_{l_1} ({J_{l_1}}_n) \notin B_{m_2, \epsilon}\) and \(dist (s_{l_1} ({J_{l_1}}_n), m_1) = \delta{l_1}\) for an \(n \le N_{l_1}\), and let \(s (1) = s_{l_1} ({J_{l_1}}_n)\), so, \(dist (s (1), m_1) = \delta{l_1} \lt (1 / 2)^1\) and \(f \circ s (1) \notin B_{m_2, \epsilon}\).
In general, there would be a \(\delta_{l_j} \lt (1 / 2)^j\), as before.
That would mean that \(f \circ s_{l_j} ({J_{l_j}}_n) \notin B_{m_2, \epsilon}\) and \(dist (s_{l_j} ({J_{l_j}}_n), m_1) = \delta{l_j}\) for an \(n \le N_{l_j}\), and let \(s (j) = s_{l_j} ({J_{l_j}}_n)\), so, \(dist (s (j), m_1) = \delta{l_j} \lt (1 / 2)^j\) and \(f \circ s (j) \notin B_{m_2, \epsilon}\).
Thus \(s\) had been defined.
\(s\) converged to \(m_1\), because \(dist (s (j), m_1) \lt (1 / 2)^j\).
But \(f \circ s (j) \notin B_{m_2, \epsilon}\) for each \(j \in \mathbb{N}\), so, \(f \circ s\) would not converge to \(m_2\), a contradiction against the supposition.
So, \(0 \lt \delta\).
Step 2-3:
Then, \(f (B_{m_1, \delta} \setminus \{m_1\}) \subseteq B_{m_2, \epsilon}\), because if there was a \(m'_1 \in B_{m_1, \delta} \setminus \{m_1\}\) such that \(f (m'_1) \notin B_{m_2, \epsilon}\), there would be an \(s_l\) such that \(s_l ({J_l}_{N_l}) = m'_1\), which would mean that \(\delta \le \delta_l \le dist (m_1, m'_1) \lt \delta\), a contradiction.
That means that \(f\) converges to \(m_2\) with respect to \(m_1\).
Step 2-4:
The convergence is unique, because \(M_2\) is Hausdorff, by the proposition that the topological space induced by any metric is Hausdorff, and Note for the definition of convergence of map from topological space minus point into topological space with respect to point applies.
So, \(lim_{m'_1 \to m_1} f (m'_1) = m_2\).
Step 3:
Let us suppose that \(f\) converges to \(m_2\) with respect to \(m_1\).
Step 4:
Let \(s: J \to M_1 \setminus \{m_1\}\) be any sequence that converges to \(m_1\).
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is a \(\delta \in \mathbb{R}\) such that \(0 \lt \delta\) and \(f (B_{m_1, \delta} \setminus \{m_1\}) \subseteq B_{m_2, \epsilon}\), because \(f\) converges to \(m_2\) with respect to \(m_1\).
There is an \(N \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N \lt n\), \(s (J_n) \in B_{m_1, \delta} \setminus \{m_1\}\), because \(s\) converges to \(m_1\).
Then, for each \(n \in \mathbb{N}\) such that \(N \lt n\), \(f \circ s (J_n) \in B_{m_2, \epsilon}\).
That means that \(f \circ s\) converges to \(m_2\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for map between metric spaces and domain point, iff for each sequence on domain that converges to point, its image converges to image of point, map is continuous at point
Topics
About:
metric space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any map between any metric spaces and any domain point, if and only if for each sequence on the domain that converges to the point, its image converges to the image of the point, the map is continuous at the point.
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:
\(M_1\): \(\in \{\text{ the metric spaces }\}\)
\(M_2\): \(\in \{\text{ the metric spaces }\}\)
\(f\): \(: M_1 \to M_2\)
\(m_1\): \(\in M_1\)
//
Statements:
\(\forall s: J \to M_1 \in \{\text{ the sequences that converge to } m_1\} (f \circ s: J \to M_2 \in \{\text{ the sequences that converge to } f (m_1)\})\)
\(\iff\)
\(f \in \{\text{ the continuous maps }\}\)
//
2: Proof
Whole Strategy: Step 1: suppose that \(f \circ s\) converges to \(f (m_1)\) for each \(s\); Step 2: for each \(\epsilon\), take \(\delta\) such that \(f (B_{m_1, \delta}) \subseteq B_{f (m_1), \epsilon}\); Step 3: suppose that \(f\) is continuous; Step 4: see that \(f \circ s\) converges to \(f (m_1)\) for each \(s\).
Step 1:
Let us suppose that for each \(s: J \to M_1\) that converges to \(m_1\), \(f \circ s: J \to M_2\) converges to \(f (m_1)\).
Step 2:
Step 2 Strategy: for each \(\epsilon\), take \(\delta\) such that \(f (B_{m_1, \delta}) \subseteq B_{f (m_1), \epsilon}\), as follows: Step 2-1: take the set of the infinite sequences that converge to \(m_1\), \(\{s_l \vert l \in L\}\), for each \(l \in L\), take \(N_l\) such that for each \(N_l \lt n\), \(f \circ s_l ({J_l}_n) \in B_{f (m_1), \epsilon}\), take \(\delta_l := Min (\{dist (m_1, s_l ({J_l}_n)) \vert n \le N_l \land f \circ s_l ({J_l}_n) \notin B_{f (m_1), \epsilon}\})\), and take \(\delta := Inf (\{\delta_l \vert l \in L^`\})\), where \(L^` \subseteq L\); Step 2-2: see that \(0 \lt \delta\); Step 2-3: see that \(f (B_{m_1, \delta}) \subseteq B_{f (m_1), \epsilon}\).
Step 2-1:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
Let \(\{s_l: J_l \to M_1 \vert l \in L\}\) be the set of the infinite sequences that converge to \(m_1\), where \(L\) is a possibly uncountable index set and \(J_l \subseteq \mathbb{N}\).
\(L\) is not empty, because there is at least \(s_l: \mathbb{N} \to M_1\) constant to \(m_1\).
Let \(l \in L\) be any.
There is an \(N_l \subseteq \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N_l \lt n\), \(f \circ s_l ({J_l}_n) \in B_{f (m_1), \epsilon}\), because \(f \circ s_l\) converges to \(f (m_1)\).
Let \(L^` := \{l \in L \vert \{n \in \mathbb{N} \vert n \le N_l \land f \circ s_l ({J_l}_n) \notin B_{f (m_1), \epsilon}\} \neq \emptyset\}\).
When \(L^` = \emptyset\), let \(\delta \in \mathbb{R}\) be any such that \(0 \lt \delta\). Then, \(f (B_{m_1, \delta}) \subseteq B_{f (m_1), \epsilon}\), because if there was an \(m'_1 \in B_{m_1, \delta}\) such that \(f (m'_1) \notin B_{f (m_1), \epsilon}\), there would be an \(s_l: \mathbb{N} \to M_1\) such that \(s_l (0) = m'_1\) and \(s_l (n) = m_1\) for each \(0 \lt n\), which would converge to \(m_1\) and \(l \in L^`\): \(1 \le N_l\) and \(s_l (0) \notin B_{f (m_1), \epsilon}\), a contradiction.
Let us suppose that \(L^` \neq \emptyset\), hereafter.
Let \(l \in L^`\) be any.
Let us take \(\delta_l := Min (\{dist (m_1, s_l ({J_l}_n)) \vert n \in \mathbb{N} \text{ such that } n \le N_l \land f \circ s_l ({J_l}_n) \notin B_{f (m_1), \epsilon}\})\), which exists, because \(\{n \in \mathbb{N} \vert n \le N_l \land f \circ s_l ({J_l}_n) \notin B_{f (m_1), \epsilon}\}\) is not empty.
Let us take \(\delta := Inf (\{\delta_l \vert l \in L^`\})\), which exists because \(\{\delta_l \vert l \in L^`\}\) is not empty and lower bounded by \(0\).
Step 2-2:
While \(0 \le \delta\), let us see that \(0 \lt \delta\).
Let us suppose that \(0 = \delta\).
Let us take \(s: \mathbb{N} \to M_1\) as follows.
There would be a \(\delta_{l_0} \lt (1 / 2)^0\), by the proposition that for any linearly-ordered set and any subset, any element of the set is the infimum of the subset if and only if the element is equal to or smaller than each element of the subset and for each element of the set larger than the element, there is an element of the subset smaller: \(\delta \lt (1 / 2)^0\).
That would mean that \(f \circ s_{l_0} ({J_{l_0}}_n) \notin B_{f (m_1), \epsilon}\) and \(dist (s_{l_0} ({J_{l_0}}_n), m_1) = \delta{l_0}\) for an \(n \le N_{l_0}\), and let \(s (0) = s_{l_0} ({J_{l_0}}_n)\), so, \(dist (s (0), m_1) = \delta{l_0} \lt (1 / 2)^0\) and \(f \circ s (0) \notin B_{f (m_1), \epsilon}\).
Then, there would be a \(\delta_{l_1} \lt (1 / 2)^1\), as before.
That would mean that \(f \circ s_{l_1} ({J_{l_1}}_n) \notin B_{f (m_1), \epsilon}\) and \(dist (s_{l_1} ({J_{l_1}}_n), m_1) = \delta{l_1}\) for an \(n \le N_{l_1}\), and let \(s (1) = s_{l_1} ({J_{l_1}}_n)\), so, \(dist (s (1), m_1) = \delta{l_1} \lt (1 / 2)^1\) and \(f \circ s (1) \notin B_{f (m_1), \epsilon}\).
In general, there would be a \(\delta_{l_j} \lt (1 / 2)^j\), as before.
That would mean that \(f \circ s_{l_j} ({J_{l_j}}_n) \notin B_{f (m_1), \epsilon}\) and \(dist (s_{l_j} ({J_{l_j}}_n), m_1) = \delta{l_j}\) for an \(n \le N_{l_j}\), and let \(s (j) = s_{l_j} ({J_{l_j}}_n)\), so, \(dist (s (j), m_1) = \delta{l_j} \lt (1 / 2)^j\) and \(f \circ s (j) \notin B_{f (m_1), \epsilon}\).
Thus \(s\) had been defined.
\(s\) converged to \(m_1\), because \(dist (s (j), m_1) \lt (1 / 2)^j\).
But \(f \circ s (j) \notin B_{f (m_1), \epsilon}\) for each \(j \in \mathbb{N}\), so, \(f \circ s\) would not converge to \(f (m_1)\), a contradiction against the supposition.
So, \(0 \lt \delta\).
Step 2-3:
Then, \(f (B_{m_1, \delta}) \subseteq B_{f (m_1), \epsilon}\), because if there was a \(m'_1 \in B_{m_1, \delta}\) such that \(f (m'_1) \notin B_{f (m_1), \epsilon}\), there would be an \(s_l\) such that \(s_l ({J_l}_{N_l}) = m'_1\), which would mean that \(\delta \le \delta_l \le dist (m_1, m'_1) \lt \delta\), a contradiction.
That means that \(f\) is continuous.
Step 3:
Let us suppose that \(f\) is continuous.
Step 4:
Let \(s: J \to M_1\) be any sequence that converges to \(m_1\).
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is a \(\delta \in \mathbb{R}\) such that \(0 \lt \delta\) and \(f (B_{m_1, \delta}) \subseteq B_{f (m_1), \epsilon}\), because \(f\) is continuous.
There is an \(N \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N \lt n\), \(s (J_n) \in B_{m_1, \delta}\), because \(s\) converges to \(m_1\).
Then, for each \(n \in \mathbb{N}\) such that \(N \lt n\), \(f \circ s (J_n) \in B_{f (m_1), \epsilon}\).
That means that \(f \circ s\) converges to \(f (m_1)\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for \(1\)-dimensional Euclidean topological space and open subset, open subset is disjoint union of countable open intervals
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for the \(1\)-dimensional Euclidean topological space and any open subset, the open subset is the union of some disjoint countable open intervals.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(U\): \(\in \{\text{ the open subsets of } T\}\)
//
Statements:
\(\exists J \in \{\text{ the countable index sets }\}, \exists \{I_j \in \{\text{ the open intervals }\} \vert j \in J\} (\forall j, j' \in J \text{ such that } j \neq j' (I_j \cap I_{j'} = \emptyset) \land U = \cup_{j \in J} I_j)\)
//
2: Proof
Whole Strategy: Step 1: for each \(u \in U\), take the maximum open interval around \(u\) contained in \(U\), \(I_u\), and take \(\{I_u \vert u \in U\}\); Step 2: see that \(\{I_u \vert u \in U\}\) is disjoint, take a rational number in each \(I_u\), take an injection, \(f: \{I_u \vert u \in U\} \to \mathbb{N}\), and a bijection, \(f': \mathbb{N} \to \{I_u \vert u \in U\}\), take \(J = \mathbb{N}\) and \(f' (J) = \{I_j \vert j \in J\}\), and see that \(U = \cup_{j \in J} I_j\).
Step 1:
For each \(u \in U\), take the maximum open interval around \(u\) contained in \(U\), \(I_u \subseteq \mathbb{R}\), which is determined uniquely, because that is the union of the set of the open intervals around \(u\) contained in \(U\): the set is nonempty, because as \(U\) is open, there is at least \(1\) \((u - \epsilon, u + \epsilon) \subseteq U\), and the union is indeed an interval, by the proposition that the union of any possibly uncountable dichotomically nondisjoint set of \(\mathbb{R}\) intervals is a \(\mathbb{R}\) interval: the set is dichotomically nondisjoint, because \(u\) is contained in each element, and the union is an open interval, because it is open as the union of some open subsets, and the union contains \(u\) and is contained in \(U\), and the union is the maximum, because any open interval around \(u\) contained in \(U\) is an element of the set and is contained in the union.
Let us take \(\{I_u \vert u \in U\}\).
Note that for some \(u \neq u'\), \(I_u = I_{u'}\) is possible, and they are a single element, so, \(\{I_u \vert u \in U\}\) does not necessarily really have the points of \(U\) number elements.
Step 2:
Let us see that \(\{I_u \vert u \in U\}\) is disjoint.
Let \(I_u, I_{u'} \in \{I_u \vert u \in U\}\) be any such that \(I_u \neq I_{u'}\).
Let us suppose that \(I_u \cap I_{u'} \neq \emptyset\).
There would be an \(r \in I_u \cap I_{u'}\).
There would be \(I_r\) and \(I_u, I_{u'} \subseteq I_r\), because \(I_u\) and \(I_{u'}\) were some open intervals around \(r\) contained in \(U\) and \(I_r\) was the union of all the such open intervals, which would imply that \(u, u' \in I_r\), so, \(I_r \subseteq I_u, I_{u'}\), because \(I_r\) was an open interval around \(u\) contained in \(U\) while \(I_u\) was the union of all the such open intervals, and \(I_r\) was an open interval around \(u'\) contained in \(U\) while \(I_{u'}\) was the union of all the such open intervals.
So, \(I_r \subseteq I_u, I_{u'} \subseteq I_r\), so, \(I_u = I_{u'} = I_r\), a contradiction against \(I_u \neq I_{u'}\).
So, \(I_u \cap I_{u'} = \emptyset\).
So, \(\{I_u \vert u \in U\}\) is disjoint.
There is the map, \(g: \{I_u \vert u \in U\} \to Pow (\mathbb{Q})\), that maps \(I_u\) to the set of the points of \(\mathbb{Q}\) contained in \(I_u\).
Each \(g (I_u)\) is nonempty, because \(I_u\) contains at least \(1\) rational number.
So, by the axiom of choice, there is a map, \(g': \{I_u \vert u \in U\} \to \mathbb{Q}\), where \(g' (I_j) \in g (I_u)\), which is an injection, because \(\{I_u \vert u \in U\}\) is disjoint.
As \(\mathbb{Q}\) is countable, there is a bijection, \(h: \mathbb{N} \to \mathbb{Q}\).
Let \(f: \{I_u \vert u \in U\} \to \mathbb{N} = h^{-1} \circ g'\), which is an injection, by the proposition that any finite composition of injections is an injection.
So, there is a bijection, \(f': \mathbb{N} \to \{I_u \vert u \in U\}\), by the proposition that for any infinite set, if there is any injection from the set into the natural numbers set, there is a bijection from the natural numbers set onto the set.
Let \(J := \mathbb{N}\).
Let \(f' (J) = \{I_j = f' (j) \vert j \in J\}\).
\(\{I_j \vert j \in J\} = \{I_u \vert u \in U\}\), because \(f'\) is a bijection onto \(\{I_u \vert u \in U\}\).
\(\{I_j \vert j \in J\}\) is disjoint, because \(\{I_u \vert u \in U\}\) is so.
Let us see that \(U = \cup_{j \in J} I_j\).
For each \(u \in U\), \(u \in I_u\), but there is a \(j \in J\) such that \(I_j = I_u\), because \(\{I_j \vert j \in J\} = \{I_u \vert u \in U\}\), so, \(u \in I_j\).
So, \(u \in \cup_{j \in J} I_j\).
So, \(U \subseteq \cup_{j \in J} I_j\).
For each \(r \in \cup_{j \in J} I_j\), \(r \in I_j\) for a \(j \in J\), but \(I_j = I_u\) for a \(u \in U\), and \(r \in I_j = I_u \subseteq U\).
So, \(r \in U\).
So, \(\cup_{j \in J} I_j \subseteq U\).
So, \(U = \cup_{j \in J} I_j\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for infinite set, if there is injection from set into natural numbers set, there is bijection from natural numbers set onto set
The table of contents of this article
Main Body
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for continuous map between topological spaces, boundary of preimage of subset is contained in but not necessarily equal to preimage of boundary of subset
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any continuous map between any topological spaces, the boundary of the preimage of any subset of the codomain is contained in but not necessarily equal to the preimage of 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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(S_2\): \(\subseteq T_2\)
//
Statements:
\(Bou (f^{-1} (S_2)) \subseteq f^{-1} (Bou (S_2))\)
\(\land\)
Not necessarily "\(Bou (f^{-1} (S_2)) = f^{-1} (Bou (S_2))\)"
//
2: Proof
Whole Strategy: Step 1: see that \(Bou (f^{-1} (S_2)) = \overline{f^{-1} (S_2)} \setminus Int (f^{-1} (S_2))\) and \(Bou (S_2) = \overline{S_2} \setminus Int (S_2)\); Step 2: apply the proposition that for any map, the preimage of any subset minus any subset is the preimage of the 1st subset minus the preimage of the 2nd subset, the proposition that for any continuous map between any topological spaces, the closure of the map preimage of any subset of the codomain is contained in but not necessarily equal to the preimage of the closure of the subset, and 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; Step 3: see an example that "\(Bou (f^{-1} (S_2)) = f^{-1} (Bou (S_2))\)" does not hold.
Step 1:
\(Bou (f^{-1} (S_2)) = \overline{f^{-1} (S_2)} \setminus Int (f^{-1} (S_2))\), by the proposition that for any topological space and any subset, the space is the disjoint union of the interior of the subset, the boundary of the subset, and the complement of the closure of the subset.
\(Bou (S_2) = \overline{S_2} \setminus Int (S_2)\), likewise.
Step 2:
\(f^{-1} (Bou (S_2)) = f^{-1} (\overline{S_2} \setminus Int (S_2)) = f^{-1} (\overline{S_2}) \setminus f^{-1} (Int (S_2))\), by the proposition that for any map, the preimage of any subset minus any subset is the preimage of the 1st subset minus the preimage of the 2nd subset.
\(\overline{f^{-1} (S_2)} \subseteq f^{-1} (\overline{S_2})\), by the proposition that for any continuous map between any topological spaces, the closure of the map preimage of any subset of the codomain is contained in but not necessarily equal to the preimage of the closure of the subset.
\(f^{-1} (Int (S_2)) \subseteq Int (f^{-1} (S_2))\), by 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.
So, \(\overline{f^{-1} (S_2)} \setminus Int (f^{-1} (S_2)) \subseteq f^{-1} (\overline{S_2}) \setminus f^{-1} (Int (S_2))\).
But the left hand side is \(Bou (f^{-1} (S_2))\), by Step 1, while the right hand side is \(f^{-1} (Bou (S_2))\), as has been seen above.
So, \(Bou (f^{-1} (S_2)) \subseteq f^{-1} (Bou (S_2))\).
Step 3:
Let us see an example that "\(Bou (f^{-1} (S_2)) = f^{-1} (Bou (S_2))\)" 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.
\(f^{-1} (S_2) = [-1, 1]\) and \(Bou (f^{-1} (S_2)) = \{-1, 1\}\).
But \(Bou (S_2) = \{0, 1\}\) and \(f^{-1} (Bou (S_2)) = \{-1, 0, 1\}\).
So, \(Bou (f^{-1} (S_2)) = \{-1, 1\} \neq \{-1, 0, 1\} = f^{-1} (Bou (S_2))\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
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
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the continuous maps }\}\)
\(S_2\): \(\subseteq T_2\)
//
Statements:
\(f^{-1} (Int (S_2)) \subseteq Int (f^{-1} (S_2))\)
\(\land\)
Not necessarily "\(f^{-1} (Int (S_2)) = Int (f^{-1} (S_2))\)"
//
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, \(f^{-1} (Int (S_2)) = (-1, 0) \cup (0, 1) \neq (-1, 1) = Int (f^{-1} (S_2))\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
