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3.96MB jiangxiubao 2010-11-05 15:45:42
• Exercise 22.6 Recall that $$\mathbb{R}_{K}$$ denotes the real line in the $$K$$-topology. Let $$Y$$ be the quotient space obtained from $$\mathbb{R}_K$$ by collapsing the set $$K$$ to a point; let $$p...  Exercise 22.6 Recall that \(\mathbb{R}_{K}$$ denotes the real line in the $$K$$-topology. Let $$Y$$ be the quotient space obtained from $$\mathbb{R}_K$$ by collapsing the set $$K$$ to a point; let $$p: \mathbb{R}_K \rightarrow Y$$ be the quotient map.
(a) Show that $$Y$$ satisfies the $$T_1$$ axiom, but is not Hausdorff.
(b) Show that $$p \times p: \mathbb{R}_K \times \mathbb{R}_K \rightarrow Y \times Y$$ is not a quotient map.
Comment This exercise shows that the product map of two quotient maps is not necessarily a quotient map.
Proof: (a) At first, we will clarify the forms of open sets in the quotient space $$Y$$, which are defined as the images of saturated open sets in $$\mathbb{R}_K$$ under the quotient map $$p$$. Assume the set $$K$$ coalesces to $$\alpha$$, $$Y$$ can be written as: $$Y = (\mathbb{R} - K) \cup \{\alpha\}$$. For any $$x$$ in $$\mathbb{R} - K$$, $$p^{-1}(x) = x$$ and $$p^{-1}(\alpha) = K$$. Then the saturated open sets in $$\mathbb{R}_K$$ have the following two forms:
open set $$U$$ of $$\mathbb{R}_K$$ which contains $$K$$;$$U - K$$ with $$U$$ being arbitrary open set in $$\mathbb{R}_K$$.
Then their images under the quotient map $$p$$ are
$$(U - K) \cup \{\alpha\}$$ with $$K \subsetneq U$$$$U - K$$
which comprise the quotient topology on $$Y$$. To prove $$Y$$ satisfies the $$T_1$$-axiom, by referring to Theorem 17.8, we only need to show that one-point set $$\{x_0\}$$ is closed. Then finite union of such closed singletons is also closed. To achieve this, there are two cases to be discussed.
If $$x_0 = \alpha$$, for any point $$x \in Y$$ and $$x \neq x_0$$, i.e. $$x \in \mathbb{R} - K$$, there exists an open set $$U - K$$ in $$Y$$ containing $$x$$, which does not contain $$x_0$$. Therefore, for all $$x \in \mathbb{R} - K$$, it does not belong to the closure of $$\{\alpha\}$$. Hence $$\{\alpha\}$$ is closed.  If $$x_0 \in \mathbb{R} - K$$, there are further two sub-cases:
For any $$x \in \mathbb{R} - K$$ and $$x \neq x_0$$, because $$\mathbb{R}_K$$ is Hausdorff, there exists open sets $$U$$ and $$V$$ in $$\mathbb{R}_K$$, such that $$x_0 \in U$$, $$x \in V$$ and $$U \cap V = \Phi$$. Then $$x_0 \in (U - K)$$, $$x \in (V - K)$$ and $$(U - K) \cap (V - K) = \Phi$$, where both $$U - K$$ and $$V - K$$ are open in $$Y$$. Hence $$\{x_0\} \cap (V - K) = \Phi$$.  For $$x = \alpha$$, the open set containing $$x$$ has the form $$(U - K) \cup \{\alpha\}$$ where $$U$$ is an open set in $$\mathbb{R}_K$$ containing $$K$$. Then,
when $$x_0 \in (-\infty, 0]$$, let $$U = (0, 2)$$;when $$x_0 \in (0, 1]$$, let $$U = (0,x_0) \cup (x_0, \frac{3}{2})$$;when $$x_0 \in (1, +\infty)$$, let $$U = (0,x_0)$$,such that $$K \subset U$$ and $$\{x_0\} \cap ((U - K) \cup \{\alpha\}) = \Phi$$. Combining the above two sub-cases, we have for any $$x \neq x_0$$ in $$Y$$, it does not belong to the closure of $$\{x_0\}$$. Hence $$\{x_0\}$$ is closed.
Summarize the above cases, one-point set in $$Y$$ is closed. Hence $$Y$$ satisfies the $$T_1$$-axiom.
Next, we will show $$Y​$$ is not Hausdorff.
Let $$x_1, x_2 \in Y$$, $$x_1 = \alpha$$ and $$x_2 = 0$$. For any open set in $$Y$$ containing 0 but not $$\alpha$$, it must have the form $$V - K$$ with $$V$$ being open in $$\mathbb{R}_K$$. Then there exists an open interval $$(a_2, b_2)$$ with $$a_2 < 0$$ and $$b_2 > 0$$ such that $$0 \in (a_2, b_2)$$ and $$(a_2, b_2) \subset V$$. We can find an $$n_0 \in \mathbb{Z}_+$$ such that $$\frac{1}{n_0} < b_2$$ and hence $$\frac{1}{n_0} \in (a_2, b_2)$$. Meanwhile, any open set containing $$\alpha$$ has the form $$(U - K ) \cup \{\alpha\}$$ with $$U$$ being open in $$\mathbb{R}_K$$ and $$K \subsetneq U$$. Then there exists an open interval $$(a_1,b_1)$$ such that $$\frac{1}{n_0} \in (a_1, b_1)$$ and $$(a_1, b_1) \subset U$$. Therefore, $$(a_1,b_1) \cap (a_2,b_2) \neq \Phi$$ and $$U \cap V \neq \Phi$$, especially, $$(U-K)\cap(V-K)\neq\Phi$$. Hence, $$((U-K)\cup\{\alpha\}) \cap (V-K) \neq \Phi$$. Therefore, for any open set containing 0, there is no open set containing $$\alpha$$ which has no intersection with it. So $$Y$$ is not Hausdorff.
(b) To prove this part, Exercise 13 in Section 17 should be adopted, which is presented below:

$$X$$ is Hausdorff if and only if the diagonal $$\Delta = \{x \times x \vert x \in X \}$$ is closed in $$X \times X$$.

If $$X$$ is Hausdorff, for any $$x_1, x_2 \in X$$ and $$x_1 \neq x_2$$, there exist $$U$$ and $$V$$ open in $$X$$ such that $$x_1 \in U$$, $$x_2 \in V$$ and $$U \cap V = \Phi$$. Because $$U$$ and $$V$$ have no common points, $$(U \times V) \cap \Delta = \Phi$$. Then according to Theorem 17.5, $$(x_1, x_2)$$ does not belong to the closure of $$\Delta$$. Because $$x_1$$ and $$x_2$$ are arbitrary two different points in $$X$$, $$\Delta$$ is closed.On the contrary, if $$\Delta$$ is closed, for all $$x_1, x_2 \in X$$ and $$x_1 \neq x_2$$, there exists an open set $$W$$ in $$X \times X$$ containing $$(x_1,x_2)$$ such that $$W \cap \Delta = \Phi$$. Then there exists a basis element $$U \times V$$ in $$X \times X$$ such that $$(x_1, x_2) \subset U \times V \subset W$$. Hence $$x_1 \in U$$ and $$x_2 \in V$$. Because $$(U \times V) \cap \Delta = \Phi$$, $$U \cap V = \Phi$$. Because $$x_1$$ and $$x_2$$ are arbitrary two different points in $$X$$, $$X$$ is Hausdorff.
With the proved S17E13 and the obtained conclusion in part (a) that $$Y$$ is no Hausdorff, we know that the diagonal set $$\Delta$$ is not closed in $$Y \times Y$$. Meanwhile, because its preimage $$(p \times p)^{-1}(\Delta) = \{x \times x \vert x \in \mathbb{R}\}$$ is closed in $$\mathbb{R}_K \times \mathbb{R}_K$$, the product map $$p \times p$$ is not a quotient map.
Finally, the following figure illustrates the original space $$\mathbb{R}_K$$ and the quotient space $$Y$$. The transformation from $$\mathbb{R}_K$$ to $$Y$$ can be considered as merging a countable number of knots on a rope.

PS: Because the world we are living in is Hausdorff, Diagon Alley is always closed.

转载于:https://www.cnblogs.com/peabody/p/10428601.html
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weixin_34019144 2019-02-24 23:04:00
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5.41MB u014694280 2014-04-13 20:00:02
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9.73MB john9072 2009-06-18 19:28:20
• 5.46MB Danguishu 2020-08-06 21:45:04
• Theorem 16.3 If $$A$$ is a subspace of $$X$$ and $$B$$ is a subspace of $$Y$$, then the product topology on $$A \times B$$ is the same as the topology $$A \times B$$ inherits as a subspace of $$X \tim...  Theorem 16.3 If \(A$$ is a subspace of $$X$$ and $$B$$ is a subspace of $$Y$$, then the product topology on $$A \times B$$ is the same as the topology $$A \times B$$ inherits as a subspace of $$X \times Y$$.
Comment: To prove the identity of two topologies, we can either show they mutually contain each other or prove the equivalence of their bases. Because a topological basis has smaller number of elements or cardinality than the corresponding topology, proof via basis is more efficient.
Proof: Let $$\mathcal{C}$$ be the topological basis of $$X$$ and $$\mathcal{D}$$ be the basis of $$Y$$. Because $$A \subset X$$ and $$B \subset Y$$, the subspace topological bases of them are $$\mathcal{B}_A = \{C \cap A \vert \forall C \in \mathcal{C} \}$$ and $$\mathcal{B}_B = \{D \cap B \vert \forall D \in \mathcal{D} \}$$ respectively according to Lemma 16.1.
Due to Lemma 15.1, the basis of the product topology on $$A \times B$$ is
$\mathcal{B}_{A \times B} = \{ (C \cap A) \times (D \cap B) \vert \forall C \in \mathcal{C}, \forall D \in \mathcal{D} \}.$
Meanwhile, the basis of the product topology on $$X \times Y$$ is
$\mathcal{B}_{X \times Y} = \{ C \times D \vert \forall C \in \mathcal{C}, \forall D \in \mathcal{D} \}.$
Restricting $$\mathcal{B}_{X \times Y}$$ to the subset $$A \times B$$, the basis of the subspace topology on $$A \times B$$ is
\begin{aligned} \tilde{\mathcal{B}}_{A \times B} &= \{ (C \times D) \cap (A \times B) \vert \forall C \in \mathcal{C}, \forall D \in \mathcal{D} \} \\ &= \{ (C \cap A) \times (D \cap B) \vert \forall C \in \mathcal{C}, \forall D \in \mathcal{D} \}, \end{aligned}
which is the same as that of the product topology on $$A \times B$$. Hence, this theorem is proved.
The above process of proof can be illustrated as below.

Remark: The above two routes for generating topology on $$A \times B$$ must lead to the same result, otherwise, the theory itself is inappropriately proposed. A theory must be at least self-consistent before its debut in reality.

转载于:https://www.cnblogs.com/peabody/p/10117156.html
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weixin_33860722 2018-12-13 23:18:00
• (Cofinite topology) Let $$\mathcal{T}_c$$ be the cofinite topology of the space $$X$$. Then for all $$U \in \mathcal{T}_c$$, either $$U$$ is empty or its complement $$U^c$$ is finite. Next, we'll ...

Exercise 1. Let $$X$$ be a space. Let $$\mathcal{D}$$ be a collection of subsets of $$X$$ that is maximal with respect to the finite intersection property (FIP).
(a) Show that $$x \in \bar{D}$$ for every $$D \in \mathcal{D}$$ if and only if every neighborhood of $$x$$ belongs to $$\mathcal{D}$$. Which implication uses maximality of $$\mathcal{D}$$?
Proof:
Prove in the forward direction Let $$x \in \bar{D}$$ for every $$D \in \mathcal{D}$$. Let $$U$$ be any neighborhood of $$x$$ in $$X$$. According to Theorem 17.5 (a) in Section 17, if $$x \in \bar{D}$$, we have $$U \cap D \neq \varPhi$$. This means any neighborhood $$U$$ of $$X$$ intersects every element in the maximal collection $$\mathcal{D}$$. According to Lemma 37.2 (b), $$U \in \mathcal{D}$$. The maximality of $$\mathcal{D}$$ is used when applying this lemma.  Prove in the backward direction If there exists a $$D_0 \in \mathcal{D}$$ such that $$x \notin \bar{D}_0$$, $$x$$ belongs to the complement of $$\bar{D}_0$$, which is open in $$X$$. According to the given condition $$U \in D$$ for all $$D \in \mathcal{D}$$, $$\bar{D}_0^c$$ also belongs to $$\mathcal{D}$$. Then $$\bar{D}_0^c \cap D_0 = \varPhi$$ contradicts the fact that $$\mathcal{D}$$ has the FIP.
(b) Let $$D \in \mathcal{D}$$. Show that if $$A \supset D$$, then $$A \in \mathcal{D}$$.
Proof: Because $$\mathcal{D}$$ has the FIP, for all $$D' \in \mathcal{D}$$, $$D \cap D' \neq \varPhi$$. Because $$D$$ is contained in $$A$$, $$A \cap D' \neq \varPhi$$. According to Lemma 37.2 (b), $$A \in \mathcal{D}$$.
(c) Show that if $$X$$ satisfies the $$T_1$$ axiom, there is at most one point belonging to the intersection of all elements in $$\mathcal{D}$$, i.e., $$\bigcap_{D \in \mathcal{D}} \bar{D}$$.
Proof: Assume that there are at least two points $$x_1$$ and $$x_2$$ in $$\bigcap_{D \in \mathcal{D}} \bar{D}$$. If $$X$$ is a Hausdorff space, there are disjoint open sets $$U_1$$ and $$U_2$$ in $$X$$ containing $$x_1$$ and $$x_2$$ respectively. According to part (a) of this exercise, we have $$U_1 \in \mathcal{D}$$ and $$U_2 \in \mathcal{D}$$. Then, $$U_1$$ and $$U_2$$ being disjoint contradicts the fact that $$\mathcal{D}$$ has the FIP.
Unfortunately, the given condition in this exercise, i.e. $$X$$ satisfies the $$T_1$$ axiom, is weaker than the above assumption that $$X$$ is Hausdorff. Hence the above proof does not work. However, there seems no obvious or direct proof for the claim in the exercise. This may imply that the original statement is erroneous.
According to the discussion here, a counter example involving the cofinite topology $$\mathcal{T}_c$$ on the set of natural numbers $$\mathbb{N}$$ is given. It further shows that the intersection of all the elements in the maximal collection $$\mathcal{D}$$ is actually $$\mathbb{N}$$ itself. This contradicts the claim in the exercise. In the following, the construction of the counter example will be given.
Definition of the cofinite topology
Definition (Cofinite topology) Let $$\mathcal{T}_c$$ be the cofinite topology of the space $$X$$. Then for all $$U \in \mathcal{T}_c$$, either $$U$$ is empty or its complement $$U^c$$ is finite.
Next, we'll show $$\mathcal{T}_c$$ satisfying the conditions in the above definition really defines a topology on $$X$$.
It is obvious that $$\varPhi$$ belongs to $$\mathcal{T}_c$$.  When $$U = X$$, $$U^c = \varPhi$$, which is finite. Hence $$X$$ belongs to $$\mathcal{T}_c$$.  Check the closeness of the union operation. Let $$\{U_i\}_{i \in I}$$ be a collection of open sets in $$\mathcal{T}_c$$. If some $$U_i$$ in the collection is empty, it has no contribution to the union. Hence we assume all the $$U_i$$ in the collection are non-empty. Then we have $\left( \bigcup_{i \in I} U_i \right)^c = \bigcap_{i \in I} U_i^c,$ where each $$U_i^c$$ is finite. The above intersection of $$\{U_i^c\}_{i \in I}$$ is a subset of finite set, which is also finite. Therefore $$\bigcup_{i \in I} U_i \in \mathcal{T}_c$$.  Check the closeness of the finite intersection operation. For a finite collection of open sets in $$\mathcal{T}_c$$, we have $\left( \bigcap_{k = 1}^n U_k \right)^c = \bigcup_{k = 1}^n U_k^c.$ Because each $$U_i^c$$ is a finite set, the union of a finite number of finite sets is still finite. Hence $$\bigcap_{k = 1}^n U_k \in \mathcal{T}_c$$.
Due to the above analysis, $$\mathcal{T}_c$$ is really a topology for $$X$$. We also know that because every finite set in $$X$$ assigned with the topology $$\mathcal{T}_c$$ is closed, $$X$$ satisfies the $$T_1$$ axiom.
Counter example derived from the cofinite topology on $$\mathbb{N}$$
Let the set of natural numbers $$\mathbb{N}$$ be assigned with the cofinite topology $$\mathcal{T}_c$$. $$\mathbb{N}$$ satisfies the $$T_1$$ axiom. Let $$\mathcal{C}$$ be a collection of all those subsets in $$\mathbb{N}$$, each of which has a finite complement. This means all the open sets in $$\mathcal{T}_c$$ except $$\varPhi$$ are included in $$\mathcal{C}$$. Accordingly, the following can be obtained.
For all $$U \in \mathcal{C}$$, because $$\mathbb{N} - U$$ is finite while $$\mathbb{N}$$ is infinite, $$U$$ is an infinite subset of $$\mathbb{N}$$.  Let $$\{U_k\}_{k = 1}^n$$ be a finite collection arbitrarily selected from $$\mathcal{C}$$. Then we have $\left( \bigcap_{k = 1}^n U_k \right)^c = \bigcup_{k = 1}^n U_k^c.$ Because $$U_k^c$$ for each $$k$$ from $$1$$ to $$n$$ is a non-empty finite set, their finite union is still finite. Because $$\mathbb{N}$$ is infinite, $$\bigcap_{k = 1}^n U_k$$ must be infinite, which is a non-empty open set. Therefore $$\mathcal{C}$$ has the FIP.
Next, by applying the Zorn's Lemma, a maximal collection $$\mathcal{D}$$ exists, which contains $$\mathcal{C}$$ as its sub-collection and also has the FIP. For all $$D \in \mathcal{D}$$, $$D$$ must have infinite number of elements. Otherwise, if $$D = \{d_i\}_{i = 1}^m$$, we can select a sub-collection $$\{C_i\}_{i = 1}^m$$ from $$\mathcal{C}$$, such that $$d_i \notin C_i$$. Then $$D \cap C_1 \cap \cdots \cap C_m = \varPhi$$, which contradicts the fact that $$\mathcal{D}$$ has the FIP.
Select an arbitrary $$x$$ in $$D^c$$, for any open set $$U$$ in $$\mathcal{T}_c$$ containing $$x$$, it has non-empty intersection with $$D$$ because $$D$$ is an infinite set. This means any point $$x$$ in $$D^c$$ is a limiting point of $$D$$, so $$\bar{D} = D \cup D^c = \mathbb{N}$$. Hence $$\bigcap_{D \in \mathcal{D}} \bar{D} = \mathbb{N}$$, which obviously has more than one point.

转载于:https://www.cnblogs.com/peabody/p/8988031.html
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weixin_34342207 2018-05-03 22:56:00
• Let $$U = [0, 1) \times (1, 2)$$ be an open set of $$A$$ in the subspace topology, which is not saturated. $$q(U) = [0, 1)$$ is not open in $$\mathbb{R}$$. Hence $$q$$ is not an open map. Let $$U = ...  Exercise 22.3 Let \(\pi_1: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$$ be projection on the first coordinate. Let $$A$$ be the subspace of $$\mathbb{R}\times\mathbb{R}$$ consisting of all points $$x \times y$$ for which either $$x \geq 0$$ or $$y = 0$$ (or both); let $$q: A \rightarrow \mathbb{R}$$ be obtained by restricting $$\pi_1$$. Show that $$q$$ is a quotient map that is neither open nor closed.
Proof (a) Show $$q$$ is a quotient map.
The projection map $$\pi_1$$ is continuous because the pre-image of any open set $$U$$ in $$\mathbb{R}$$ under $$\pi_1$$ is $$U \times \mathbb{R}$$, which is open in the product space $$\mathbb{R}\times\mathbb{R}$$. Then its restriction $$q$$ is also continuous due to Theorem 18.2.
According to the illustrated domain of $$q$$ in Figure 1 which is marked in light grey, it is obvious that $$q$$ is surjective. It also shows the three types of saturated open sets in $$A$$ with respect to $$q$$, which are marked in red:
$$(a,b) \times \{0\}$$ with $$a < 0$$ and $$b \leq 0$$ and its image under $$q$$ is $$(a, b)$$.$$(a,b) \times \mathbb{R}$$ with $$a \geq 0$$ and $$b > 0$$ and its image under $$q$$ is $$(a, b)$$.$$(a, 0) \times \{0\} \cup [0,b) \times \mathbb{R}$$ with $$a < 0$$ and $$b > 0$$. Because a map preserves set union operation, its image under $$q$$ is $$(a, b)$$.
It can be seen that for the three types of saturated open sets, their images are all open in $$\mathbb{R}​$$. Meanwhile, arbitrary union of the above three types saturated open sets is also a saturated open set with its image open in $$\mathbb{R}​$$. Therefore, $$q​$$ is a quotient map.

Figure 1. Illustration of the domain of $$q$$ and saturated open sets in $$A$$.
(b) Show $$q$$ is neither an open nor a closed map.
Let $$U = [0, 1) \times (1, 2)$$ be an open set of $$A$$ in the subspace topology, which is not saturated. $$q(U) = [0, 1)$$ is not open in $$\mathbb{R}$$. Hence $$q$$ is not an open map.
Let $$U = \{(x,y) \vert xy = 1 \;\text{and}\; x > 0 \}$$ which is closed in $$\mathbb{R} \times \mathbb{R}$$. According to Theorem 17.2, $$U$$ is also closed in the subspace $$A$$. Then $$q(U)=(0,+\infty)$$, which is not closed in $$\mathbb{R}$$. Hence $$q$$ is not a closed map.
Comment This exercise shows that a function being open or closed map is a sufficient but not a necessary condition for the function to be a quotient map.

转载于:https://www.cnblogs.com/peabody/p/10357768.html
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weixin_33860553 2019-02-09 18:10:00
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