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Further Hints for Exercise 2.B
A subset $A$ of the real line $\mathbb{R}$ is nowhere dense in $\mathbb{R}$ if for any nonempty open subset $U$ of $\mathbb{R}$, there is a nonempty open subset $V$ of $U$ such that $V \cap A=\varnothing$. If we replace open sets by open intervals, we have the same notion.
Show that the... | {
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Since $X=[0,1]^\omega$ is compact, it follows from Fact E.2 that the product space $X=[0,1]^\omega$ is a Baire space.
Fact E.4
Let $X=[0,1]^\omega$ and $Y=(0,1)^\omega$. The product space $Y=(0,1)^\omega$ is a dense $G_\delta$-subset of $X=[0,1]^\omega$. Furthermore, $X-Y$ is a dense subset of $X$.
It follows from th... | {
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Case 2. $k(r) = 0$ for some $r \in I$.
Since $k \notin W$, $k \ne W_x$ for all $x \in I$. In particular, $k \ne W_r$. This means that $k(t) \ne W_r(t)$ for some $t \in I$. Define the open set $G$ as follows:
$G=\{ b \in Y: b(r)=0 \text{ and } b(t)=k(t) \}$
Clearly $k \in G$. Observe that $W_r \notin G$ since $W_r(t) ... | {
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Proofs of Key Steps for Exercise 2.D
Suppose that $X$ is Lindelof and that $Y$ is compact. Let $\mathcal{U}$ be an open cover of $X \times Y$. For each $x \in X$, let $\mathcal{U}_x \subset \mathcal{U}$ be finite such that $\mathcal{U}_x$ is a cover of $\{ x \} \times Y$. Putting it another way, $\{ x \} \times Y \sub... | {
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Suppose that $\bigcap_{i=1}^\infty O_i=\varnothing$. Choose $x_1 \in O_1$. There must exist some $n_1$ such that $x_1 \notin O_{n_1}$. Choose $x_2 \in O_{n_1}$. There must exist some $n_2>n_1$ such that $x_2 \notin O_{n_2}$. Continue in this manner we can choose inductively an infinite set $A=\{ x_1,x_2,x_3,\cdots \} \... | {
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Suppose $Y$ is $\sigma$-compact. Let $Y=\bigcup_{n=1}^\infty B_n$ where each $B_n$ is compact. Each $B_n$ is obviously a closed subset of $X$. We claim that each $B_n$ is a closed nowhere dense subset of $X$. To see this, let $U$ be a non-empty open subset of $X$. Since $X-Y$ is dense in $X$, $U$ contains a point $p$ w... | {
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Concluding Remarks
Exercise 2.A
The exercise is to show that the product of uncountably many $\sigma$-compact spaces does not need to be Lindelof. The approach suggested in the hints is to show that $\mathbb{R}^{c}$ has uncountable extent where $c... | {
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Exercise 2.E
Like Exercise 2.B, this exercise is also to show a certain space is not $\sigma$-compact. In this case, the suggested space is $\mathbb{R}^{\omega}$, the product of countably many copies of the real line. The hints given use a Baire category argument, as outlined in Fact E.1 through Fact E.4. The product s... | {
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Dan Ma math
Daniel Ma mathematics
$\copyright$ 2019 – Dan Ma
Lindelof Exercise 1
A space $X$ is called a $\sigma$-compact space if it is the union of countably many compact subspaces. Clearly, any $\sigma$-compact space is Lindelof. It is well known that the product of Lindelof spaces does not need to be Lindelof. ... | {
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Then the product $\prod_{i=1}^\infty C_i$ is Lindelof.
Note that in the topological sum $C_{i,1} \oplus C_{i,2} \oplus C_{i,3} \oplus \cdots$, the spaces $C_{i,1},C_{i,2},C_{i,3},\cdots$ are considered pairwise disjoint. The open sets in the sum are simply unions of the open sets in the individual spaces. Another way ... | {
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For convenience, each point $p_i$ is called a point at infinity.
Note that Theorem 2 follows from Lemma 3 and Theorem 4. In order to establish Theorem 1 from Theorem 2, observe that the Lindelof property is preserved by any continuous mapping and that there is a natural continuous map from the product space in Theorem... | {
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Each set $F_i$ is compact since it is closed in $Z$. The intersection of finitely many $F_i$ is also compact. Thus the $\cap G$ in the definition of $Y$ in the above claim is compact. There can be only countably many $\cap G$ in the definition of $Y$. Thus $Y$ is a $\sigma$-compact space that is covered by the open cov... | {
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Reference
1. Engelking R., General Topology, Revised and Completed edition, Elsevier Science Publishers B. V., Heldermann Verlag, Berlin, 1989.
2. Hager A. W., Approximation of real continuous functions on Lindelof spaces, Proc. Amer. Math. Soc., 22, 156-163, 1969.
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Dan Ma topology
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The space $\omega_1$ is the space of all countable ordinals with the order topology. Recall $C$ is the product space $I^I$. The product space $\omega_1 \times C$ is Example 106 in [2]. This product is not normal. The non-normality of $\omega_1 \times C$ is based on this theorem: for any compact space $Y$, the product $... | {
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First, we define a discrete subspace. For each $x$ with $0, define $f_x: I \rightarrow I$ as follows:
$\displaystyle f_x(y) = \left\{ \begin{array}{ll} \displaystyle 0 &\ \ \ \ \ \ 0 \le y < x \\ \text{ } & \text{ } \\ \displaystyle \frac{1}{2} &\ \ \ \ \ y=x \\ \text{ } & \text{ } \\ \displaystyle 1 &\ \ \ \ \ \ x
L... | {
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For each $F$ and $G$ as detailed above, we define a function $L(F,G):I \rightarrow I$ as follows:
$\displaystyle L(F,G)(x) = \left\{ \begin{array}{ll} \displaystyle t_1 &\ \ \ \ \ 0 \le x < q_1 \\ \text{ } & \text{ } \\ \displaystyle t_2 &\ \ \ \ \ q_1 \le x < q_2 \\ \text{ } & \text{ } \\ \displaystyle \vdots &\ \ \ ... | {
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A natural question is: is there a compact non-metric space $V$ such that both $V$ and $V \times V$ are hereditarily normal and $V \times V \times V$ is not hereditarily normal? In other words, is there an example where the hereditarily normality fails at dimension 3? If we do not assume extra set-theoretic axioms beyon... | {
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It is also well known that a regular space of countable spread which is not hereditarily separable contains an L-space and a regular space of countable spread which is not hereditarily Lindelof contains an S-space. Thus an absolute example of a space satisfying (Statement) A would contain a proof of the existence of S ... | {
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One amazing aspect about set-theoretic topology is that one sometimes does not have to stray far from basic topological notions to encounter pathological objects such as S-spaces and L-spaces. The definition of a topological space is of course a basic definition. Separable spaces and Lindelof spaces are basic notions t... | {
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Diagram 1 – Properties surrounding countable spread
The implications (the arrows) in Diagram 1 can be verified easily. Central to the discussion at hand, both hereditarily separable and hereditarily Lindelof imply countable spread. The best way to see this is that if a space has an uncountable discrete subspace, that ... | {
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Now focus on the arrows emanating from countable spread in Diagram 2. These arrows are about the basic fact discussed earlier. From Diagram 1, we know that hereditarily separable implies countable spread. Can the implication be reversed? Any L-space would be an example showing that the implication cannot be reversed. N... | {
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To that end, we use the concepts of right separated space and left separated space. Recall that an initial segment of a well-ordered set $(X,<)$ is a set of the form $\{y \in X: y where $x \in X$. A space $X$ is a right separated space if $X$ can be well-ordered in such a way that every initial segment is open. A right... | {
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$\Longleftarrow$ of the first bullet point.
Suppose that $X$ is not hereditarily separable. Let $Y \subset X$ be a subspace that is not separable. We now inductively derive an uncountable left separated subspace of $Y$. Choose $y_0 \in Y$. For each $\alpha<\omega_1$, let $A_\alpha=\{ y_\beta \in Y: \beta <\alpha \}$. T... | {
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$\Longleftarrow$ of the second bullet point.
Suppose that $X$ is not hereditarily Lindelof. Let $Y \subset X$ be a subspace that is not Lindelof. Let $\mathcal{U}$ be an open cover of $Y$ that has no countable subcover. We now inductively derive a right separated subspace of $Y$ of type $\omega_1$.
Choose $U_0 \in \ma... | {
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To see this, fix $\alpha<\kappa$. The initial segment $A_\alpha=\{ w_\beta: \beta<\alpha \}$ is closed in $X$ since $X$ is a left separated space. On the other hand, the initial segment $\{ w_\beta: \beta < \alpha+1 \}$ is open in $X$ since $X$ is a right separated space. Then $B_{\alpha}=\{ w_\beta: \beta \ge \alpha+1... | {
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To start, pick any $y_\gamma$ in $Y$ and relabel it $t_0$. The final segment $\{y_\beta \in Y: t_0 <_L \ y_\beta \}$ must intersect the final segment $\{x_\beta \in X: t_0 <_R \ x_\beta \}$ in uncountably many points. Choose the least such point (according to $<_R$) and call it $t_1$. It is clear how $t_{\delta+1}$ is ... | {
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Suppose that $X$ is of countable spread and that $X$ is not hereditarily Lindelof. By Theorem A, $X$ has an uncountable right separated subspace $Y$ (assume it is of type $\omega_1$). By Theorem C, $Y$ is an S-space or $Y$ contains an S-space. In either way, $X$ contains an S-space.
Reference
1. Eisworth T., Nyikos P... | {
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Lemma 1
Let $X$ be a space. For any open cover $\mathcal{U}$ of $X$, there exists a discrete subspace $A$ of $X$ such that $X=\text{St}(A,\mathcal{U})$. Furthermore, the set $A$ can be chosen in such a way that it is also a closed subset of the space $X$.
Any space that satisfies the condition in Lemma 1 is said to be... | {
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Lemma 1a naturally leads to other star covering properties. For example, a space $X$ is said to be a star countable space if for any open cover $\mathcal{U}$ of $X$, there exists a countable subspace $A$ of $X$ such that $\{ \text{St}(x,\mathcal{U}): x \in A \}$ is a cover of $X$. A space $X$ is said to be a star Linde... | {
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Then it follows from Lemma 1 that any space $X$ that has countable extent is star countable. Any star countable space is obviously a star Lindelof space. The following diagram displays these relationships.
According to the diagram, the star countable and star Lindelof are both downstream from the countable spread prop... | {
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We claim that the induction process must stop at some $\alpha<\kappa^+$. In other words, at some $\alpha<\kappa^+$, the star of the previous points must be the entire space and we run out of points to choose. Otherwise, we would have obtained a subset of $X$ with cardinality $\kappa^+$, a contradiction. Choose the leas... | {
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This post discusses Michael line from the point of view of the three conjectures of Kiiti Morita.
K. Morita defined the notion of P-spaces in [7]. The definition of P-spaces is discussed here in considerable details. K. Morita also proved that a space $X$ is a normal P-space if and only if the product $X \times Y$ is ... | {
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Because $M \times \mathbb{P}$ is not normal, the Michael line $M$ is not a normal P-space. Prior to E. Michael’s 1963 article, we have to reach back to 1955 to find an example of a non-normal product where one factor is a metric space. In 1955, M. E. Rudin used a Souslin line to construct a Dowker space, which is a nor... | {
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The contrapositive statement of Morita’s conjecture II is that for any non-metrizable space $X$, there exists a normal P-space $Y$ such that $X \times Y$ is not normal. The pairing is more specific than for conjecture I. Any non-metrizable space is paired with a normal P-space to form a non-normal product. As illustrat... | {
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Michael line as an example of a non-normal P-space is a great tool to help us walk through the three conjectures of Morita. Are there other examples of non-normal P-spaces? Dowker spaces mentioned above (normal spaces whose products with the closed unit interval are not normal) are non-normal P-spaces. Note that conjec... | {
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Consider this well known result: for any infinite compact space $X$, the product $\omega_1 \times X$ is normal if and only if the space $X$ has countable tightness (see Theorem 1 here). Thus any compact space with uncountable tightness is paired with $\omega_1$, the space of all countable ordinals, to form a non-normal... | {
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Examples of Non-Normal P-Spaces
Another non-normal product is $X_B \times B$ where $B \subset \mathbb{R}$ is a Bernstein set and $X_B$ is the space with the real line as the underlying set such that points in $B$ are isolated and points in $\mathbb{R}-B$ retain the usual open sets. The set $B \subset \mathbb{R}$ is sa... | {
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The next set of non-normal P-spaces requires set theory. A Michael space is a Lindelof space whose product with $\mathbb{P}$, the space of irrational numbers, is not normal. Michael problem is the question: is there a Michael space in ZFC? It is known that a Michael space can be constructed using continuum hypothesis [... | {
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Reference
1. Alster K.,On the product of a Lindelof space and the space of irrationals under Martin’s Axiom, Proc. Amer. Math. Soc., Vol. 110, 543-547, 1990.
2. Balogh Z.,Normality of product spaces and Morita’s conjectures, Topology Appl., Vol. 115, 333-341, 2001.
3. Chiba K., Przymusinski T., Rudin M. E.Nonshrinking... | {
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These statements are no longer conjectures. Partial results appeared after the conjectures were proposed in 1976. The complete resolution of the conjectures came in 2001 in a paper by Zoli Balogh [5]. Though it is more appropriate to call these statements theorems, it is still convenient to call them conjectures. Just ... | {
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Duality Conjectures
Here’s three theorems that are duals to the conjectures.
Theorem 1
Let $X$ be a space. The product space $X \times Y$ is normal for every discrete space $Y$ if and only if $X$ is normal.
Theorem 2
Let $X$ be a space. The product space $X \times Y$ is normal for every metrizable space $Y$ if and o... | {
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$\Longrightarrow$ This direction uses Dowker’s theorem. We give a contrapositive proof. Suppose that $X$ is not both normal and countably paracompact. Case 1. $X$ is not normal. Then $X \times \{ y \}$ is not normal where $\{ y \}$ is any one-point discrete space. Case 2. $X$ is normal and not countably paracompact. Th... | {
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Conjecture II implies Conjecture I
We give a contrapositive proof of Conjecture I. Suppose that $X$ is not discrete. We wish to find a normal space $Y$ such that $X \times Y$ is not normal. Consider two cases for $X$. Case 1. $X$ is not metrizable. By Conjecture II, $X \times Y$ is not normal for some normal P-space $Y... | {
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Conjecture 9 was part of a set of 14 conjectures stated in [14]. It is also discussed in [7]. In [6], conjecture 9 was shown to be equivalent to Morita’s second conjecture. In [5], Balogh used his technique for constructing a Dowker space of cardinality continuum to obtain a space as described in conjecture 9.
The res... | {
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Another interesting observation about conjecture II is that normal P-spaces are not productive with respect to normality. More specifically, for any non-metrizable normal P-space $X$, conjecture II tells us that there exists another normal P-space $Y$ such that $X \times Y$ is not normal.
Now we narrow the focus to sp... | {
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In finding a normal pair $Y$ for a normal space $X$, if we do not care about $Y$ having a high degree of normal productiveness (e.g. normal P or normal countably paracompact), we can always let $Y$ be a Dowker space or $\kappa$-Dowker space. In fact, if the starting space $X$ is a metric space, the normal pair for a no... | {
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Reference
1. Atsuji M.,On normality of the product of two spaces, General Topology and Its Relation to Modern Analysis and Algebra (Proc. Fourth Prague Topology sympos., 1976), Part B, 25–27, 1977.
2. Atsuji M.,Normality of product spaces I, in: K. Morita, J. Nagata (Eds.), Topics in General
Topology, North-Holland, A... | {
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$\text{ }$
Dan Ma math
Daniel Ma mathematics
$\copyright$ 2018 – Dan Ma
Morita’s normal P-space
In this post we discuss K. Morita’s notion of P-space, which is a useful and interesting concept in the study of normality of product spaces.
The Definition
In [1] and [2], Morita defined the notion of P-spaces. First... | {
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By switching closed sets and open sets and by switching decreasing collection and increasing collection, the following is an alternative but equivalent definition of P-spaces.
The space $X$ is a P-space if for any cardinal $\kappa \ge 1$ and for any increasing collection $\left\{U_\sigma \subset X: \sigma \in \Gamma \... | {
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The above condition implies the following condition.
For any decreasing sequence $F_1 \supset F_2 \supset F_3 \cdots$ of closed subsets of $X$ such that $\bigcap_{n=1}^\infty F_n=\varnothing$, there exist $U_1,U_2,U_3,\cdots$, open subsets of $X$, such that $F_n \subset U_n$ for all $n$ and such that $\bigcap_{n=1}^\i... | {
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Not surprisingly, the notion of P-spaces is about normality in product spaces where one factor is a metric space. In fact, this is precisely the characterization of P-spaces (see Theorem 1 and Theorem 2 below).
A Characterization of P-Space
Morita gave the following characterization of P-spaces among normal spaces. T... | {
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Theorem 4
Let $\kappa$ be the positive integers $2,3,4,\cdots$ or $\kappa=\omega$, the countably infinite cardinal. Let $X$ be a space. Then $X$ is a P(2)-space if and only if $X$ is a P($\kappa$)-space.
To give a context for Theorem 4, note that if $X$ is a P($\kappa$)-space, then $X$ is a P($\tau$)-space for any car... | {
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The $\Sigma$-product of real lines $\Sigma_{\alpha<\tau} \mathbb{R}$ is a normal P-space. For any metric space $Y$, the product $(\Sigma_{\alpha<\tau} \mathbb{R}) \times Y$ is a $\Sigma$-product of metric spaces. By a well known result, the $\Sigma$-product of metric spaces is normal.
Examples of Non-Normal P-Spaces
... | {
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Consider the Sorgenfrey line. It is perfectly normal. Thus the product of the Sorgenfrey line with any metric space is also perfectly normal, hence normal. It is well known that the square of the Sorgenfrey line is not normal.
The space $\omega_1$ of all countable ordinals is a normal and countably compact space, henc... | {
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Daniel Ma mathematics
$\copyright$ 2018 – Dan Ma
In between G-delta diagonal and submetrizable
This post discusses the property of having a $G_\delta$-diagonal and related diagonal properties. The focus is on the diagonal properties in between $G_\delta$-diagonal and submetrizability. The discussion is followed by a... | {
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A space $X$ is submetrizable if there is a metrizable topology that is weaker than the topology for $X$. Then the diagonal $\Delta$ would be a $G_\delta$-set with respect to the weaker metrizable topology of $X \times X$ and thus with respect to the orginal topology of $X$. This means that the class of spaces having $G... | {
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If $A=\{ x \}$, we write $St(x, \mathcal{U})$ instead of $St(\{ x \}, \mathcal{U})$. Then $St(x, \mathcal{U})$ refers to the union of all sets in $\mathcal{U}$ that contain the point $x$. The set $St(x, \mathcal{U})$ is then called the star of the point $x$ with respect to the collection $\mathcal{U}$.
Note that the s... | {
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It is informative to compare the property of $G_\delta$-diagonal with the definition of Moore spaces. A development for the space $X$ is a sequence $\mathcal{D}_0,\mathcal{D}_1,\mathcal{D}_2,\cdots$ of open covers of $X$ such that for each $x \in X$, $\{ St(x, \mathcal{D}_n): n=0,1,2,\cdots \}$ is a local base at the p... | {
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Theorem 2
Let $X$ be a space. Then the following statements are equivalent.
1. The space $X$ has a regular $G_\delta$-diagonal.
2. There exists a sequence $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$ of open covers of $X$ such that for every two distinct points $x,y \in X$, there exist open sets $U$ and $V$ with... | {
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Rank-k Diagonals
Metric spaces and submetrizable spaces have regular $G_\delta$-diagonals. We discuss this fact after introducing another set of diagonal properties. First some notations. For any family $\mathcal{U}$ of subsets of the space $X$ and for any $x \in X$, define $St^1(x, \mathcal{U})=St(x, \mathcal{U})$. F... | {
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1. The sequence $\mathcal{U}_0,\mathcal{U}_1,\mathcal{U}_2,\cdots$ is a rank-$k$ diagonal sequence for the space $X$.
2. For any two distinct points $x$ and $y$ of $X$, there is an integer $n$ such that $y \notin St^k(x,\mathcal{U}_n)$.
3. For any two distinct points $x$ and $y$ of $X$, there is an integer $n$ such tha... | {
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To prove Theorem 5, let $\{ \mathcal{U}_n \}$ be a development for the space $X$. Let $x$ and $y$ be two distinct points of $X$. We claim that there exists some $n$ such that $y \notin St^2(x,\mathcal{U}_n)$. Suppose not. This means that for each $n$, $y \in St^2(x,\mathcal{U}_n)$. This also means that $St(x,\mathcal{U... | {
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We instead prove Theorem 7 topologically. To this end, we use an appropriate metrization theorem. The following theorem is a good candidate.
Alexandrov-Urysohn Metrization Theorem. A space $X$ is metrizable if and only if the space $X$ has a development $\{ \mathcal{U}_n \}$ such that for any $U_1,U_2 \in \mathcal{U}_... | {
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We show this claim is true for $k=2$. Observe that there cannot exist $U_1, U_2 \in \mathcal{U}_{m+1}$ such that $x \in U_1$, $y \in U_2$ and $U_1 \cap U_2 \ne \varnothing$. If there exists such a pair, then $U_1 \cup U_2$ would be contained in $\text{St}(x,\mathcal{U}_m)$ and $\text{St}(y,\mathcal{U}_m)$, a contradict... | {
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Examples and Questions
The preceding discussion focuses on properties that are in between $G_\delta$-diagonal and submetrizability. In fact, one of the properties has infinitely many levels (rank-$k$ diagonal for integers $k \ge 1$). We would like to have a diagram showing the relative strengths of these properties. B... | {
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In the diagram, “rank-k diagonal” means that the space has a rank-$k$ diagonal where $k \ge 1$ is an integer, which in terms means that the space has a rank-$k$ diagonal sequence as defined above. Thus rank-$k$ diagonal is not to be confused with the rank of a diagonal. The rank of the diagonal of a given space is the ... | {
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Example 2.9 in [2] is a Tychonoff separable Moore space $Z$ that has a rank-3 diagonal but not of higher diagonal rank. As a result of not having a rank-4 diagonal, $Z$ is not submetrizable. Thus $Z$ is an example of a space with rank-3 diagonal (hence with a regular $G_\delta$-diagonal) that is not submetrizable. Acco... | {
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Example 2.17 in [2] is a non-submetrizable Moore space that has a zero-set diagonal and has rank-3 diagonal exactly (i.e. it does not have a higher rank diagonal). This example shows that having a zero-set diagonal does not imply having a rank-4 diagonal. A natural question is then this. Does having a zero-set diagonal... | {
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Reference
1. Arhangelskii A. V., Burke D. K., Spaces with a regular $G_\delta$-diagonal, Topology and its Applications, Vol. 153, No. 11, 1917–1929, 2006.
2. Arhangelskii A. V., Buzyakova R. Z., The rank of the diagonal and submetrizability, Comment. Math. Univ. Carolinae, Vol. 47, No. 4, 585-597, 2006.
3. Buzyakova R... | {
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$\copyright$ 2018 – Dan Ma
Pseudocompact spaces with regular G-delta diagonals
This post complements two results discussed in two previous blog posts concerning $G_\delta$-diagonal. One result is that any compact space with a $G_\delta$-diagonal is metrizable (see here). The other result is that the compactness in th... | {
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The space $X$ is said to have a regular $G_\delta$-diagonal if the diagonal $\Delta$ is a regular $G_\delta$-set in $X \times X$, i.e. $\Delta=\bigcap_{n=1}^\infty \overline{U_n}$ where each $U_n$ is an open subset of $X \times X$ with $\Delta \subset U_n$. If $\Delta=\bigcap_{n=1}^\infty \overline{U_n}$, then $\Delta=... | {
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Choose open $V$ with $x \in V \subset \overline{V} \subset U$. Consider the sequence $\{ O_n \cap (X-\overline{V}) \}$. This is a decreasing sequence of non-empty open subsets of $X$. By Theorem 2 in this previous post, $\bigcap \overline{O_n \cap (X-\overline{V})} \ne \varnothing$. Let $y$ be a point in this non-empty... | {
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Theorem 3 follows from Theorem 1.4 in [1], which states that for any $T_0$-space $X$, $X$ is metrizable if and only if there exists a sequence $\mathcal{G}_1, \mathcal{G}_2, \mathcal{G}_3,\cdots$ of open covers of $X$ such that for each open $U \subset X$ and for each $x \in U$, there exist an open $V \subset X$ and an... | {
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We want to show that $V \cap St(x,\mathcal{U}_n)=\varnothing$, which implies that $y \notin \overline{St(x,\mathcal{U}_n)}$. Suppose $V \cap St(x,\mathcal{U}_n) \ne \varnothing$. This means that $V \cap W \ne \varnothing$ for some $W \in \mathcal{U}_n$ with $x \in W$. Then $(U \times V) \cap (W \times W) \ne \varnothin... | {
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We claim that $St(y,\mathcal{U}_k) \cap St(St(x,\mathcal{U}_k), \mathcal{U}_k)=\varnothing$. Suppose not. Choose $w \in St(y,\mathcal{U}_k) \cap St(St(x,\mathcal{U}_k), \mathcal{U}_k)$. It follows that $w \in B$ for some $B \in \mathcal{U}_k$ such that $B \cap St(x,\mathcal{U}_k) \ne \varnothing$ and $B \cap St(y,\math... | {
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Example
Any submetrizable space has a $G_\delta$-diagonal. The converse is not true. A classic example of a non-submetrizable space with a $G_\delta$-diagonal is the Mrowka space (discussed here). The Mrowka space is also called the psi-space since it is sometimes denoted by $\Psi(\mathcal{A})$ where $\mathcal{A}$ is ... | {
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# 8.5. GMRES¶
The most important use of the Arnoldi iteration is to solve the square linear system $$\mathbf{A}\mathbf{x}=\mathbf{b}$$.
In Demo 8.4.3, we attempted to replace the linear system $$\mathbf{A}\mathbf{x}=\mathbf{b}$$ by the lower-dimensional approximation
$\min_{\mathbf{x}\in \mathcal{K}_m} \| \mathbf{A}... | {
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which is of size $$(m+1)\times m$$. We call the solution of this minimization $$\mathbf{z}_m$$, and then $$\mathbf{x}_m=\mathbf{Q}_m \mathbf{z}_m$$ is the $$m$$th approximation to the solution of $$\mathbf{A}\mathbf{x}=\mathbf{b}$$.
Algorithm 8.5.1 : GMRES
Given $$n\times n$$ matrix $$\mathbf{A}$$ and $$n$$-vector $... | {
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Function 8.5.3 : gmres
GMRES for a linear system
1"""
2 gmres(A,b,m)
3
4Do m iterations of GMRES for the linear system A*x=b. Returns
5the final solution estimate x and a vector with the history of
6residual norms. (This function is for demo only, not practical use.)
7"""
8function gmres(A,b,m)
9 n = length(b... | {
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One of the practical challenges in GMRES is that as the dimension of the Krylov subspace grows, the number of new entries to be found in $$\mathbf{H}_m$$ and the total number of columns in $$\mathbf{Q}$$ also grow. Thus both the work and the storage requirements are quadratic in $$m$$, which can become intolerable in s... | {
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# calculating number of boolean functions
I would just like to clarify if I am on the right track or not. I have these questions:
Consider the Boolean functions $f(x,y,z)$ in three variables such that the table of values of $f$ contains exactly four $1$’s.
1. Calculate the total number of such functions.
2. We apply... | {
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A "block of two" on a Karnaugh map corresponds to a Boolean function of the form $xy$ (complements of variables allowed). So we are looking for Boolean functions of weight $4$ ($4$ ONEs in the truth table or on the Karnaugh map) that can be covered by $3$ "blocks of two" but not by $2$ "blocks of two", that is, the min... | {
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• Prime and irredundant covers with four implicants. There are two of them, even and odd parity. No two minterms may be adjacent on the map, and there's only two ways to achieve that.
• Prime and irredundant covers with three prime implicants. There are eight of them of the type shown by @DilipSarwate, namely $(x \wedg... | {
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Finally, if a prime cover consists of all essential primes, it is the unique prime and irredundant cover. All these are classic results. Since all covers listed above are made of essential primes, they are all unique.
L shapes: --. or --' or '-- or .-- count: 8
T shapes: -'- or -.- count: 4
Z shapes: -.. or ..- coun... | {
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Documentation
### This is machine translation
Translated by
Mouseover text to see original. Click the button below to return to the English version of the page.
# surface
Create surface object
## Syntax
surface(Z)
surface(Z,C)
surface(X,Y,Z)
surface(X,Y,Z,C)
surface(x,y,Z)
surface(...'PropertyName',PropertyValue,... | {
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surface(ax,...) creates the surface in the axes specified by ax instead of in the current axes (gca). The option ax can precede any of the input argument combinations in the previous syntaxes.
h = surface(...) returns a primitive surface object.
## Examples
collapse all
Plot the function $z=x{e}^{-{x}^{2}-{y}^{2}}$... | {
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If you do not specify separate color data (C), MATLAB® uses the matrix (Z) to determine the coloring of the surface. In this case, color is proportional to values of Z. You can specify a separate matrix to color the surface independently of the data defining the area of the surface.
You can specify properties as prope... | {
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+0
# whats is bigger 1+1 or 1 + half +half of a half + half of a half of a half infinitely
0
305
5
whats is bigger 1+1 or 1 + half +half of a half + half of a half of a half infinitely
Guest Apr 22, 2015
#2
+91451
+10
These can be quite deceptive MG :)
Which is bigger
1+1=2
OR
$$1+\frac{1}{2}+\frac{1}{4}+\fra... | {
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MathsGod1 Apr 23, 2015
### 19 Online Users
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | {
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# Where is the mistake in my evaluation of $\int\frac{x}{x^2+2x+3}\,dx$?
Here is how I did it:
First, write $$\int\frac{x}{x^2+2x+3}\,dx=\int\frac{2x+2-x-2}{x^2+2x+3}\,dx=\int\frac{2x+2}{x^2+2x+3}\,dx-\int\frac{x+2}{x^2+2x+3}\,dx.$$ Now consider the integral in the minuend. Letting $$u=x^2+2x+3$$, one finds $$du=(2x+... | {
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Recall $$\ln(a/b) = \ln(a) - \ln(b)$$, so
$$\ln\left|\dfrac{x^2+2x+3}{2} \right| = \ln\dfrac{|x^2+2x+3|}{|2|} = \ln|x^2+2x+3|-\ln2$$ so, distributing the $$-\dfrac{1}{2}$$, we obtain $$\ln|x^2+2x+3|-\dfrac{1}{2}\ln|x^2+2x+3|-\dfrac{1}{2}\ln2 - \dfrac{\sqrt{2}\arctan\frac{x+1}{\sqrt 2}}{2}+C$$ which is just $$\dfrac{1}... | {
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# rv uniformly distributed
#### Francobati
##### New member
Hello.
Let $Y=1-X^2$, where $X~ U(0,1)$. What statement is TRUE?
-$E(Y^2)=2$
- $E(Y^2)=1/2$
- $var(Y)=1/12$
- $E(Y)=E(Y^2)$
-None of the remaining statements.
Solution:
I compute: $E(Y^2)=E(1-X^2)^2=E(1+X^4-2X^2)=1+E(X^4)-2E(X^2)$, then?
#### Klaas van Aars... | {
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# If $f$ divides $g$ in $S[x]$, show that $f$ divides $g$ in $R[x]$ for $R$ a sub-ring of $S$.
Let $$R$$ be a sub-ring of a ring $$S$$. Let $$f,g$$ be non-zero polynomials in $$R[x]$$ and assume that the leading coefficient of $$f$$ is a unit in $$R$$. If $$f$$ divides $$g$$ in $$S[x]$$, show that $$f$$ divides $$g$$ ... | {
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• Ok. I see. So there couldn't be a q(x), r(x) with deg(r(x))> 0 in R[x] such that g = q f + r because that would contradict the uniqueness of euclidean division? – IntegrateThis Feb 27 at 23:57
• @IntegrateThis Exactly. I added a remark and link on the gcd perspective. – Bill Dubuque Feb 27 at 23:59
• Are $R$ and $S$ ... | {
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Suppose that $$g(x)=f(x)k(x)+r(x)$$ in $$R[x]$$. Since $$R[x] \subseteq S[x]$$, the same equation holds in $$S[x]$$. On the other hand, you had assumed that $$g(x)=f(x)q(x)$$ in $$S[x]$$; so we get that $$f(x)q(x)=f(x)k(x)+r(x)$$. Hence, $$f(x)\big( q(x) -k(x) \big) = r(x)$$.
Since the leading coefficient of $$f(x)$$ ... | {
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# [SOLVED]If Derivative is Not Zero Anywhere Then Function is Injective.
#### caffeinemachine
##### Well-known member
MHB Math Scholar
Hello MHB.
I am sorry that I haven't been able to take part in discussions lately because I have been really busy.
I am having trouble with a question.
In a past year paper of an e... | {
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#### Deveno
##### Well-known member
MHB Math Scholar
Let us suppose by way of contradiction a counter-example exists.
Thus we have two points $c < d \in (a,b)$ such that:
$f(c) = f(d)$, but $c \neq d$.
By supposition, $c$ and $d$ are, of course, interior points of $(a,b)$, and thus since $f$ is differentiable on $(... | {
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# Is there any mathematical reason for this "digit-repetition-show"?
The number $$\sqrt{308642}$$ has a crazy decimal representation : $$555.5555777777773333333511111102222222719999970133335210666544640008\cdots$$
Is there any mathematical reason for so many repetitions of the digits ?
A long block containing only a... | {
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I want to emphasize the role of the binomial series. In particular the Taylor expansion $$\sqrt{1+x}=1+\frac x2-\frac{x^2}8+\frac{x^3}{16}-\frac{5x^4}{128}+\frac{7x^5}{256}-\frac{21x^6}{1024}+\cdots$$ If we plug in $$x=2/(5000)^2=8\cdot10^{-8}$$, we get $$M:=\sqrt{1+8\cdot10^{-8}}=1+4\cdot10^{-8}-8\cdot10^{-16}+32\cdot... | {
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• @Peter It is quite common to unaccept an answer after a better answer appears. Doing so helps guide readers to the best answer, which is often not the highest voted one, due to many factors, e.g. earlier answers usually get more votes, and less technical answers usually get more votes from hot-list activity (as here)... | {
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