• Theorem 20.4 The uniform topology on $$\mathbb{R}^J$$ is finer than the product topology and coarser than the box topology; these three topologies are all different if $$J$$ is infinite. Proof: a) Pr...
Theorem 20.4 The uniform topology on $$\mathbb{R}^J$$ is finer than the product topology and coarser than the box topology; these three topologies are all different if $$J$$ is infinite.

Proof: a) Prove the uniform topology is finer than the product topology.

Analysis: Look inside an open ball in the product topology for an open ball in the uniform topology and then apply Lemma 20.2. It should be also noted that the product topology on $$\mathbb{R}^J$$ has each of its coordinate space assigned the standard topology, which is consistent with both topologies induced from the two metrics $$d$$ and $$\bar{d}$$ according to example 2 in this section and Theorem 20.1.

According to the second part of Theorem 19.2, let $$\prod_{\alpha \in J} B_{\alpha}$$ be an arbitrary basis element for the product topology on $$\mathbb{R}^J$$, where only a finite number of $$B_{\alpha}$$s are open intervals in $$\mathbb{R}$$ and not equal to $$\mathbb{R}$$. Let the indices for these $$B_{\alpha}$$s be $$\{\alpha_1, \cdots, \alpha_n\}$$ and for all $$i \in \{1, \cdots, n\}$$, $$B_{\alpha_i} = (a_i, b_i)$$. Then for all $$\vect{x} \in \prod_{\alpha \in J} B_{\alpha}$$ and for all $$\alpha \in J$$, $$x_{\alpha} \in B_{\alpha}$$. Specifically, for all $$i \in \{1, \cdots, n\}$$, $$x_{\alpha_i} \in B_{\alpha_i}$$. Let $$\varepsilon_{\alpha_i} = \min \{ x_{\alpha_i} - a_i, b_i - x_{\alpha_i} \}$$ and $$\varepsilon = \min_{1 \leq i \leq n} \{\varepsilon_{\alpha_1}, \cdots, \varepsilon_{\alpha_n}\}$$. Then we’ll check the open ball $$B_{\bar{\rho}}(\vect{x}, \varepsilon)$$ in $$\mathbb{R}^J$$ with the uniform topology is contained in the basis element $$\prod_{\alpha \in J} B_{\alpha}$$.

For all $$\vect{y} \in B_{\bar{\rho}}(\vect{x}, \varepsilon)$$, $$\bar{\rho}(\vect{x}, \vect{y}) < \varepsilon$$, i.e. $$\sup_{\forall \alpha \in J} \{\bar{d}(x_{\alpha}, y_{\alpha})\} < \varepsilon$$. Therefore, for all $$i \in \{1, \cdots, n\}$$, $$\bar{d}(x_{\alpha_i}, y_{\alpha_i}) < \varepsilon$$. Note that when $$\varepsilon > 1$$, $$B_{\bar{\rho}}(\vect{x}, \varepsilon) = \mathbb{R}^J$$, which is not what we desire. Instead, we need to define the open ball’s radius as $$\varepsilon' = \min\{\varepsilon, 1\}$$. Then we have for all $$\vect{y} \in B_{\bar{\rho}}(\vect{x}, \varepsilon')$$, $$\bar{d}(x_{\alpha_i}, y_{\alpha_i}) = d(x_{\alpha_i}, y_{\alpha_i}) < \varepsilon'$$, i.e. $$y_{\alpha_i} \in (x_{\alpha_i} - \varepsilon', x_{\alpha_i} + \varepsilon') \subset B_{\alpha_i}$$. For other coordinate indices $$\alpha \notin \{\alpha_1, \cdots, \alpha_n\}$$, because $$B_{\alpha} = \mathbb{R}$$, $$y_{\alpha} \in (x_{\alpha} - \varepsilon', x_{\alpha} + \varepsilon') \subset B_{\alpha}$$ holds trivially.

Therefore, the uniform topology is finer than the product topology.

b) Prove the uniform topology is strictly finer than the product topology, when $$J$$ is infinite.

When $$J$$ is infinite, for an open ball $$B_{\bar{\rho}}(\vect{x}, \varepsilon)$$ with $$\varepsilon \in (0, 1]$$, there are infinite number of coordinate components comprising this open ball which are not equal to $$\mathbb{R}$$. Therefore, there is no basis element for the product topology which is contained in $$B_{\bar{\rho}}(\vect{x}, \varepsilon)$$.

c) Prove the box topology is finer than the uniform topology.

For any basis element $$B_{\bar{\rho}}(\vect{x}, \varepsilon)$$ for the uniform topology, when $$\varepsilon > 1$$, $$B_{\bar{\rho}}(\vect{x}, \varepsilon) = \mathbb{R}^J$$. Then for all $$\vect{y} \in B_{\bar{\rho}}(\vect{x}, \varepsilon)$$, any basis element for the box topology containing this $$\vect{y}$$ is contained in $$B_{\bar{\rho}}(\vect{x}, \varepsilon)$$.

When $$\varepsilon \in (0, 1]$$, $$\bar{d}$$ is equivalent to $$d$$ on $$\mathbb{R}$$. Then for all $$\vect{y} \in B_{\bar{\rho}}(\vect{x}, \varepsilon)$$, we have

$\sup_{\alpha \in J} \{ \bar{d}(x_{\alpha}, y_{\alpha}) \} = \sup_{\alpha \in J} \{ d(x_{\alpha}, y_{\alpha}) \} < \varepsilon.$

Therefore, for all $$\alpha \in J$$, $$y_{\alpha} \in (x_{\alpha} - \varepsilon, x_{\alpha} + \varepsilon)$$. Then we may tend to say that $$\prod_{\alpha \in J} (x_{\alpha} - \varepsilon, x_{\alpha} + \varepsilon)$$ is a basis element for the box topology containing $$\vect{y}$$, which is contained in $$B_{\bar{\rho}}(\vect{x}, \varepsilon)$$. However, this is not true. Because $$\vect{y}$$ can be thus selected such that as $$\alpha$$ changes in $$J$$, $$\bar{d}(x_{\alpha}, y_{\alpha})$$ can be arbitrarily close to $$\varepsilon$$, which leads to $$\sup_{\alpha \in J} \{ \bar{d}(x_{\alpha}, y_{\alpha}) \} = \varepsilon$$. This makes $$\vect{y} \notin B_{\bar{\rho}}(\vect{x}, \varepsilon)$$ and $$\prod_{\alpha \in J} (x_{\alpha} - \varepsilon, x_{\alpha} + \varepsilon)$$ is not contained in $$B_{\bar{\rho}}(\vect{x}, \varepsilon)$$. Such example can be given for $$\mathbb{R}^{\omega}$$, where we let $$\vect{y} = \{y_n = x_n + \varepsilon - \frac{\varepsilon}{n}\}_{n \in \mathbb{Z}_+}$$. When $$n \rightarrow \infty$$, $$\bar{d}(x_n, y_n) \rightarrow \varepsilon$$.

With this point clarified, a smaller basis element should be selected for the box topology, such as $$\prod_{\alpha \in J} (x_{\alpha} - \frac{\varepsilon}{2}, x_{\alpha} + \frac{\varepsilon}{2})$$. For all $$\vect{y}$$ in this basis element, $$\sup_{\alpha \in J} \{ \bar{d}(x_{\alpha}, y_{\alpha}) \} \leq \frac{\varepsilon}{2} < \varepsilon$$. Hence $$\prod_{\alpha \in J} (x_{\alpha} - \frac{\varepsilon}{2}, x_{\alpha} + \frac{\varepsilon}{2}) \subset B_{\bar{\rho}}(\vect{x}, \varepsilon)$$ and the box topology is finer than the uniform topology.

Remark: The proof in the book for this part inherently adopts the definition of open set via topological basis introduced in section 13.

d) Prove the box topology is strictly finer than the uniform topology, when $$J$$ is infinite.

Analysis: Because the open ball in the uniform topology sets an upper bound on the dimension of each coordinate component, it can be envisioned that if we construct a basis element for the box topology with the dimension for each coordinate component approaching to zero, it cannot cover any open ball in the uniform topology with a fixed radius no matter how small it is.

Let’s consider the case in $$\mathbb{R}^{\omega}$$. Select a basis element for the box topology as $$\prod_{n = 1}^{\infty} (x_n - \frac{c}{n}, x_n + \frac{c}{n})$$ with $$(c > 0)$$. Then for all $$\varepsilon > 0$$, there exists $$\vect{y}_0 \in B_{\bar{\rho}}(\vect{x}, \varepsilon)$$ such that $$\vect{y}_0 \notin \prod_{n = 1}^{\infty} (x_n - \frac{c}{n}, x_n + \frac{c}{n})$$. For example, we can select $$\vect{y}_0 = (x_n + \frac{\varepsilon}{2})_{n \geq 1}$$. Then there exists an $$n_0 \in \mathbb{Z}_+$$ such that when $$n > n_0$$, $$\frac{c}{n} < \frac{\varepsilon}{n}$$ and $$y_n \notin (x_n - \frac{c}{n}, x_n + \frac{c}{n})$$. Hence, the box topology is strictly finer than the uniform topology.
转载于:https://www.cnblogs.com/peabody/p/10223052.html
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• 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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• Theorem 18.4 in James Munkres “Topology” states that if a function $$f : A \rightarrow X \times Y$$ is continuous, its coordinate functions $$f_1 : A \rightarrow X$$ and $$f_2 : A \rightarrow Y$$ ...
Theorem 18.4 in James Munkres “Topology” states that if a function $$f : A \rightarrow X \times Y$$ is continuous, its coordinate functions $$f_1 : A \rightarrow X$$ and $$f_2 : A \rightarrow Y$$ are also continuous, and the converse is also true. This is what we have been familiar with, such as a continuous parametric curve $$f: [0, 1] \rightarrow \mathbb{R}^3$$ defined as $$f(t) = (x(t), y(t), z(t))^T$$ with its three components being continuous. However, if a function $$g: A \times B \rightarrow X$$ is separately continuous in each of its components, i.e. both $$g_1: A \rightarrow X$$ and $$g_2 : B \rightarrow X$$ are continuous, $$g$$ is not necessarily continuous.
Here, the said “separately continuous in each of its components” means arbitrarily selecting the value of one component variable from its domain and fix it, then the original function depending only on the other component is continuous. In the above, the function $$g$$ can be envisaged as a curved surface in 3D space. With $$g_1$$ being continuous, the intersection profiles between this curved surface and those planes perpendicular to the coordinate axis for $$B$$ are continuous. Similarly, because $$g_2$$ is continuous, the intersection profiles obtained from those planes perpendicular to the coordinate axis for $$A$$ are also continuous. The continuity of intersection curves is only ensured in these two special directions, so it is not guaranteed that the original function $$g$$ is continuous.
In Exercise 12 of Section 18, an example is given as
$F(x \times y) = \begin{cases} \frac{xy}{x^2 + y^2} & (x \neq 0, y \neq 0) \\ 0 & (x = 0, y = 0) \end{cases},$
where $$F$$ is continuous separately in each of its component variables but is not continuous by itself. This is function is visualized below.

Fix $$y$$ at $$y_0$$, we have $$F_{y_0}(x) = F(x \times y_0)$$. When $$y_0 \neq 0$$, $$F_{y_0}(x)$$ is continuous with respect to $$x$$ because it is only a composition of continuous real valued functions via simple arithmetic. When $$y_0 = 0$$, if $$x \neq 0$$, $$F_0(x) = 0$$; if $$x =0$$, $$F_0(x)$$ is also 0 due to the definition of $$F(x \times y)$$. Therefore, $$F_0(x)$$ is a constant function, which is continuous due to Theorem 18.2 (a). Similarly, $$F_{x_0}(y)$$ is also continuous with respect to $$y$$.
However, if we let $$x = y$$ and approach $$(x, y) = (x, x)$$ to $$(0, 0)$$, it can be seen that $$F(x \times x)$$ is not continuous, because
when $$x \neq 0$$, $$F(x \times x) = \frac{x^2}{x^2 + x^2} = \frac{1}{2}$$;
when $$x = 0$$, $$F(x \times x) = 0$$.

If we let $$x = -y$$ and approach $$(x ,y) = (x, -x)$$ to $$(0, 0)$$, $$F = -\frac{1}{2}$$ when $$x \neq 0$$ and $$F = 0$$ when $$x = 0$$.
Then, if we select an open set such as $$(-\frac{1}{4}, \frac{1}{4})$$ around the function value $$0$$ in $$\mathbb{R}$$, its pre-image $$U$$ in $$\mathbb{R} \times \mathbb{R}$$ should include the point $$(0, 0)$$ and exclude the rays $$(x, x)$$ and $$(x, -x)$$ with $$x \in \mathbb{R}$$ and $$x \neq 0$$. Due to these excluded rays, there is no neighborhood of $$(0, 0)$$ in $$\mathbb{R} \times \mathbb{R}$$ that is contained completely in $$U$$. Therefore, $$U$$ is not an open set and $$F(x \times y)$$ is not continuous.
From the above analysis, some lessons can be learned.
Pure analysis can be made and general conclusions can be obtained before entering into the real world with a solid example.
A tangible counter example is a sound proof for negation of a proposition. Just one is enough!

转载于:https://www.cnblogs.com/peabody/p/10140171.html
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• Because the box topology is finer than the product topology, the projection map is also continuous when the box topology is adopted for $$\prod X_{\alpha}$$. Therefore, this part of the theorem is ...
Theorem 19.6 Let $$f: A \rightarrow \prod_{\alpha \in J} X_{\alpha}$$ be given by the equation

$f(a) = (f_{\alpha}(a))_{\alpha \in J},$

where $$f_{\alpha}: A \rightarrow X_{\alpha}$$ for each $$\alpha$$. Let $$\prod X_{\alpha}$$ have the product topology. Then the function $$f$$ is continuous if and only if each function $$f_{\alpha}$$ is continuous.

Comment: This is an extension of Theorem 18.4, where only two component spaces are involved.

Proof: a) First, we prove the projection map is continuous, which is defined on the Cartesian space constructed from a $$J$$-tuple of component spaces .

For all $$\beta \in J$$, let $$\pi_{\beta}: \prod X_{\alpha} \rightarrow X_{\beta}$$ be the projection map. For arbitrary open set $$V_{\beta}$$ in $$X_{\beta}$$, its pre-image under $$\pi_{\beta}$$ is $$\pi_{\beta}^{-1}(V_{\beta})$$, which is a subbasis element for the product topology on $$\prod X_{\alpha}$$. Therefore, $$\pi_{\beta}^{-1}(V_{\beta})$$ is open and the projection map $$\pi_{\beta}$$ is continuous.

Next, we notice that for all $$\alpha \in J$$, the coordinate function $$f_{\alpha}: A \rightarrow X_{\alpha}$$ is a composition of the two continuous functions $$f$$ and $$\pi_{\alpha}$$, i.e. $$f_{\alpha} = \pi_{\alpha} \circ f$$. Then according to Theorem 18.2 (c), $$f_{\alpha}$$ is continuous.

Remark: Because the box topology is finer than the product topology, the projection map is also continuous when the box topology is adopted for $$\prod X_{\alpha}$$. Therefore, this part of the theorem is true for both product topology and box topology.

b) Analysis: To prove the continuity of a function, showing that the pre-image of any subbasis element in the range space is open in the domain space is more efficient than using basis element or raw open set in the range space. In addition, the subbasis element for the product topology on $$\prod X_{\alpha}$$ has the form $$\pi_{\beta}^{-1}(U_{\beta})$$ with $$U_{\beta}$$ being a single coordinate component and open in $$X_{\beta}$$. This is the clue of the proof.

For all $$\beta \in J$$ and arbitrary open set $$U_{\beta}$$ in $$X_{\beta}$$, we have $$f_{\beta}^{-1}(U_{\beta}) = f^{-1} \circ \pi_{\beta}^{-1}(U_{\beta})$$. Because $$f_{\beta}$$ is continuous and $$U_{\beta}$$ is open, $$f_{\beta}^{-1}(U_{\beta})$$ is open. In addition, $$\pi_{\beta}^{-1}(U_{\beta})$$ is an arbitrary subbasis element for $$\prod X_{\alpha}$$ with the product topology, whose pre-image under $$f$$ is just the open set $$f_{\beta}^{-1}(U_{\beta})$$, therefore $$f$$ is continuous.

Remark: Part b) of this theorem really depends on the adopted topology for $$\prod X_{\alpha}$$, which can be understood as below.

At first, we will show that for all $$\vect{U} = \prod U_{\alpha}$$ being a subset of $$\prod X_{\alpha}$$, $$f^{-1}(\vect{U}) = \bigcap_{\alpha \in J} f_{\alpha}^{-1}(U_{\alpha})$$.

For all $$x \in f^{-1}(\vect{U})$$, because $$f(x) \in \vect{U}$$, then for all $$\alpha \in J$$, $$f_{\alpha}(x) \in U_{\alpha}$$, hence $$x \in \bigcap_{\alpha \in J} f_{\alpha}^{-1}(U_{\alpha})$$ and $$f^{-1}(\vect{U}) \subset \bigcap_{\alpha \in J} f_{\alpha}^{-1}(U_{\alpha})$$.

On the other hand, for all $$x \in \bigcap_{\alpha \in J} f_{\alpha}^{-1}(U_{\alpha})$$, we have for all $$\alpha \in J$$, $$f_{\alpha}(x) \in U_{\alpha}$$. Therefore, $$f(x) \in \vect{U}$$ and $$x \in f^{-1}(\vect{U})$$. Hence $$\bigcap_{\alpha \in J} f_{\alpha}^{-1}(U_{\alpha}) \subset f^{-1}(\vect{U})$$.

Next, if we assign the product topology to $$\prod X_{\alpha}$$, for any $$\vect{U} = \prod U_{\alpha}$$ with $$U_{\alpha}$$ open in $$X_{\alpha}$$ and only a finite number of them not equal to $$X_{\alpha}$$, it is a basis element of the product topology. Let the set of all indices with which $$U_{\alpha} \neq X_{\alpha}$$ be $$\{\alpha_1, \cdots, \alpha_n\}$$ and also notice that when $$U_{\alpha} = X_{\alpha}$$, $$f_{\alpha}^{-1}(U_{\alpha}) = A$$, we have

$f^{-1}(\vect{U}) = \bigcap_{\alpha \in J} f_{\alpha}^{-1}(U_{\alpha}) = \bigcap_{i=1}^n f_{\alpha_i}^{-1}(U_{\alpha_i}), \tag{*} \label{eq:intersection}$

where those $$f_{\alpha}^{-1}(U_{\alpha})$$ with $$\alpha \notin \{\alpha_1, \cdots, \alpha_n\}$$ do not contribute to the intersection. This indicates that $$f^{-1}(\vect{U})$$ is a finite intersection of open sets which is still open. Hence $$f$$ is continuous.

However, if the box topology is adopted for $$\prod X_{\alpha}$$, qualitatively speaking, because the topology for the range space becomes finer, according to our previous post, it makes a function to be continuous more difficult. Specifically in this theorem, $$f^{-1}(\vect{U})$$ in \eqref{eq:intersection} can be an intersection of infinite number of open sets $$U_{\alpha}$$ not equal to $$X_{\alpha}$$. Thus $$f^{-1}(\vect{U})$$ may not be open anymore.

After understanding this point, it is not difficult to construct a counter example for part b) as below.

Let $$f: \mathbb{R} \rightarrow \mathbb{R}^{\omega}$$ be defined as $$f(t) = (t, t, \cdots)$$. Select a basis element $$\vect{U}$$ in $$\mathbb{R}^{\omega}$$ such that the intersection of all its coordinate components is not open. For example, $$\vect{U} = \prod_{n=1}^{\infty} (-\frac{1}{n}, \frac{1}{n})$$, which is a neighborhood of $$f(0) = (0, 0, \cdots)$$.

For any basis element $$(a, b)$$ in $$\mathbb{R}$$ containing $$0$$, with $$a < 0$$ and $$b > 0$$, by letting $$\delta = \min\{-a, b\}$$, we have $$(-\delta, \delta) \subset (a, b)$$ and $$0 \in (-\delta, \delta)$$. The image of $$(-\delta, \delta)$$ under $$f$$ is $$\prod_{n=1}^{\infty} (-\delta, \delta)$$. Then there exist an $$n_0 \in \mathbb{Z}_+$$ such that $$(-\delta, \delta)$$ is not contained in $$(-\frac{1}{n_0}, \frac{1}{n_0})$$. Therefore, $$\pi_{n_0}(f((-\delta, \delta)))$$ is not contained in $$\pi_{n_0}(\vect{U})$$ and $$\pi_{n_0}(f((a, b)))$$ is not contained in $$\pi_{n_0}(\vect{U})$$. Hence the image of $$(a, b)$$ under $$f$$ is not contained in $$\vect{U}$$. This contradicts Theorem 18.1 (4) and $$f$$ is not continuous.

转载于:https://www.cnblogs.com/peabody/p/10165090.html
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