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Topologies on product spaces of $\mathbb{R}$ and their relationships

作者:互联网

In this post, I will summarise several topologies established on the product spaces of \(\mathbb{R}\), i.e. \(\mathbb{R}^n\), \(\mathbb{R}^{\omega}\) and \(\mathbb{R}^J\), as well as their relationships.

Topologies on product spaces of \(\mathbb{R}\)

  1. Topology induced from the euclidean metric \(d\) on \(\mathbb{R}^n\), where for all \(\vect{x}, \vect{y} \in \mathbb{R}^n\),
    \[
    d(\vect{x}, \vect{y}) = \left( \sum_{i=1}^n (x_i - y_i)^2 \right)^{\frac{1}{2}}.
    \]
  2. Topology induced from the square metric \(\rho\) on \(\mathbb{R}^n\), where for all \(\vect{x}, \vect{y} \in \mathbb{R}^n\),
    \[
    \rho(\vect{x}, \vect{y}) = \max_{1 \leq i \leq n} \abs{x_i - y_i}.
    \]
  3. Product topology on \(\mathbb{R}^J\): its basis has the form \(\vect{B} = \prod_{\alpha \in J} U_{\alpha}\), where each \(U_{\alpha}\) is an open set in \(\mathbb{R}\) and only a finite number of them are not equal to \(\mathbb{R}\).

    Specifically, when \(J = \mathbb{Z}_+\), the product topology on \(\mathbb{R}^{\omega}\) can be constructed.

  4. Box topology on \(\mathbb{R}^J\): its basis has the form \(\vect{B} = \prod_{\alpha \in J} U_{\alpha}\), where each \(U_{\alpha}\) is an open set in \(\mathbb{R}\).

    Specifically, when \(J = \mathbb{Z}_+\), the box topology on \(\mathbb{R}^{\omega}\) can be constructed.

  5. Uniform topology on \(\mathbb{R}^J\): it is induced by the uniform metric \(\bar{\rho}\) on \(\mathbb{R}^J\), where for all \(\vect{x}, \vect{y} \in \mathbb{R}^J\),
    \[
    \bar{\rho}(\vect{x}, \vect{y}) = \sup_{\alpha \in J} \{ \bar{d}(x_{\alpha}, y_{\alpha}) \}
    \]
    with \(\bar{d}\) being the standard bounded metric on \(\mathbb{R}\).

    Specifically, when \(J = \mathbb{Z}_+\), the uniform topology on \(\mathbb{R}^{\omega}\) can be obtained.

    When \(J = n\), the topology induced from the metric \(\bar{\rho}\) on \(\mathbb{R}^n\) is equivalent to the topology induced from the square metric \(\rho\).

  6. Topology induced from the metric \(D\) on \(\mathbb{R}^{\omega}\), where for all \(\vect{x}, \vect{y} \in \mathbb{R}^{\omega}\),
    \[
    D(\vect{x}, \vect{y}) = \sup_{i \in \mathbb{Z}_+} \left\{ \frac{\bar{d}(x_i, y_i)}{i} \right\},
    \]
    which is transformed from the uniform metric \(\bar{\rho}\) by suppressing its high frequency component.

    Specifically, when \(J = n\), the topology induced from the metric \(D\) is equivalent to the topology induced from the metric \(\bar{\rho}\) and hence is also equivalent to the topology induced from the square metric \(\rho\).

N.B. In the definitions of product topology and box topology for \(\mathbb{R}^J\) as above, the openness of \(U_{\alpha}\) in \(\mathbb{R}\) is with respect to the standard topology on \(\mathbb{R}\), which does not require a metric to be induced from but only depends on the order relation on \(\mathbb{R}\).

Relationships between topologies on product spaces of \(\mathbb{R}\)

According to Theorem 20.3 and Theorem 20.4, the following points about the relationships between topologies on product spaces of \(\mathbb{R}\) are summarised.

  1. On \(\mathbb{R}^n\): Topology induced from \(\rho\) \(\Leftrightarrow\) Uniform topology induced from \(\bar{\rho}\) \(\Leftrightarrow\) Topology induced from \(D\) \(\Leftrightarrow\) Product topology \(\Leftrightarrow\) Box topology.
  2. On \(\mathbb{R}^{\omega}\): Topology induced from \(D\) \(\Leftrightarrow\) Product topology \(\subsetneq\) Uniform topology induced from \(\bar{\rho}\) \(\subsetneq\) Box topology.
  3. On \(\mathbb{R}^J\): Product topology \(\subsetneq\) Uniform topology induced from \(\bar{\rho}\) \(\subsetneq\) Box topology.

It can be seen that the finite dimensional Euclidean space \(\mathbb{R}^n\) has the most elegant property, where all topologies are equivalent.

标签:mathbb,topology,vect,induced,rho,metric,bar
来源: https://www.cnblogs.com/peabody/p/10260040.html