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

2019-01-12 16:01:33 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 $\mathbb{R}$

Topology induced from the euclidean metric $d$ on $\mathbb{R}^n$, where for all $\vect{x}, \vect{y} \in \mathbb{R}^n$,
\[
d(\vect{x}, \vect{y}) = \left( \sum_{i=1}^n (x_i - y_i)^2 \right)^{\frac{1}{2}}.
\]
Topology induced from the square metric $\rho$ on $\mathbb{R}^n$, where for all $\vect{x}, \vect{y} \in \mathbb{R}^n$,
\[
\rho(\vect{x}, \vect{y}) = \max_{1 \leq i \leq n} \abs{x_i - y_i}.
\]
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.
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.
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 $\rho$.
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.

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.
On $\mathbb{R}^{\omega}$: Topology induced from $D$ $\Leftrightarrow$ Product topology $\subsetneq$ Uniform topology induced from $\bar{\rho}$ $\subsetneq$ Box topology.
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

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

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

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