----- > [!proposition] Proposition. ([[euclidean, sup, and product topology are equal in finite dimensions]]) > The [[topological space|topologies]] on $\mathbb{R}^{n}$ induced by the [[Euclidean metric]] $d$ and the [[sup Metric]] $\rho$ are the same as the [[product topology]] on $\mathbb{R}^{n}$. > [!proof]- Proof. ([[euclidean, sup, and product topology are equal in finite dimensions]]) > > We use the [[basis-nestling characterization of comparing topologies]]. We know that, in general, $\rho(x,y) \leq d(x,y) \leq \sqrt{ n }\rho (x,y)$ > for all $x,y \in \mathbb{R}^{n}$. The first inequality shows that $B_{d}(x, \varepsilon) \subset B_{\rho}(x, \varepsilon)$ > for all $x \in \mathbb{R}^{n}$ and $\varepsilon>0$, for if $d(x,y) < \varepsilon$ then $\rho(x,y) < \varepsilon$ also. Similarly, the second shows that $B_{\rho}(x,\varepsilon) \subset B_{d }\left( x, \frac{\varepsilon}{\sqrt{ n }} \right).$ > Thus $\tau_{d}=\tau_{\rho}$. To show that each equals the [[product topology]] on $\mathbb{R}^{n}$, consider a basis element $B=\prod_{i=1}^{n}U_{i}$ of the [[product topology]], where [[product of factor bases is basis of product topology|WLOG]] each $U_{i}$ is an [[standard topology on the real line|open interval]] of the real line: $B=\prod_{i=1}^{n}(a_{i}, b_{i}).$ > > Let $x^{(0)} \in B$ be arbitrary, and set $\varepsilon:=\min \{ |a_{i} - b_{i}| \}_{i=1}^{n}$. Then $B_{\rho}(x_{0}, \varepsilon) \subset B$, for if $x=(x_{1},\dots,x_{n}) \in B_{\rho}(x_{0}, \varepsilon)$ then $\max \{ |x_{i}-x^{(0)}| \}_{i=1}^{n} < \min \{ |a_{i} - b_{i}| \}_{i=1}^{n}$ > and therefore > > An arbitrary basis element of $\tau_{\rho}$ looks like a cube $B_{\rho}(x^{(0)}, \varepsilon)$ for some $x^{(0)} \in \mathbb{R}^{n}$ and $\varepsilon>0$. It is clear that $B_\rho(x^{(0)}, \varepsilon) = \prod_{i=1}^{n} (x_{i}- \varepsilon, x_{i}+\varepsilon) \in \mathscr{B}_{\text{product}}.$ > So every basis element of $\tau_{\rho}$ belongs to $\tau_{\text{product}}$, from it which it follows that every element of $\tau_{\rho}$ (as a union of such elements) belongs to $\tau_{\text{product}}$, so $\tau_{\rho} \subset \tau_{\text{product}}$. > > Next we employ the [[basis-nestling characterization of comparing topologies]] to show $\tau_{\text{product}} \subset \tau_{\rho}$. Let $B_{\rho}(\overline{x}, \varepsilon)$ be an arbitrary basis element for $\tau_{\rho}$ where $\overline{x}=(\overline{x}_{1 }, \dots, \overline{x}_{2}) \in \mathbb{R}^{n}$ and $\varepsilon>0$. Then, for any $x=(x_{1},\dots,x_{n}) \in B_{\rho}(\overline{x}, \varepsilon)$, it is clear that $x \in \prod_{i=1}^{n}(x_{i}, \overline{x}_{i}) \subset B_{\rho}(\overline{x}, \varepsilon)$ which implies that $\tau_{\text{product}} \subset \tau_{\rho}$. > ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```