I'm reading a proof that says that the topologies on $\mathbb{R}^n$ induced by the euclidean metric $d$ and the square metric $p$ are the same as the product topology on $\mathbb{R}^n$. It goes like this:

Let $x = (x_1,\cdots, x_n)$ and $y=(y_1,\cdots, y_n)$ be two points of $\mathbb{R}^n$. Them:

$$p(x,y)\le d(x,y)\le \sqrt{n}p(x,y)$$

The first inequalioty shows that

$$B_d(x,\epsilon)\subset B_p(x,\epsilon)$$ since if $d(x,y)< \epsilon$ then $p(x,y)< \epsilon$

then the proof continues...

Well, why the inclusion of the balls is in that direction? For me, since $p(x,y)\le d(x,y)$, then the ball with respect to $p$ should be included in theone with respect to $d$

  • 1
    $\begingroup$ If $p(x,y) \leqslant d(x,y)$, then it's easier for two points to have a $p$-distance smaller than $\epsilon$ than to have a $d$-distance below $\epsilon$. So the $p$-ball of the same radius is larger (not necessarily strictly) than the $d$-ball of the same radius. $\endgroup$ – Daniel Fischer Nov 15 '16 at 22:50
  • $\begingroup$ @DanielFischer how do you know that? Geometrically? I can't see it. $\endgroup$ – Guerlando OCs Nov 16 '16 at 0:04
  • 1
    $\begingroup$ Let's say we have two runners, $A$ and $B$. $p(x,y)$ is the time runner $A$ needs to run from $x$ to $y$, and $d(x,y)$ is the time runner $B$ needs. $A$ is the faster runner (but for some tracks they may not be able to actually run faster than $B$, say because they both sink in mud), so $p(x,y) \leqslant d(x,y)$. Then $B_p(x,\epsilon)$ is the set of points $A$ could reach from point $x$ in time $\epsilon$, and $B_d(x,\epsilon)$ is the set of points $B$ could reach in time $\epsilon$. Since $A$ is faster, they can cover the greater area in a given time. $\endgroup$ – Daniel Fischer Nov 16 '16 at 0:39
  • $\begingroup$ @DanielFischer I completely understood the analogy, but how did you know about the 'velocities'? Did you plot the metrics? How do you know it's indeed valid by only visualizing it geometrically? $\endgroup$ – Guerlando OCs Nov 16 '16 at 1:34

From the geometry, it is sort of obvious that $B_{d}\left(x, \epsilon\right) \subset B_{p}\left(x, \epsilon\right)$

Sketch of the geometry

However, you asked for a proof. First let us show $y \in B_{d}\left(x, \epsilon\right) \Rightarrow y \in B_{p}\left(x, \epsilon\right)$:

$$ y \in B_{d}\left(x, \epsilon\right) \\ \Leftrightarrow d(x, y) < \epsilon \\ \Rightarrow p(x, y) < \epsilon \\ \Leftrightarrow y \in B_{p}\left(x, \epsilon\right) $$

where the third line is the inequality $p(x, \epsilon) \le d(x, \epsilon)$. To show that $y \in B_{p}\left(x, \epsilon\right) \nRightarrow y \in B_{p}\left(x, \epsilon\right)$, consider the point $y=\left( x_1 + \alpha\epsilon, x_2 + \alpha \epsilon, \dots \right)^T$, then $p(x, y)=\alpha\epsilon$, while $d(x, y)=\sqrt{n}\alpha\epsilon$, so for $\frac{1}{\sqrt{n}} < \alpha < 1$, $y \not\in B_d\left(x, \epsilon\right)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.