# Question about proof that euclidean metric and square metric are the product topology

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 $d$

• 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. – Daniel Fischer Nov 15 '16 at 22:50
• @DanielFischer how do you know that? Geometrically? I can't see it. – Guerlando OCs Nov 16 '16 at 0:04
• 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. – Daniel Fischer Nov 16 '16 at 0:39
• @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? – Guerlando OCs Nov 16 '16 at 1:34

From the geometry, it is sort of obvious that $B_{d}\left(x, \epsilon\right) \subset B_{p}\left(x, \epsilon\right)$
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 $\frac{1}{\sqrt{n}} < \alpha < 1$, $y \not\in B_d\left(x, \epsilon\right)$.