So I have shown that any $x,y\in\mathbb{R}^n$ that $d_\infty(x,y) \le d_2(x,y)\le d_1(x,y)\le nd_\infty(x,y)$, where $$d_\infty=\max\limits_{j\in\{1,\dots,n\}}\lvert x_j-y_j\rvert,$$ $$d_2(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2+\cdots+(x_n-y_n)^2},$$ $$d_1(x,y)=|x_1-y_1|+|x_2-y_2|+\cdots+|x_n-y_n|.$$ Will refer to this result as (1).

I have also shown that for metrics that satisfy $d(x_1,x_2)\le kd'(x_1,x_2)$ for some positive constant k and $x_1,x_2 \in X$ that any subset of $X$ that is $d$-open is also $d'$-open. Will refer to this result as (2).

I now have to show that the open sets in $\mathbb{R}^n$ are the same for the metrics $d_1,d_2,d_\infty$.

What I am wondering is that if the following answer is good enough: Let $(\tau,d)$ denote the family of all $d$-open subsets of $X=\mathbb{R}^n$. From (2) we have that $(\tau,d_1)=(\tau,d_\infty)$. From (1) we have that $d_\infty(x,y)\le d_2(x,y)\le d_1(x,y)$, so that must mean that $(\tau,d_1)=(\tau,d_\infty)=(\tau,d_2)$. So all open sets in $\mathbb{R}^n$ are the same for the three metrics.

Thank you for your time


There's something wrong in your argument.

You have written:

From (2) we have that $(\tau,d_1)=(\tau,d_\infty)$.

But (2), as stated, starting with the hypothesis $d \le k d'$, does not let you conclude that $(\tau,d)=(\tau,d')$, it only says that $(\tau,d) \subset (\tau,d')$ because that's what it means to say "any subset of $X$ that is $d$ open is also $d'$ open".

So (2), when coupled with $d_1 \le n d_\infty$, only gives you the inclusion $$(\tau,d_1) \subset (\tau,d_\infty) $$ Similarly, (2) when coupled with $d_\infty \le d_2$, gives you the inclusion $$(\tau,d_\infty) \subset (\tau,d_2) $$ Perhaps you can see where this is going now, and can complete the proof.

  • $\begingroup$ Ahh, so (1) and (2) combined gives that $(\tau,d_1)⊂(\tau,d_\infty)$ and $(\tau,d_\infty)⊂(\tau,d_2)$ and $(\tau,d_2)⊂(\tau,d_1)$, so that means that $(\tau,d_1)=(\tau,d_\infty)=(\tau,d_2)$. Is that correct then? $\endgroup$ – Nikolaj Nov 9 '18 at 11:23
  • 1
    $\begingroup$ That's correct. $\endgroup$ – Lee Mosher Nov 9 '18 at 13:38

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