I am working on this problem from Simmons' "Introduction to Topology and Modern Analysis" (Problem 10-6).

Let $X_1, X_2, \dots, X_n$ be a finite class of metric spaces with metrics $d_1, d_2, \dots, d_n$. If $X = X_1 \times X_2 \times \dots \times X_n$, and $d$ and $\bar{d}$ are metrics defined as follows:

$$d(\{x_i\}, \{y_i\}) = \max d_i(x_i, y_i)$$ $$\bar{d}(\{x_i\}, \{y_i\}) = \sum_{i=1}^{n} d_i(x_i, y_i)$$

then show that the two metric spaces $(X, d)$ and $(X, \bar{d})$ have precisely the same open sets.

(Here, $\{x_i\}$ denotes an n-tuple from $X = X_1 \times X_2 \times \dots \times X_n$)

Here is my approach:

I know a theorem that two metrics are equivalent iff. every open ball in the first metric contains an open ball from the second metric (centred at the same point) and vice-versa.

So, let $B_d(\{x_i\}, r)$ be and open ball in $(X, d)$. Here, $B_d(p, r)$ denotes an open ball with respect to metric $d$, centered at p and with radius $r$. Let $\{y_i\} \in B_d(\{x_i\}, r)$, now, we wish to show that $\{y_i\} \in B_{\bar{d}}(\{x_i\}, s)$ for some $s > 0$ (possibly a function of $r$?).

Now, $$\{y_i\} \in B_d(\{x_i\}, r) \implies \max d_i(x_i, y_i) < r \implies \sum_{i=1}^{n} d_i(x_i, y_i) < nr \\ \implies \{y_i\} \in B_{\bar{d}}(\{x_i\}, nr)$$

Similarly, let $\{y_i\} \in B_{\bar{d}}(\{x_i\}, r)$, then,

$$\{y_i\} \in B_{\bar{d}}(\{x_i\}, r) \implies \sum_{i=1}^{n} d_i(x_i, y_i) < r \implies \max d_i(x_i, y_i) < r \\ \implies \{y_i\} \in B_d(\{x_i\}, r)$$

So, we conclude that the metrics $d$ and $\bar{d}$ are equivalent.

I would really appreciate some comments/correction on this proof. In particular, I would like to know that when we show an open ball (with respect to first metric) having radius $r$ is contained in an open ball (with respect to second metric) having radius $s$, then, can $s$ be a function of $r$ or must it be constant depending only on the center? At this point I haven't yet studied homeomorphisms or Lipschitz equivalence of norms, so please avoid using those concepts.

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    $\begingroup$ For the first part you wish to show that there exists $s>0$ such that $B_d(x,s)\subset B_{\bar d}(x,r)$ when $r>0.$ So you should say "Let $s= r/n,$,,, (etc.)"...BTW Metrics that generate the same topology are called equivalent. If for metrics $ d,e$ there exists positive $K, L$ such that $K e(x,y)\leq d(x,y)\leq Le(x,y)$ for all $x,y$ then $d,e$ are called uniformly equivalent . Uniform equivalence is sufficient, but not necessary, for $d,e$ to be equivalent . In the Q, let $e=\bar d$ and $K=1$ and $L=n. $ $\endgroup$ – DanielWainfleet Feb 17 '18 at 9:58
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    $\begingroup$ In general $s$ may depend on both $r$ and on the center of the open ball. For example :The function $f(x)=\tan x$ is a homeomorphism from $(-\pi /2,\pi /2)$ to $\Bbb R.$ So on $(-\pi /2,\pi /2)$ we can take the metric $d(x,y)=|x-y|$ or the equivalent metric $e(x,y)=|f(x)-f(y)|.$ $\endgroup$ – DanielWainfleet Feb 17 '18 at 10:09
  • $\begingroup$ @DanielWainfleet Thanks for the reply. I understand the first part of your comment. Since, the theorem states that two metric are equivalent if "every open ball in the first metric contains an open ball from the second metric" and not "every open ball in the first metric is contained by an open ball in the second metric". So, we should first show that $B_d(x, s) \subset B_{\bar{d}}(x, r)$ by choosing $s = r/n$. Now, regarding uniform equivalence, is it not necessary because it is derived from uniform equivalence of norms and not every metric is induced by a norm? $\endgroup$ – feynhat Feb 17 '18 at 14:01
  • $\begingroup$ Uniform equivalence of metrics is as I stated. It also happens to be true that any two norms on $\Bbb R^n$ or on $\Bbb C^n$ are uniformly equivalent as metrics, for a finite $n.$.... It is easy to show that uniformly equivalent metrics are equivalent using the method of your answer to the Q. If you had this result already then you could apply it to $d$ and $\bar d$ as in my 1st comment.....BTW. If $d,e$ are unformly equivalent and $d$ is bounded then (obviously) $e$ is bounded...BTW. Any metric $d$ is equivalent to the bounded metric $e(x,y)=\min (1,d(x,y)).$ $\endgroup$ – DanielWainfleet Feb 17 '18 at 15:34

Note that $\|x\|_\infty \le \|x\|_1 \le n \|x\|_\infty $.

Hence $ {d}(\{x_i\}, \{y_i\}) \le \bar{d}(\{x_i\}, \{y_i\}) \le n{d}(\{x_i\}, \{y_i\}) $

Just to clarify, if $x = (x_1,...,x_n)$ then $d(x,y) = \|(d_1(x_1,y_1),...,d_n(x_n,y_n))\|_\infty$ and $\bar{d}(x,y) = \|(d_1(x_1,y_1),...,d_n(x_n,y_n))\|_1$.

  • $\begingroup$ But the problem never stated that the metrics $d_1, d_2, \dots, d_n$ are induced by norms. So, I don't understand how that inequality for sup norm and 1-norm is useful in this case. $\endgroup$ – feynhat Feb 17 '18 at 9:09
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    $\begingroup$ The metrics can be written as a norm of the vector with components $d_i ( x_i , y_i ) $. $\endgroup$ – copper.hat Feb 17 '18 at 14:29
  • $\begingroup$ I am not stating that the metric is a norm, but in the case they can be written as a norm of a vector of metrics so we can use the norm equivalence to obtain the desired result. $\endgroup$ – copper.hat Feb 17 '18 at 14:31
  • $\begingroup$ Now it makes sense. You applied that norm inequality on the vector $(d_1(x_1, y_1), d_2(x_2, y_2), \dots, d_n(x_n, y_n))$, and then used uniform equivalence, right? $\endgroup$ – feynhat Feb 17 '18 at 15:09
  • $\begingroup$ @feynhat: That is correct. $\endgroup$ – copper.hat Feb 17 '18 at 15:14

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