1
$\begingroup$

Definition (equivalent metrics) Two metrics on a set $X$ are called equivalent if for all $x \in X$ $$ \forall \varepsilon > 0 \ \exists r > 0: U_1(x,r) \subset U_2(x,\varepsilon) \quad\text{and} \quad U_2(x,r) \subset U_1(x,\varepsilon), $$ where $U_i(x,r) := \{ y \in X: d_i(x,y) < r\}$ for $i \in \{1,2\}$.


Task: Show that two metrics $d_1$ and $d_2$ are equivalent if and only if $d_1$ and $d_2$- convergent sequences coincide.

Is there a constructive proof for the direction of "$\impliedby$"?

This is my approach: "$\implies$" should be correct and hopefully my approach with contraction for "$\impliedby$", too, but feel free to correct any mistakes, might they be mathematically or about proof-writing.

Proof. "$\implies \DeclareMathOperator{\N}{\mathbb{N}} \DeclareMathOperator{\eps}{\varepsilon}$": Let $x \in X$ and $\eps > 0$ and $(x_n)_{n \in \N} \subset X$ $d_1$-convergent to $x$. This means there exits an $N_{\eps} \in \N$ such that \begin{equation*} \{ x_n: n \ge N_{\eps} \} \subset U_1(x,r) \subset U_2(x, \eps), \end{equation*} where the second inclusion and the existence of such an $r > 0$ is due to $d_1$ and $d_2$ being equivalent.

"$\impliedby$": Let $x \in X$ and $\eps>0$ and define $U^{(n)} := U_1\left(x,\frac{1}{n}\right)$. Any sequence with $x_n\in U^{(n)}$ for all $n \in \N$ converges to $x$ in $(X,d_1)$.

Towards contradiction assume that there exists no $r > 0$ such that $U_1(x,r) \subset U_2(x,\eps)$ holds. Then we can find a sequence with $x_n\in U^{(n)}$ with $x_n\notin U_2(x,\eps)$ but that would imply that $x_n$ is not convergent to $x$ in $(X,d_2)$, a contradiction.

Therefore, we find a $r_1>0$ satisfying $U_1(x,r_1)\subset U_2(x,\eps)$. For reasons of symmetry we can find a $r_2$ satisfying $U_2(x,r_2)\subset U_1(x,\eps)$. Choosing $r=\min\{r_1,r_2\}$ finishes the proof.

Remark: I know that a equivalent charaterisation of equivalent metrics is that for all $x,y \in X$ there exists constants $C >0$ and $c := C^{-1} > 0$ such that $$c d_1(x,y) \le d_2(x,y) \le C d_1(x,y).$$ The equivalence between this characterisation and the convergence property is clear, and not what I am trying to prove.

$\endgroup$
1
$\begingroup$

$\impliedby$ First of all the following remark is useful:

The metrics $d_1$ and $d_2$ induce two topologies $\tau_1$ and $\tau_2$ that are equal if and only if $d_1$ and $d_2$ are equivalent.

$\rightarrow$

Let $x\in X$ and $\epsilon>0$. You can consider the ball $B^{d_1}(x,\epsilon)$ that is an open set on $\tau_1=\tau_2$ so, by definition of Topology induced by a metric, it is a union of some balls with respect to $\tau_2$:

$B^{d_1}(x,\epsilon)= \bigcup_{y\in A, s\in I} B^{d_2}(y, s)$ for an appropriate subset $A\subset X$ and $I\subset \mathbb{R}^+_0$

In particular there exists $z\in X$ and $S>0$ such that

$B^{d_2}(z, S)\subseteq B^{d_1}(x,\epsilon)$ and $x\in B^{d_2}(z, S)$

Let $r_1:=S-d_2(x,z)>0$, then

$B^{d_2}(x, r_1)\subseteq B^{d_2}(z, S)\subseteq B^{d_1}(x,\epsilon)$

So $B^{d_2}(x,r_1)\subseteq B^{d_1}(x,\epsilon)$

Now we can do the same reasoning considering te ball $B^{d_2}(x,\epsilon)$, that is an open set of the Topology $\tau_2=\tau_1$, so, by definition of Topology induced by a metric, we have

$B^{d_2}(x,\epsilon)= \bigcup_{y\in B, s\in J} B^{d_1}(y, s)$ for an appropriate subset $B\subset X$ and $J\subset \mathbb{R}^+_0$

In particular there exists $z\in X$ and $S>0$ such that

$B^{d_1}(z, S)\subseteq B^{d_2}(x,\epsilon)$ and $x\in B^{d_1}(z, S)$

Let $r_1=S-d_1(x,z)>0$, then

$B^{d_1}(x, r_2)\subseteq B^{d_1}(z, S)\subseteq B^{d_2}(x,\epsilon)$

so

$B^{d_1}(x,r_2)\subseteq B^{d_2}(x,\epsilon)$

To sum up we have

$B^{d_2}(x,r_1)\subseteq B^{d_1}(x,\epsilon)$ and $B^{d_1}(x,r_2)\subseteq B^{d_2}(x,\epsilon)$

If we choose $r:=\min{r_1,r_2}>0$, then we have

$B^{d_2}(x,r)\subseteq B^{d_1}(x,\epsilon)$ and $B^{d_1}(x,r)\subseteq B^{d_2}(x,\epsilon)$

This is exactly the condition of equivalence between two metrics.

$\leftarrow$

We suppose that two metrics must be equivalent and we want to prove that the topologies induced by that metrics are equal.

We know that the set of all balls of a metric space is a base for the topology induced by the metric, so we must only prove that each ball of $(X,d_1)$ is an open set with respect to $\tau_2$ and viceversa.

Let $B^{d_1}(x,\epsilon)$ be a ball of the metric space $(X,d_1)$. Let $y\in B^{d_1}(x,\epsilon)$ and $\epsilon_y:=\epsilon-d_1(x,y)$. Then it is clear that

$B^{d_1}(y,\epsilon_y)\subseteq B^{d_1}(x,\epsilon)$

By definition of equivalence of two metrics there exists $r_y>0$ such that

$B^{d_2}(y,r_y)\subseteq B^{d_1}(y,\epsilon_y)$

and

$B^{d_1}(y,r_y)\subseteq B^{d_2}(y,\epsilon_y)$

So for each $y\in B^{d_1}(x,\epsilon)$ there exists $r_y>0$ such that

$B^{d_2}(y,r_y)\subseteq B^{d_1}(y,\epsilon_y)\subseteq B^{d_1}(x,\epsilon) $

This means that

$ B^{d_1}(x,\epsilon)= \bigcup_{y\in B^{d_1}(x,\epsilon} B^{d_2}(y,r_y)\in \tau_2$

So we have that $\tau_1\subseteq \tau_2$.

In a similar way you can prove that $\tau_2\subseteq \tau_1$.

Now we return to our problem.

We suppose that $d1, d2$ convergent sequences coincide and we want to prove that the two metrics are equivalent. The previous remark tells us that we can simply prove that $\tau_1=\tau_2$.

You can observe the following remark:

Let $X$ be a first countable space and $A\subseteq X$ a subset of $X$. If $a\in X$ is a limit point for your set $A$, then there exists a sequence $\{a_n\}_n\subseteq A$ such that $a_n\to a$. This result is not true in general.

Proof: Let $\{U_n\}_n$ be a countable fundamental neighborhood system for the point $a$. For each $n\in \mathbb{N}$ you can define

$U’_n:=\bigcap_{m\leq n}U_m$

In this case you have that $\{U’_n\}_n$ is again a countable fundamental neighborhood system for the point $a$. So we can choose for each $n\in \mathbb{N}$ a point $a_n\in U’_n\cap A\neq \emptyset$. Then $a_n\to a$ in fact if you consider an open set $B$ of $a$, there exists $U’_N\subseteq B$ so for each $n\geq N$

$a_n\in U’_n\subseteq U’_N\subseteq B$.

This result is not true in general because for example if you consider a non countable set $X$ with co-countable $\tau$ topology, then there is non trivial sequence that converges to some point $a$. In fact if by contradiction there is a non trivial sequence $\{a_n\}_n$ that converges to $a$, then the sequence $\{a_n\}_n$ is a countable set and the open set $X/\{a_n\}_n$ is an open set of the point $a$ that does not contains elements of the sequence and it is not possible.

It is true in general that each metric space is first countable because for each point $a$

$\{B(a,\frac{1}{n})\}_n$ is a countable fundamental neighborhood system.

Let $C$ be a closed set of $(X,d_1)$ and let $a$ be a limit point for $C$ with respect to the space $(X,d_2)$.

By the previous remark, there exists a sequence $\{a_n\}_n\subseteq C$ convergent to $a$ in $(X,d_2)$. By hypothesis the $d_1-d_2$ convergent sequence coincide, so $a_n\to a$ in $(X,d_1)$ but $C$ is closed in this space, so $a\in C$.

This means that

$cl_{d_2}(C)=C$ so $C$ is a closed set in $(X,d_2)$.

Hence we have $\tau_1\subseteq \tau_2$.

In a similar way you can prove that if $C$ is a closed set in $(X,d_2)$, then it is also a closed set in $(X,d_1)$.

This means that $\tau_2\subseteq \tau_1$.

So $\tau_1=\tau_2$ that tells us the two metrics are equivalent.

$\endgroup$
  • $\begingroup$ @ViktorGlombik But when you prove that the two Topology are the same, you have immediately that your metrics satisfy your condition. In fact if you consider an open ball with respect one metric, then it is a union of a ball with respect to the other metric..If you want i write it in the answer. $\endgroup$ – Federico Fallucca Sep 7 at 16:17
  • $\begingroup$ @ViktorGlombik I don’t understand your question sorry. Can you repeat please? $\endgroup$ – Federico Fallucca Sep 7 at 16:36
  • $\begingroup$ @ViktorGlombik No no, there exists some y and some $r>0$ such that the union of that balls is your ball. It is true because the Topology induced by a metric is the Topology induced by a base of balls $\endgroup$ – Federico Fallucca Sep 7 at 19:16
  • $\begingroup$ If it is not clear to you, I try to explain better $\endgroup$ – Federico Fallucca Sep 7 at 19:18
  • $\begingroup$ It is not, please do. $\endgroup$ – Viktor Glombik Sep 8 at 1:58

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.