1
$\begingroup$

If $d_1$ and $d_2$ are metrics on the same set $X$ which satisfy the hypothesis that for any point $x \in X$ and $\epsilon >0$ there is a $\delta >0$ such that

$d_1(x,y)<\delta \implies d_2(x,y) < \epsilon$

and

$d_2(x,y)<\delta \implies d_1(x,y) < \epsilon$

then these metrics define the same open sets in $X$.



Here is my proof

Because we can choose $\delta \leq \epsilon$,

we can restate the hypothesis

$B_1(x;\delta) \subset B_2(x;\epsilon)$

and

$B_2(x;\delta) \subset B_1(x;\epsilon)$.


Let $U$ be a $d_1$ open set where $U=\cup_{x \in U} B_1(x;\epsilon)$ ,

so by assumption there is a $\delta$ such that $B_2(x;\delta) \subset B_1(x;\epsilon)$ .

This means each $x \in U$ contains a $d_2$ open ball, so by definition $U$ is also open in $d_2$.

By symmetric, we can prove $V=\cup B_2(x;\epsilon)$ open in $d_1$.

$\endgroup$
3
  • $\begingroup$ what open set definition are u using. $\endgroup$ Jan 30, 2017 at 16:04
  • $\begingroup$ @Riemann-bitcoin. : the topology induced by a metric is a standard concept, you should look it up if you don't know about it :) $\endgroup$ Jan 30, 2017 at 16:06
  • $\begingroup$ I have added my proof by referring answers, could someone give me a check? $\endgroup$
    – matchz
    Jan 30, 2017 at 19:39

2 Answers 2

2
$\begingroup$

Here is a brief outline of the proof: it is sufficient to show that every $d_1$-open set is $d_2$-open set (why?)

Let $V$ be a $d_1$-open set. For each $x\in V$ we have a $d_1$-open ball $B_1(x,\varepsilon)\subseteq V$. By assumption there is $\delta$ such that $B_2(x,\delta)\subseteq V$ (how to get an inclusion?) so $V$ satisfies the definition of $d_2$-openness.

$\endgroup$
0
1
$\begingroup$

The collection of open sets is called the topology of $X$, and the topology of $X$ is defined by its collection of neighborhoods. The two properties you have described above show that a neighborhood under the first (resp. second) metric is a neighborhood under the second (resp. first) metric, so you have nothing more to show!

Hope that helps,

$\endgroup$

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.