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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 $d_1$.

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    what open set definition are u using. – 2017-01-30
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    @Riemann-bitcoin. : the topology induced by a metric is a standard concept, you should look it up if you don't know about it :) – 2017-01-30
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    I have added my proof by referring answers, could someone give me a check? – 2017-01-30

2 Answers 2

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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.

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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,