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Let $X$ be the set of positive integers. Let $d_1$ be the usual metric space on $X$ and $d_2$ be the discrete metric on $X$. Define $d_3:X\times X \rightarrow R$ by $d_3(m,n)=|\frac{1}{m}-\frac{1}{n}|$ for $m,n\in X$. Prove that $d_3$ is also a metric on $X$ and $d_1,d_2,d_3$ all induce the same topology on $X$.

I've proved that $d_3$ is also metric.. But I could not proved that all of these induces the same topology. I know that, in order to prove that, these metrices induces that the same topology... I have to prove that all are equivalent metrices. I am stuck at this part.

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  • $\begingroup$ Hint: Try to show this metrics are equivalent. What can you deduce from that? $\endgroup$
    – noctusraid
    Feb 10, 2016 at 14:14
  • $\begingroup$ I am stuck at this part... $\endgroup$
    – Learner
    Feb 10, 2016 at 14:15
  • $\begingroup$ whoops, didn't ready the last paragraph properly, my bad. Do you mean $d_3(m,n)= \Vert \frac{1}{m} - \frac{1}{n} \Vert$ ? If yes: There is an obvious upper bound for this metric. $\endgroup$
    – noctusraid
    Feb 10, 2016 at 14:19
  • $\begingroup$ It is true that equivalent metrics (in the sense that $c d(x,y) \leq d'(x,y) \leq C d(x,y)$ for constants $c$ and $C$ and all $x$ and $y$) induce the same topology but two metrics can be not equivalent and still induce the same topology. For example, you cannot have $d_1(x,y) \leq C d_2(x,y)$ for some $C$ and all $x$ and $y$ because $d_2$ is bounded and $d_1$ is not. $\endgroup$ Feb 10, 2016 at 14:19
  • $\begingroup$ yes @noctusraid $\endgroup$
    – Learner
    Feb 10, 2016 at 14:19

3 Answers 3

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I'll give one equivalence and edit this answer if OP struggles with the others:

Claim: $d_2 \tilde{} \space d_3$ with the notation as in the question.

Proof:

Observe that $d_3(m,n)= \vert \frac{1}{m}-\frac{1}{n} \vert \leq 1$ for all positive integers $m,n$. Therefore we have:

$$d_3(m,n) \leq d_2(m,n) \leq 2d_3(m,n)$$ since (w.l.o.g $n >m$):

$$ 2\vert \frac{1}{m}-\frac{1}{n} \vert=\frac{2}{m}-\frac{2}{n}=\frac{2(n-m)}{mn}= 1+\frac{2n-(mn+2m)}{mn}$$ and because

$$1 \geq \frac{m}{2}+ \frac{m}{n} \implies 2n \geq mn+2m$$

we can conclude the equivalence.

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$|1/m-1/n|$ <$|m-n|$ So these two are equivalent. Since $m,n$ are Positive Integers.

and $ |m-n|\geq d_2$ Since $m,n$ are positive integers so least value is zero. All all the Equivalent metrices.

Therefore, all metrics are equivalent. and hence induces the same topology.

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  • $\begingroup$ you are not allowed negative $C$'s for your constants when showing equivalence of metrics. They need to be strictly positive. $\endgroup$
    – noctusraid
    Feb 10, 2016 at 14:56
  • $\begingroup$ @noctusraid oh ! they I have to rethink about it... Thanks $\endgroup$
    – Learner
    Feb 10, 2016 at 14:59
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All metrics induce the discrete topology (i.e. all sets are open). Forget about any equivalence of metrics, this is a stronger requirement (which does not hold in this case) that is not needed for equivalence of topologies

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    $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$
    – Nizar
    Feb 11, 2016 at 7:43
  • $\begingroup$ It is not a complete answer, yes. However, it tells the author that their previous approaches (as provided in comments to OP and two previous answers) are doomed to fail, and gives him a hint that is straightforward to verify and that implies the solution. I did neither criticize the question nor did I ask for clarification. $\endgroup$
    – Bananach
    Feb 11, 2016 at 7:46

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