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Give an example of two metrics on $\mathbb R$ that have the same open sets but are not equivalent.

Consider $d_1(x,y) = |x-y|$ and $d_2(x,y) = \min\left(|x-y|,1\right)$. I proved that they're not equivalent but I'm stuck on proving they have the same open sets. This is what I have. Let's check these two metrics have the same open sets. To prove $S_1 \subset S_2$, choose an arbitrary open set $S$ in $S_1$. We know that for every $y \in S \implies B_r(y,d_1) \subset S$. We want to show there exists an $r'$ such that $B_{r'}(y,d_2) \subset S$.

I've been stuck for a while on this, I was thinking of grabbing an arbitrary $x \in B_{r'}(y,d_2)$ and then showing that $x \in B_{r}(y,d_1)$, so $x \in S$, but I don't think this works. I also don't know whether to set $r' = r$?

Any help would be greatly appreciated.

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  • $\begingroup$ How did you prove they are not equivalent? $\endgroup$
    – Zima
    Commented Apr 25, 2023 at 21:27
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    $\begingroup$ @Zima Now, let's prove these metrics aren't equivalent. Suppose they were, so there exists positive constants $c_1,c_2$ such that $d_1(x,y) \leq c_2d_2(x,y)$ and $d_2(x,y) \leq c_1d_1(x,y) \Rightarrow $ $$ d_2(x,y) \leq c_1d_1(x,y) \leq c_1c_2d_2(x,y)$$ Now, picking an $n \in \mathbb N$, we have $d_1(n,0) = n$ and $d_2(n,0) = 1$, meaning we have $$1 \leq c_1n \leq c_1c_2$$ However, we can make $n$ arbitrarily large, so this inequality fails and we have a contradiction. $\endgroup$
    – beginner
    Commented Apr 25, 2023 at 21:29
  • $\begingroup$ @lulu Call two metrics $d_1$ and $d_2$ on the same set $M$ equivalent if there exist positive constants $c_1,c_2$ such that $d_1(x,y) \leq c_2d_2(x,y)$ and $d_2(x,y) \leq c_1d_1(x,y)$ for all $x$ and $y$ in $M$. $\endgroup$
    – beginner
    Commented Apr 25, 2023 at 21:30
  • $\begingroup$ They generate the same topology within the unit ball (being the exact same metric there), so generate the same local basis within a ball of radius one at every point by translation, so generate the same topology. There's nothing special here about $\mathbb R$; the same result holds for for any Banach space, or more generally any (locally convex?) topological vector space if one defines the seminorm inequalities correctly. $\endgroup$ Commented Apr 25, 2023 at 21:35
  • $\begingroup$ @BrevanEllefsen When I grab an arbitrary ball, can I shrink the radius $r$ to be less than 1 and still have it be an open ball? $\endgroup$
    – beginner
    Commented Apr 25, 2023 at 22:04

2 Answers 2

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In both cases the collection of all open balls with radius $<1$ forms a basis of the topology, i.e. every open set can be written as the union of such. Those balls agree by definition. Thus both metrics literally define the same topology.

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  • $\begingroup$ Can you explain a bit more with the language of the problem? I haven't taken topology so I'm a bit confused by terms like "basis of the topology." $\endgroup$
    – beginner
    Commented Apr 25, 2023 at 22:03
  • $\begingroup$ When I grab an arbitrary ball, can I shrink the radius $r$ to be less than 1 and still have it be an open ball? $\endgroup$
    – beginner
    Commented Apr 25, 2023 at 22:04
  • $\begingroup$ It was a duplicate (see last comment above). $\endgroup$ Commented May 5, 2023 at 10:19
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You have to show that:

  1. If a set $A$ is open for d1, then it is open for d2

  2. If a set $A$ is open for d2, then it is open for d1

Notice that if $d_1(x,y)=|{x-y}|$ and $d_2=min\{|{x-y}|,1\}$, then $d_2=min\{d_1(x,y),1\}$, and this implies that $d_2(x,y)\le d_1(x,y)$

With this information proving $2)$ is easy: suppose a set $A$ is open with $d_2$, and consider $$B_r(x_0,d_1)=\{x\in \mathbb{R}\ s.t.\ d_1(x,x_0)<r\}\hspace{1cm}B_t(x_0,d_2)=\{x\in \mathbb{R}\ s.t.\ d_2(x,x_0)<t\}$$ Now let $x\in B_r(x_0,d_1)$, then $d_1(x,x_0)<r$, but $d_2(x,x_0)\le d_1(x,x_0)$, so $d_2(x,x_0)<r$. This implies that $x\in B_r(x_0,d_2)$, and so, given that $x$ was arbitrary, we have $$B_r(x_0,d_1)\subseteq B_r(x_0,d_2)\subseteq A$$ and so $A$ is open with $d_1$ too.

For part $1 )$ suppose $A$ is open with $d_1$, and let $x\in B_r(x_0,d_2)$. Then $d_2(x,x_0)<r$, but this splits into two cases: either $d_1(x,x_0)<r$ or $1<r$, depending on the value of $d_2(x,x_0)$.

If we are in the case $d_1(x,x_0)<r$ then we are basically done, since this implies $x\in B_r(x_0,d_1)$. So now, if you manage to prove that if we are in the case $1<r$, this implies that $d_1(x,x_0)<r$, then

$$B_r(x_0,d_2)\subseteq B_r(x_0,d_1)\subseteq A$$ which finishes the proof.

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  • $\begingroup$ It was a duplicate (see last comment above). $\endgroup$ Commented May 5, 2023 at 10:19

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