Show these two metrics define the same open sets. [duplicate]

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.

• How did you prove they are not equivalent?
– Zima
Commented Apr 25, 2023 at 21:27
• @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. Commented Apr 25, 2023 at 21:29
• @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 $M$. Commented Apr 25, 2023 at 21:30
• 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. Commented Apr 25, 2023 at 21:35
• @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? Commented Apr 25, 2023 at 22:04

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.

• 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." Commented Apr 25, 2023 at 22:03
• 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? Commented Apr 25, 2023 at 22:04
• It was a duplicate (see last comment above). Commented May 5, 2023 at 10:19

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 $$B_r(x_0,d_1)=\{x\in \mathbb{R}\ s.t.\ d_1(x,x_0) Now let $$x\in B_r(x_0,d_1)$$, then $$d_1(x,x_0), but $$d_2(x,x_0)\le d_1(x,x_0)$$, so $$d_2(x,x_0). This implies that $$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 $$d_2(x,x_0), but this splits into two cases: either $$d_1(x,x_0) or $$1, depending on the value of $$d_2(x,x_0)$$.

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

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

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