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What does it mean that two different metrics may define the same collection of open sets?

The assumption is that a given set is equipped with two different metrics to form two different metric spaces.

Does it simply mean that an open subset of a set is independent of a metric?

If so, why??? if not, why???

User
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    Consider the following metrics defined on $\mathbb R$. Let $d_1(x,y) = |x - y|$. Let $d_2(x,y) = 2|x - y|$. It should be fairly clear that these define the same open sets, as they define the same open balls (If $d_1(x,y) < r$, then $d_2(x,y) < 2r$, and so $B_{d_1}(x,r) = B_{d_2}(x,2r)$). However, there exist metrics on $\mathbb R$ that define different open sets -- most dramatically so, the discrete metric, where $d(x,y)$ is $1$ iff $x = y$, and $0$ otherwise. – qaphla Aug 13 '14 at 23:55
  • "Does it simply mean that an open subset of a set is independent of a metric?" No, because this does not mean anything. – Andrés E. Caicedo Aug 13 '14 at 23:59
  • What do you mean by "this does not mean anything?" – User Aug 14 '14 at 00:02
  • What does it mean for a set to be "independent" of a metric? Is this a concept that only makes sense for open sets? Are these sets subsets of an arbitrary set? Of a topological space? Of a metric space? Of a metrizable space? – Andrés E. Caicedo Aug 14 '14 at 00:06
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    @qaphla There seems to be a mistake in your comment. "where $d(x,y)$ is $0$ iff $x=y$, and 1 otherwise" should be the correct one. – User Aug 14 '14 at 00:29

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Let $X$ be a set and $d_1$ and $d_2$ metrics on $X$. That $d_1$ and $d_2$ generate the same collection of open sets means that each set $A\subset X$ is open with respect to $d_1$ if and only if it is open with respect to $d_2$. One example of this case is given in a comment. Conversely, consider $X=\mathbb R$ and $d_1(x,y):=|x-y|$ and $$ d_2(x,y):=\begin{cases}0,&x=y \\ 1,&x\neq y\end{cases}. $$ One can show that each set $A\subset\mathbb R$ is open with respect to $d_2$ (try to prove this as an exercise), but $A=\{0\}$ for example is not open with respect to $d_1$.

sranthrop
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If two metrics are bounded by positive constant multiples of each other, i.e. $cd_1(c,y) \leq d_2(x,y) \leq Cd_1(x,y)$ then they will generate the same open sets. To see that not all metrics generate the same open sets, consider the standard metric on ${\mathbb R}$ compared to the metric that assigns distance 1 between any two distinct numbers.

user2566092
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just to look at a slightly less detailed level

one way of seeing a topology (on a set $X$) is in terms of a basis of open sets. the open sets defined by a basis $B$ are arbitrary unions of sets belonging to the basis.

suppose you have two topologies for $X$ with bases $B_1$ and $B_2$. then the topologies are equivalent if and only if the bases interpenetrate - in the sense that for any $b_1 \in B_1$ we can find a $b_2 \in B_2$ such that $b_2 \subset b_1$ and vice versa.

in metric spaces we define the topology in terms of a basis of open balls of arbitrary small radius. so given a $b_1$ we may join with Hamlet's dyslexic brother in asking "$b_2$ or not $b_2$? that is the question!" (and vice versa, of course)

David Holden
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