Metric topolgy comprehension question

[gottfried]

Given the proposition: A subset U of a metric space is an open set iff U is a union of open balls.

Then we are told that it follows from the above proposition that two metrics are equivalent if each open ball of either of the metric topolgies is an open set of the other topology.

This is really confusing. Firstly if a metric topology is essentially a set of sets without a materic how do we define an open ball in it? Does a metric topology somehow inherit a metric from it's corresponding metric space?

Any thoughts or points that might help me undestand this would be appreciated thanks.

 

[HallsofIvy]

A topology for a set is a collection of sets having certain properties:
a) The topology contains the empty set and the entire set.
b) The union of any collection of set in the topology is also in the topology.
c) The intersection of any finite collection of sets in the topology is also in the topology.
(The sets in the topology are called "open sets" so "sets in the topology" can be replaced by "open sets"- the union of any collection of open sets is open and the intersection of any finite collection of open sets is open.)

In a metric space that topology is defined in a specific way: All unions and finite intersections of open balls. The open balls, of course, are defined using the metric: N_\delta(x_0)= \{ x| d(x, x_0)< \delta\}.

Every metric defines a topology but there exist topologies that cannot be defined by a metric. Example: given any set X take X itself and the empty set as the only members of the topology.

 

[Fredrik]

Let X be a set. A topology on X is a set of subsets of X (that's required to satisfy a number of conditions...that I see HallsofIvy posted while I was typing this). If there's a metric defined on X, then you can use that metric to define a topology T by saying that a subset of X belongs to T if and only if it's open with respect to the metric. This specific topology, which was defined using the metric, is called "the metric topology".

There's no definition of "open ball" that doesn't involve a metric. If there are several metrics on X, then there may be several metric topologies.

How do you define "two metrics are equivalent"? Does it mean that the metric topology defined using the first metric is the same as the metric topology defined using the other metric?

 

[gottfried]

Thanks a lot for the responses. Certainly helpful.

Two metrics are equivalent if, as you say, the metric topologies they induce are the same.

HallsofIvy said "In a metric space that topology is defined in a specific way: All unions and finite intersections of open balls." Does this mean that a metric topology is essentially a collection of open balls? knowing this and given the statement that"two metrics are equivalent if each open ball of either of the metric topolgies is an open set of the other topology" would it be correct to say that each open ball B_d(x,r) in the metric space (X,d) is an element of the metric topolgy. For the metric d to be equivalent to a metric m this ball must also be open with respect to m. Meaning that B_m(X,r) must be an element of the metric topology induced by m.

Hopefully I've articulated myself well enough for you to follow.

 

[HallsofIvy]

Thanks a lot for the responses. Certainly helpful.

Two metrics are equivalent if, as you say, the metric topologies they induce are the same.

HallsofIvy said "In a metric space that topology is defined in a specific way: All unions and finite intersections of open balls." Does this mean that a metric topology is essentially a collection of open balls?

What I said before was that any topology is, by definition, a collection of open sets, not necessarily balls. For example, in R, with the "usual topology", d(x, y)= |x- y|, N_1(0) is the open interval from -1 to 1, (-1, 1). N_1(3) is the open interval from 3-1= 2 to 3+ 1= 4. (2, 4). Both are also open sets and so in the topology. Their union, (-1, 1)\cup (2, 4) is also an open set but is not an "open ball". We say that a topology is "generated" by the collection of open balls (and so by the metric).

knowing this and given the statement that"two metrics are equivalent if each open ball of either of the metric topolgies is an open set of the other topology" would it be correct to say that each open ball B_d(x,r) in the metric space (X,d) is an element of the metric topolgy. For the metric d to be equivalent to a metric m this ball must also be open with respect to m. Meaning that B_m(X,r) must be an element of the metric topology induced by m.

Hopefully I've articulated myself well enough for you to follow.

Two metrics, d₁(x, y) and d<sub>2[/b](x, y) are equivalent if and only if they generate the same topology- the same collection of open sets.

For example, on the real line, the "standard metric" is d(x, y)= |x- y|. It should be easy to see that if, instead, we define d(x, y)= 3|x- y| we get exactly the same open sets. (In the same sense that using a yard stick rather than a foot ruler to measure distances doesn't change the basic geometry.)</sub>

 

[Fredrik]

I see that Halls was faster than me this time too...

HallsofIvy said "In a metric space that topology is defined in a specific way: All unions and finite intersections of open balls." Does this mean that a metric topology is essentially a collection of open balls?

A metric topology is a collection of unions of finite intersections of open balls. It's the smallest collection of subsets of X that satisfies the definition of a topology and contains all the subsets of X that are open with respect to the metric.

Since any finite intersection of open balls is a union of open balls, it's also true that a metric topology is a collection of unions of open balls.

would it be correct to say that each open ball B_d(x,r) in the metric space (X,d) is an element of the metric topolgy.

Yes.

For the metric d to be equivalent to a metric m this ball must also be open with respect to m. Meaning that B_m(X,r) must be an element of the metric topology induced by m.

Yes.

 

[gottfried]

Thanks guys.