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HW Question, either professor is wrong or I am.

by Poopsilon
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Poopsilon
#1
Oct30-11, 04:31 PM
P: 294
So I've got this homework problem that I think I've found a counter-example to, so either my counter-example is wrong or the professor is, here is the problem:

Let (X,d) be a metric space and let E be a nonempty subset of X. For each x ∈ X, let d(x,E) = inf{d(x,y) : y ∈ E, with y≠x}.

Show that {x ∈ X : d(x,E)<r} is open for each r ∈ ℝ.


So now for my counter-example:

Let (X,d) = {1,2,4} with the usual metric and ordering and define our topology τ on (X,d) to be τ = {∅, {1}, {4}, {1,2,4}, {1,4}}. Now simply take E = X and notice that d(1,E)=1, d(2,E)=1, and d(4,E)=2. Thus take r=1.5 and note that this means that {x ∈ X : d(x,E)<1.5} = {1,2}, which is closed according to our topology τ.

Help would be much appreciated, thanks =].
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Cant or Wont
#2
Oct30-11, 04:59 PM
P: 12
Seems to me the disagreement between you and your professor is that you have declared what you would like your topology [itex] \mathcal{T} [/itex] to be instead of taking the topology that your metric [itex] d [/itex] actually induces on [itex] \{1,2,4 \} [/itex] ?
Poopsilon
#3
Oct30-11, 06:40 PM
P: 294
Yes that would make sense, and inducing said metric would I believe allow me to solve it, but is that what people usually assume when discussing an arbitrary metric space? That you can basically find open neighborhoods of any radius you want?

Cant or Wont
#4
Oct30-11, 07:00 PM
P: 12
HW Question, either professor is wrong or I am.

I'm not even halfway through my first Topology course so I wouldn't like to say what people usually assume! But i'm given to understand that when we talk about a Topology induced by a matric on [itex]X[/itex] then the open sets [itex]U[/itex] are those that if [itex] a \in U [/itex] there is an r > 0 s.t. [itex] B \left( a,r \right) \subseteq U [/itex]. A ball of radius r.

So in your example you see that for [itex] \{ 1,4 \} [/itex] to be open you would need an r > 3? But a ball of radius greater than 3 centred on 4 would end up including [itex] \{1,2,4 \} [/itex].

But not every Topology is metrisable. So again take {1,2,4} and put the trivial topology on it and see if you can induce this topology with some metric.
Poopsilon
#5
Oct30-11, 07:13 PM
P: 294
I'm skeptical that I could, in fact I don't see how a metric could ever induce only a finite number of open sets. But there is nothing wrong with putting a metric on a topology even if that topology can't be induced by that metric, or any metric for that matter, correct? Such as in my example above.
Citan Uzuki
#6
Oct30-11, 08:21 PM
P: 274
Whenever we talk about a topology on a metric space, we are always talking about the topology induced by the metric. The same space with some other topology is not a metric space.
Poopsilon
#7
Oct30-11, 09:28 PM
P: 294
So what you're saying is that a topology can only be turned into a metric space if there is some metric that induces this topology from the underlying set?
Deveno
#8
Oct30-11, 10:00 PM
Sci Advisor
P: 906
i believe that any finite metric space induces the discrete topology as the metric topology (all points are "isolated"), so that would specifically exclude your example topology, no matter HOW you defined the metric.

to see this, suppose we have a generic 3-point set {a,b,c} (which may, or may not, be real numbers).

we are going to have to assign a positive real number for d(a,b), it really doesn't matter "which" one we pick, but let's call it x.

similarly , we have to pick a positive real number for d(a,c), call it y.

well, for any ε-ball of radius min{x,y} or less, we have Bε(a) = {a}.

the same logic applies to b and c, so all singletons are open. this means the topology is the entire power set 2{a,b,c}, that is: the discrete topology.
Citan Uzuki
#9
Oct31-11, 01:12 AM
P: 274
Quote Quote by Poopsilon View Post
So what you're saying is that a topology can only be turned into a metric space if there is some metric that induces this topology from the underlying set?
Yes.
Bacle2
#10
Nov1-11, 05:59 PM
Sci Advisor
P: 1,170
You can also check the fact that d(x,E) is a continuous function into ℝ+, and your sets would then be the inverse images of (0,r) in ℝ
lavinia
#11
Nov1-11, 09:30 PM
Sci Advisor
P: 1,716
Quote Quote by Poopsilon View Post

So now for my counter-example:

Let (X,d) = {1,2,4} with the usual metric and ordering and define our topology τ on (X,d) to be τ = {∅, {1}, {4}, {1,2,4}, {1,4}}. Now simply take E = X and notice that d(1,E)=1, d(2,E)=1, and d(4,E)=2. Thus take r=1.5 and note that this means that {x ∈ X : d(x,E)<1.5} = {1,2}, which is closed according to our topology τ.
The topology determined by your metric is the discrete topology so every subset is open.
Bacle2
#12
Nov1-11, 09:55 PM
Sci Advisor
P: 1,170
As Deveno and Uzuki posted, your proposed topology does not work, since it is not Hausdorff--which every metric topology must be (given d(x,y)=a. the two open balls
B(x,a/4) and B(y,a/4) are disjoint from each other): the points 1,2 and 2,4, cannot be separated in your scheme for a topology, since there are no open sets containing 2 but
not 1 nor containing 2 but not 4. You will then have to throw {2} into your proposed
topology, which will then, by closure under unions, be the discrete topology.
Poopsilon
#13
Nov2-11, 03:36 AM
P: 294
Thanks for the info everyone, I keep coming upon these topological considerations while studying analysis, would the first couple chapters of Munkres' Topology take care of that?
Jamma
#14
Nov5-11, 06:43 AM
P: 429
Of coure you need the induced metric, you can't just invent a new one, otherwise the question is meaningless. You could just take the trivial indiscrete topology, for example.


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