# Easy topology question

1. Jun 20, 2008

### oxeimon

So I'm trying to teach myself some topology, and the first thing I noticed, was that a metric space is a topological space under the topology of all open balls..

But then, consider the intersection of two open balls, can someone prove to me that the result is another open ball?

Or do they mean, that the topology is the set of all open balls, finite intersections of open balls, and arbitrary unions of open balls?

edit: How do I use latex notation on these forums?

2. Jun 20, 2008

### quasar987

The metric topology is not the set of all open balls, but rather, the topology generated by the set of all open balls. Meaning, is open in the metric topology any set U having the property

"For any x in U, there is an open ball containing x and itself entirely contained in U."

Alternatively, but this is more difficult to express and more vague also, the metric topology is the topology obtained, as you say, by taking the set of all open balls, finite intersections of open balls, and arbitrary unions of open balls, but also, by taking all arbitrary union of finite intersections of open balls.

Last edited: Jun 20, 2008
3. Jun 20, 2008

### nicksauce

To use latex use the tags [ tex] [/tex] (without the space in the first one).

4. Jun 20, 2008

### Crosson

Yes, that is what they mean.

Take care to note the difference between open sets and open balls in a metric space, just in case they said "a metric space is a topological space under the topology of all sets that are open with respect to that metric."

5. Jun 20, 2008

### oxeimon

On another note, can someone prove that for any two points $$x \ne y$$, we have that the intersection $$B(x)\cap B(y)$$ is not itself an $$\epsilon$$-ball?

why is the epsilon so high up? I used "\epsilon" ...

Last edited: Jun 20, 2008
6. Jun 20, 2008

### ice109

ahh i typed out a good responce and then didn't post it. ok here goes.

take a universal set X

take a collection of subsets of X

1. the collection is closed under arbitrary (countably infinite) union and finite intersection meaning the union of any numbers of sets in the collection is a set in the collection and the intersection of finitely many sets in the collection.

2. the empty set and X are in the collection.

then the collection is called a topology. to be a set in this collection, in the topology, is the definition of being an open set.

take a collection of subsets of X

1. the union of all these subsets = X, they "cover" X
2. inside the intersection of any 2 of these sets is another set in the collection

this collection is called a basis. sets inside the basis are called basic sets

it can be proved that a topology defined by, meaning here's how we tell you which sets are in the topology: all sets such that for every element in the set there is a basic set containing that element in the set.

you can now see how open balls serve as a basis, which if you prove that all bases generates topologies, generates a "metric topology".

for example:

basis on R1: all sets (a,b) a<b a,b in R.
now the topology this basis generates is the usual topology.

another example

basis on R1: all sets (x-e,x+e) x,e in R
now the topology this basis generates is the "euclidean metric" topology on R1.

7. Jun 20, 2008

### oxeimon

Aren't these the same topologies? If $$x = (a+b)/2, e = (a-b)/2$$, then $$(a,b) = (x-e,x+e)$$

So I guess the "usual topology" on $$\mathbb{R}$$ is just the "euclidean metric"?

8. Jun 20, 2008

### Crosson

When you want to include latex in a paragraph, as opposed to being on its own line, use

[ itex] [/itex]

without that first space as before ("i" stands for "inline").

9. Jun 20, 2008

### ice109

they're not the "same"; they're homeomorphic. and you've misused the words a little. the "euclidean metric" is a function, with properties that meet those of a metric. the topology generated by the "euclidean metric" is the "euclidean metric" topology, which turns out to be homeomorphic to the usual topology. note that "euclidean metric" is a phrase i made up and not standard nomenclature.

also your proof that the two topologies are homeomorphic isn't sufficient because you've only proven for xs that are in the middle of the (a,b) interval that there exists an open ball around them. to prove that the two tops(topologies) are homeomorphic you need to show that any open set in 1 is open in the other and vice versa. this amounts to provingthat for all xs in an open set in one top there exists an openball (and open interval for the reverse direction) in the other.

anyway your intuition is correct and if you don't care about proofs don't worry about it. i wager you're a physics student since you know latex and are interested in topology. am i correct?

Last edited: Jun 20, 2008