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September 3rd, 2012, 12:17 PM   #1
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Topology

Show that the countable collection {(a,b) x (c,d) such that a < b and c < d, and a,b,c,d are rational} is a basis for the standard topology on R^2.

In class, we learned this theorem...

Let X be a set and T be a topology on X. A basis for T is a collection of sets B, such that
1) Every b in B is open
2) IF U is in T and x is in U, then there exists a b in B such that x is in B which is contained in (or equal to) U.

I am guessing that I have to use this theorem to answer the question...the only thing that confuses me is...can I use this theorem to prove that a set is a basis for a topology (when the basis and the topology are the same set...since we know that {(a,b) x (c,d) such that a < b and c < d, and a,b,c,d are rational} is a standard topology on R^2 right?)?

Thanks in advance.
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September 3rd, 2012, 01:43 PM   #2
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Re: Topology

Quote:
Originally Posted by Artus
Show that the countable collection {(a,b) x (c,d) such that a < b and c < d, and a,b,c,d are rational} is a basis for the standard topology on R^2.

In class, we learned this theorem...

Let X be a set adn T be a topology on X. A basis for T is a colection of sets B, such that
1) Every b in B is open
2) IF U is in T and x is in U, then there exists a b in B such that x is in B which is contained in (or equal to) U.
So, you have to show that your countable collection above satisfies the hypotheses of this theorem. In other words you have to let B be your collection of sets, and show that B has the properties in (1) and (2).

Quote:
Originally Posted by Artus
we know that {(a,b) x (c,d) such that a < b and c < d, and a,b,c,d are rational} is a standard topology on R^2 right?)?
If you think this through, you'll see that it's false. Are you saying that every open set in R^2 has to be of that form? How about, say, (sqrt(2), (pi) X (sqrt(2), pi)? Is that an open set in R^2? Is it in your countable collection? Hint: Yes to the first, No to the second. Verify these for yourself.

I think a good approach to this problem is to do it in one dimension. Do it in R with intervals (a,b) and show that those are not ALL the open sets, but are a basis for the open sets.
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September 4th, 2012, 10:57 AM   #3
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Re: Topology

Quote:
Originally Posted by Maschke
Quote:
Originally Posted by Artus
Show that the countable collection {(a,b) x (c,d) such that a < b and c < d, and a,b,c,d are rational} is a basis for the standard topology on R^2.

In class, we learned this theorem...

Let X be a set adn T be a topology on X. A basis for T is a colection of sets B, such that
1) Every b in B is open
2) IF U is in T and x is in U, then there exists a b in B such that x is in B which is contained in (or equal to) U.
So, you have to show that your countable collection above satisfies the hypotheses of this theorem. In other words you have to let B be your collection of sets, and show that B has the properties in (1) and (2).

Quote:
Originally Posted by Artus
we know that {(a,b) x (c,d) such that a < b and c < d, and a,b,c,d are rational} is a standard topology on R^2 right?)?
If you think this through, you'll see that it's false. Are you saying that every open set in R^2 has to be of that form? How about, say, (sqrt(2), (pi) X (sqrt(2), pi)? Is that an open set in R^2? Is it in your countable collection? Hint: Yes to the first, No to the second. Verify these for yourself.

I think a good approach to this problem is to do it in one dimension. Do it in R with intervals (a,b) and show that those are not ALL the open sets, but are a basis for the open sets.
Thanks for answering...what I actually meant by "we know that {(a,b) x (c,d) such that a < b and c < d, and a,b,c,d are rational} is a standard topology on R^2 right?)?" is not that all open sets have to be in that form. From what I read about standard topologies...a standard topology is a set that, when you choose a point inside it, you can always have an open set around that point that is entirely in the topology. Doesn't that happen with the countable collection that is mentioned in this question?
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September 4th, 2012, 11:37 AM   #4
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Re: Topology

Quote:
Originally Posted by Maschke
Quote:
Originally Posted by Artus
Show that the countable collection {(a,b) x (c,d) such that a < b and c < d, and a,b,c,d are rational} is a basis for the standard topology on R^2.

In class, we learned this theorem...

Let X be a set adn T be a topology on X. A basis for T is a colection of sets B, such that
1) Every b in B is open
2) IF U is in T and x is in U, then there exists a b in B such that x is in B which is contained in (or equal to) U.
So, you have to show that your countable collection above satisfies the hypotheses of this theorem. In other words you have to let B be your collection of sets, and show that B has the properties in (1) and (2).

Quote:
Originally Posted by Artus
we know that {(a,b) x (c,d) such that a < b and c < d, and a,b,c,d are rational} is a standard topology on R^2 right?)?
If you think this through, you'll see that it's false. Are you saying that every open set in R^2 has to be of that form? How about, say, (sqrt(2), (pi) X (sqrt(2), pi)? Is that an open set in R^2? Is it in your countable collection? Hint: Yes to the first, No to the second. Verify these for yourself.

I think a good approach to this problem is to do it in one dimension. Do it in R with intervals (a,b) and show that those are not ALL the open sets, but are a basis for the open sets.


I tried to prove it...and I was just wondering if you could check to see if my answer is correct (if you don't mind)...

We know that the standard topology on R is {(a,b) x (c,d) such that a<b, c<d and a,b,c,d is in R}

I am trying to prove that the countable colection {(a,b)x(c,d) such that a<b, c<d, and a,b,c,d are rational} is a basis for that standard topology

Let T be the standard topology on R^2.

a) Every set (a,b)x(c,d) in the countable collection must also be in the standard topology, because we know that a,b,c,d rational (which is contained in the reals). So it must also be contained in T.

b) If U is of the form (a,b)x(c,d) where a,b,c,d is in R. And if we have a point (w,v) in (a,b)x(c,d). Since the rational numbers are countably infinite, we know that there exist rational numbers e,f,g,h such that a=<e<w<f=<b and c=<g<v<h=<d. So we know that (w,v) is in (e,f)x(g,h), where (e,f)x(g,h) is in the countable collection, which is contained in (or equal to) U.

So this countable collection actually is a basis for the T.

Do you think my answer is correct? Also, the professor said something about induction, and also about the countability of the set (if I remember correctly)...so do you think I missed anything (that had to do with induction or the countability of the set) in the proof?

Thanks in advance
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September 5th, 2012, 08:10 AM   #5
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Re: Topology

Quote:
Originally Posted by Artus
Quote:
Originally Posted by Maschke
Quote:
Originally Posted by Artus
Show that the countable collection {(a,b) x (c,d) such that a < b and c < d, and a,b,c,d are rational} is a basis for the standard topology on R^2.

In class, we learned this theorem...

Let X be a set adn T be a topology on X. A basis for T is a colection of sets B, such that
1) Every b in B is open
2) IF U is in T and x is in U, then there exists a b in B such that x is in B which is contained in (or equal to) U.
So, you have to show that your countable collection above satisfies the hypotheses of this theorem. In other words you have to let B be your collection of sets, and show that B has the properties in (1) and (2).

Quote:
Originally Posted by Artus
we know that {(a,b) x (c,d) such that a < b and c < d, and a,b,c,d are rational} is a standard topology on R^2 right?)?
If you think this through, you'll see that it's false. Are you saying that every open set in R^2 has to be of that form? How about, say, (sqrt(2), (pi) X (sqrt(2), pi)? Is that an open set in R^2? Is it in your countable collection? Hint: Yes to the first, No to the second. Verify these for yourself.

I think a good approach to this problem is to do it in one dimension. Do it in R with intervals (a,b) and show that those are not ALL the open sets, but are a basis for the open sets.
Thanks for answering...what I actually meant by "we know that {(a,b) x (c,d) such that a < b and c < d, and a,b,c,d are rational} is a standard topology on R^2 right?)?" is not that all open sets have to be in that form. From what I read about standard topologies...a standard topology is a set that, when you choose a point inside it, you can always have an open set around that point that is entirely in the topology. Doesn't that happen with the countable collection that is mentioned in this question?
I'm not familiar with any meaning of "standard topology" other than ... the standard topology. It's the topology that everything thinks of as the usual topology on some set. The collection of rational intervals is not a topology on the reals, let alone a "standard" one. It is a basis for the standard topology.
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September 5th, 2012, 08:21 AM   #6
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Re: Topology

Quote:
Originally Posted by Artus


We know that the standard topology on R is {(a,b) x (c,d) such that a<b, c<d and a,b,c,d is in R}
I'm sorry I didn't carefully read the rest of your post, but I think there's some confusion over concepts or terminology.

What you claim as the "standard topology" is not even a topology. It's not closed under finite unions, for example, let alone arbitrary unions.

I think if you review the definitions of the terms topology and basis, you will get a better handle on this problem.

Also I'd repeat my earlier suggestion, which would be to do this problem in one dimension first. Is the collection of all intervals (a,b) of reals a topology on the reals? No it's not. It's not even closed under finite unions. For example the set (1,2) UNION (3,4) is an open set of reals that's not an interval. So the collection of open intervals is not a topology. It's a basis for the standard topology on the reals.
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