## Halp at Toplogy!

For the discussion of math. Duh.

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Marbas
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### Halp at Toplogy!

I am teaching myself topology and I have run into a problem that I am having some trouble with:

Show that the interiors of all circles with radii of rational length and whose centers have rational coordinates, form a basis* for the euclidean topology of the plane.

*Here in the book a different definition of basis is used instead of the official one: it is defined as a subcollection of the open sets in a topological space S such that every open set in S is a union of elements from the said subcollection.
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gmedina
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### Re: Halp at Toplogy!

Given any open set U and x in U you must find at least one circle B with radius of rational length and whose center has rational coordinates such that x is an element of B and B is a subset of U.
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dosboot
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### Re: Halp at Toplogy!

Marbas wrote:*Here in the book a different definition of basis is used instead of the official one: it is defined as a subcollection of the open sets in a topological space S such that every open set in S is a union of elements from the said subcollection.

But that is the usual definition, n'est pas?

Marbas
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### Re: Halp at Toplogy!

But that is the usual definition, n'est pas?

Nope.

First of all, in the usual definition, the basis has to cover S, not just be able to create all open sets by union.

Second, if we have two base elements, B1 and B2, then for every point in there intersection I there exists an element Bn such that Bn contains said point and is contained in I.
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aguacate
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### Re: Halp at Toplogy!

Marbas wrote:
But that is the usual definition, n'est pas?

Nope.

First of all, in the usual definition, the basis has to cover S, not just be able to create all open sets by union.

This is a necessary condition for S to be made up of a union of elements of the basis. If the basis did not cover S, then arbitrary unions of the basis would never be equal to S.

Second, if we have two base elements, B1 and B2, then for every point in there intersection I there exists an element Bn such that Bn contains said point and is contained in I.

This can be proven through the first definition. The fact is, they are equivalent definitions. It's like defining "closed set" as a set whose complement is open, then proving that it contains all its limit points, then defining it as a set that contains all its limit points and proving that its complement is open.

Marbas
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### Re: Halp at Toplogy!

aguacate wrote:
Marbas wrote:
But that is the usual definition, n'est pas?

Nope.

First of all, in the usual definition, the basis has to cover S, not just be able to create all open sets by union.

This is a necessary condition for S to be made up of a union of elements of the basis. If the basis did not cover S, then arbitrary unions of the basis would never be equal to S.

Second, if we have two base elements, B1 and B2, then for every point in there intersection I there exists an element Bn such that Bn contains said point and is contained in I.

This can be proven through the first definition. The fact is, they are equivalent definitions. It's like defining "closed set" as a set whose complement is open, then proving that it contains all its limit points, then defining it as a set that contains all its limit points and proving that its complement is open.

Oh...
That is why the book doesn't discriminate between them.

I learnded something new today.

Hurrah.

EDIT: Wait, so there are no situations where a collection of sets can cover S while being incapable of creating every open set in S by arbitrary unions? I can see now how the second part of the usual definition combined with the first part could make it equivalent to the one I thought was different, if there are no such situations where a collection of open sets in S is capable of creating every open set in S by union of its' elements without also covering S.

Can such a situation occur?

Put another way: Can there exist a topological space S such that the union of all open sets in S does not equal S?
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Token
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### Re: Halp at Toplogy!

Marbas wrote:EDIT: Wait, so there are no situations where a collection of sets can cover S while being incapable of creating every open set in S by arbitrary unions? I can see now how the second part of the usual definition combined with the first part could make it equivalent to the one I thought was different, if there are no such situations where a collection of open sets in S is capable of creating every open set in S by union of its' elements without also covering S.

Can such a situation occur?

Put another way: Can there exist a topological space S such that the union of all open sets in S does not equal S?

You seem rather confused. You've worded your question in three different ways, no two of which are equivalent. All are easily answerable:

Wait, so there are no situations where a collection of sets can cover S while being incapable of creating every open set in S by arbitrary unions?

Yes, there are. Example in R: (-∞,0), (0,∞), (-1,1).

I can see now how the second part of the usual definition combined with the first part could make it equivalent to the one I thought was different, if there are no such situations where a collection of open sets in S is capable of creating every open set in S by union of its' elements without also covering S.

Can such a situation occur?

No. S is an open set in S. If you can create every open set in S with a set of open sets, it trivially covers S, as aguacate has pointed out.

Put another way: Can there exist a topological space S such that the union of all open sets in S does not equal S?

No. Again, S is open in S.
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Marbas
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### Re: Halp at Toplogy!

Token wrote:
Marbas wrote:EDIT: Wait, so there are no situations where a collection of sets can cover S while being incapable of creating every open set in S by arbitrary unions? I can see now how the second part of the usual definition combined with the first part could make it equivalent to the one I thought was different, if there are no such situations where a collection of open sets in S is capable of creating every open set in S by union of its' elements without also covering S.

Can such a situation occur?

Put another way: Can there exist a topological space S such that the union of all open sets in S does not equal S?

You seem rather confused. You've worded your question in three different ways, no two of which are equivalent. All are easily answerable:

Wait, so there are no situations where a collection of sets can cover S while being incapable of creating every open set in S by arbitrary unions?

Yes, there are. Example in R: (-∞,0), (0,∞), (-1,1).

I can see now how the second part of the usual definition combined with the first part could make it equivalent to the one I thought was different, if there are no such situations where a collection of open sets in S is capable of creating every open set in S by union of its' elements without also covering S.

Can such a situation occur?

No. S is an open set in S. If you can create every open set in S with a set of open sets, it trivially covers S, as aguacate has pointed out.

Put another way: Can there exist a topological space S such that the union of all open sets in S does not equal S?

No. Again, S is open in S.

The first question wasn't meant to be equivalent to the second two.

Oh, and I was forgetting one of the parts of the definition of open sets, thanks for your patience.

Oh, and sorry but I have one more question:

How are the second two questions not equivalent? If there is a collection of open sets B in S, and B forms a basis for S, then isn't the union of all sets in B equivalent to the union of all open sets in S?
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aguacate
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### Re: Halp at Toplogy!

I see what you are saying, and I think the last two questions are indeed equivalent. Yes, you cannot take the union of all the elements of a basis without covering S. OK wait stop. One of the possibly confusing points of topology is that a topological space consists of TWO sets. Here I am going to use the letters that my book uses. One is the set X. One is the topology on the set X, called in my book T (cursive T actually). T is a collection of sets of elements of X. By definition of a topology, T contains X. Therefore the union of all elements of T covers X.

The union of all the basis elements will always cover X.

Token
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### Re: Halp at Toplogy!

There is a difference between all the open sets of S and an arbitrary collection (as asked in the question) of open sets, or indeed a basis.
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antonfire
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### Re: Halp at Toplogy!

Regarding the OP's original question: the easiest solution is way overkill, so look for an overkill solution. Start with an open set U. You want to construct it as a union of "rational intervals", so include everything you possibly can in your union. Then show that the original open set is indeed that union.
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