Topology of R: Basis and Rationals

[Damascus Road]

Consider the collection of sets C = {[a,b), | a<b, and and b are rational }

a.) Show that C is a basis for a topology on R.
b.) prove that the topology generated by C is not the standard topology on R.So, I know for C to be a basis, there must be some x $$\in$$ R,
and in the union of some C1 $$\bigcap$$ C2 there must be a C3, so that x $$\in$$ C3.
So, since my C = {[a,b), | a<b,
Letting a1 < a2 < x < b1 < b2
my C3 = [a2, b1) $$\in$$ C.

I'm confused how to work in the rational numbers however, must I truncate them or something? Don't know how this would fit into the logic..

thanks in advance.

 

[foxjwill]

So, since my C = {[a,b), | a<b,
Letting a1 < a2 < x < b1 < b2
my C3 = [a2, b1) $$\in$$ C.

You've got the right idea here. However, you're going to need to make the proof a bit clearer: What are you using for $C_1$ and $C_2$? Why did you choose $C_3$ to be $[a_2,b_1)$?

I'm confused how to work in the rational numbers however, must I truncate them or something?

I'm not sure what you mean by this.

 

[Damascus Road]

How about this:

Let $$C1 = [a_{1},b_{1})$$ and $$C2 = [a_{2}, b_{2})$$, then $$C1 \bigcap C2$$ gives C3.
Where $$C3 = [a_{2},b_{1})$$ ?

With regards to the truncation, I'm not sure how to prove that the topology generated by C is not in the standard topology when a and b are rational. Hence, I was thinking along the lines of some kind of truncation (to make it an integer)

 

[foxjwill]

Let $$C1 = [a_{1},b_{1})$$ and $$C2 = [a_{2}, b_{2})$$, then $$C1 \bigcap C2$$ gives C3.
Where $$C3 = [a_{2},b_{1})$$ ?

Almost. Remember what you need to prove:

Given any two basis sets $C_1$ and $C_2$ and any point $x\in C_1\cap C_2$, there exists a basis set $C_3$ such that

$$x\in C_3\subseteq C_1\cap C_2.$$​

With regards to the truncation, I'm not sure how to prove that the topology generated by C is not in the standard topology when a and b are rational. Hence, I was thinking along the lines of some kind of truncation (to make it an integer)

You're making this too difficult! Here's a hint: By definition, every basis set of a topology is open in that topology.

 

[Damascus Road]

Almost. Remember what you need to prove:

Given any two basis sets $C_1$ and $C_2$ and any point $x\in C_1\cap C_2$, there exists a basis set $C_3$ such that

$$x\in C_3\subseteq C_1\cap C_2.$$​

Ah, I think I see what I'm missing.

Let C1 = [a1,b1) and C2 = [a2,b2). Then there exists an $$x\in C_3\subseteq C_1\cap C_2.$$ Where C3 = [a2,b1).

Although this doesn't sound like I'm proving anything, but rather just saying there is an x... so maybe this is still wrong.

You're making this too difficult! Here's a hint: By definition, every basis set of a topology is open in that topology.

Sorry, I'm still confused how to go about this, and partially what it's asking. The standard topology on R is just the real line, right? I can go on with just that?

 

[foxjwill]

Let C1 = [a1,b1) and C2 = [a2,b2). Then there exists an $$x\in C_3\subseteq C_1\cap C_2.$$ Where C3 = [a2,b1).

Almost. What you need to do is show that no matter what point x you choose from the intersection $C_1\cap C_2$, you can find a basis set $C_3$ such that $x\in C_3\subseteq C_1\cap C_2$.

Sorry, I'm still confused how to go about this, and partially what it's asking. The standard topology on R is just the real line, right? I can go on with just that?

The standard topology on the real line is the topology generated by the open intervals (a,b); i.e., a set U is open in the standard topology iff it's a union of arbitrarily many open intervals.

 

[Damascus Road]

Almost. What you need to do is show that no matter what point x you choose from the intersection $C_1\cap C_2$, you can find a basis set $C_3$ such that $x\in C_3\subseteq C_1\cap C_2$.

Am I ok up until where I am? i.e. i need to add more?

My gut feeling is that it has something to do with the openness of the intersection... but I haven't found it in my notes yet.

The standard topology on the real line is the topology generated by the open intervals (a,b); i.e., a set U is open in the standard topology iff it's a union of arbitrarily many open intervals.

So I need to show that it cannot be expressed as an arbitrary union of open intervals.

 

[radou]

To show that a basis B generates a certain topology T, for any U in T, and for any x in U, you need to prove existence of a basis element from B which contains x and is contained in U.

This is a useful operative criterion which may help you.

 

[Damascus Road]

I have an example in my book: "on the real line R, let B = {(a,b) $$\subset$$ R | a < b} set of open intervals in R. Certainly every point of R is contained in an open interval and therefore is contained in a set in B."

Can I say something similar? Except my set isn't an open set of something, since that's not the same as saying it's open on R right?

 

[radou]

Your sets are open sets in the topology generated by these sets, as a start. Now, you need to see what's the relation between this topology and the standard topology on R.

Take an open interval <a, b> in R. Take an element x in <a, b>. Does there exist an element from your collection which contains x and is contained in <a, b>?

 

[Damascus Road]

Yes of course there is, since <a,b> is open. But I have [a,b)...

 

[radou]

Yes of course there is, since <a,b> is open. But I have [a,b)...

Well, precisely speaking, <a, b> being open doesn't prove the fact.

Let x be in <a, b>. Is there a rational number between x and a? Between x and b? What do you conclude?

 

[Damascus Road]

That there's a rational number in every interval?

 

[Damascus Road]

Do I need to do an analysis of each case of intersection to show c is a basis for a topology on R?

 

[radou]

Yes, you should.

To check that some collection is a basis for a topology on a set, you need the two conditions mentioned in your posts above.

To check that some collection is a basis for a specific topology on a set, you use the condition I mentioned in posts #8 and #10.