Topology & basis

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.

Answers and Replies

Related Calculus and Beyond Homework Help News on Phys.org
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.

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)

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.

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?

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.

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
Homework Helper
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.

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
Homework Helper
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>?

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

radou
Homework Helper
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?

That there's a rational number in every interval?

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

radou
Homework Helper
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.