Basis for a Topology in the Real Line

Question

Question:

Show that the set of closed intervals

            {[a,b] | a,b in rational number set, a<b}------->(1)

are not a basis for a topology in the real line R.

My approach to the question:

1. My initial thought was to prove this by proving that 'the 2 axioms for a given set to be a ', do not hold for the given set, i.e.

          Let X be a set. A basis of a topology on X is a collection B of subsets in X such that

          (B1) For every x ∈ X, there is an element B in B such that x ∈ U.

          (B2) If x ∈ B1 ∩ B2 where B1, B2 are in B, then there is B3 in B such that x ∈ B3 ⊂ B1 ∩ B2.

1. We know that rational numbers are of the form p/q, where p, q are in R(set of real numbers), q not equal to 0 and the only common factor p & q both have is 1

      That is, the given set is of the form, provided that a=p/g and b=r/s

      B = {[p/q ,r/s] | p,q,r,s are in R, q & s not equal to 0, p/q<r/s}  ---->(2)

1. Let's consider (B1)

I noticed that, [20, 21] does not belong to B, as 20=100/5 and 21=105/5 are not rational numbers.

However, sqrt(440)= 20.976176963403030939829070273599, which belongs to R falls in the region [20, 21].

1. Then, we can conclude that not all x in R doesn't belong to set P, which means (B1) does not hold for the set P.

1. With this, can we conclude that P is not a basis for a topology in the real line R?

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I'm taking a Topology course on the 3rd year my undergraduate degree.

So, I was wondering whether the approach I've taken is correct and whether a considerably strong proof for the level of my degree programme/course. I didn't come up with a general method of proving this, but rather do it using a counterexample, as above.

Any help and thoughts would be highly appreciated!

I'd be really glad if someone could guide me on doing this as a general proof.

Thank you!

Answer 1

Hi! This seems more like a topology/real analysis question than algebra 2 (it was somehow classified as such by the system).

Your approach consisting in finding a counterexample is a great first step. Often in math, to show that something is not true, we just have to find a counterexample!

i) In the above definition of B, it should say B instead of U. Also, it is better to use different symbols for the basis and its elements (I know, it is tricky with the available symbols if want to keep using a "b" letter; perhaps, we can use bold case and italics for the basis.)

 (B1) For every x ∈ X, there is an element B in B such that x ∈ B.

ii) The interval [20,21] belongs to B (every natural number is a rational number), so it cannot help us with a counterexample.

iii) We can try to disprove B1 or B2. Because the intervals in B are closed, I suspect that leveraging this will help us find a counterexample.

We can observe that to negate B2, we want to find B₁ and B₂ in B such that :

if x ∈ B₁ ∩ B₂, then there is no B₃ in B such that x ∈ B₃ ⊂ B₁ ∩ B_2.

Let's take B₁=[20,21] and B₂=[21,25], for example. They are both in B. Their intersection

[20,21] ∩ [21,25] = {21}.

All intervals in B must be intervals [a,b] with the left-end point strictly smaller than the right-end point.

But for x in [20,21] ∩ [21,25], we cannot find B₃ in B such that x ∈ B₃ ⊂ [20,21] ∩ [21,25].

Why? Because [20,21] ∩ [21,25] = {21} and {21} is the set of only one point in it and such a set cannot contain " ⊂" an interval of the type [a,b] with a<b (such sets contain more than one point).

We found a counterexample for B2, hence our set B is not a basis.

I hope this helps! Please let me know if you have any questions Good luck!