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:

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.

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)

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].

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.

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!