Question

Let A be a set consisting of the intersection of the set of rationals with the closed and bounded interval between 0 and 1. Find an open cover with no finite subcover.

What I was thinking is that you could choose some irrational in the interval, and have an infinite sequence of open sets approach the irrational from both sides. Does that seem like a reasonable and clean solution?

See below for a proper write-up of the problem.

Answer 1

Let \[A=\mathbb{Q} \cap [0, 1] \subset \mathbb{R}\]Find an open cover for A with no finite subcover.

Answer 2

Topology was never my strong suit...I suffered through it, but it was, by no means, fun.

Given that, the approach seems sound, but the only problems i'd see you having is defining the infinite sequences of open sets such that
(a) They include 0 and 1.
(b) They approach the irrational value from left and right

My first thought on defining so that the only subcover is an infinite subcover is to have something akin to {1/1, 1/2, 1/3, ...}, {2/2, 2/3, 2/4, ...}, etc.

You would have an issue with 0 there, but you would not have a cover without taking them all.

Just my thoughts. Don't know if they help.

Answer 3

I was thinking something like this
\[O = \left\{\left(-1, \sqrt{\frac{1}{2}} - \frac{1}{n}\right) \cup \left(\sqrt{\frac{1}{2}} + \frac{1}{n}, 2\right) \mid n \in Z_+ \right\}\]

Answer 4

Sorry...my computer cuts off the denominators of fractions...what is the denomiators of your ends?

Answer 5

sqrt(1/2)

Answer 6

(-1, sqrt(1/2)-1/n) union (sqrt(1/2) + 1/n, 2)

Answer 7

ok...lemme think on that for a sec :)

Answer 8

for all n in the set of positive integers

Answer 9

It looks good. Let's look at the conditions.

Can we prove that it's an open cover?
It is clearly open since the end points of each part of the union are open. It includes to points 0 and 1 which were part of the closed set A.

For any rational number, k, it will on either side of sqrt(1/2). You can construct your n by finding the difference between k and sqrt(1/2) and set n as the ceiling of the reciprocal of this value. That guarantees 1/n is less than that difference. So clearly this covers.

Prove there is no finite subcover.
Since each element of O is a proper subset of the next element of O, in order for there to be a finite subcover, there must be a distinct value for n which is a cover of A. Suppose you haveone such value. Then you'd need to find a rational closer to sqrt(1/2) than (1/n+1). If you find that value, then your value of n which provided a cover cannot be correct.

Figure out how to describe that contradictory number and i think you have a decent proof.

Answer 10

Alright, I guess its not enough to suggest that its intuitively obvious that it approaches but does not include the irrational number and the nature of it being an infinite sequence.

Well in any case thanks.

Answer 11

Lol...that approach *might* work, but mathematicians get a little touchy with the word "intuitively"

I think you can find a way to describe that number...but i'd need to look a little longer.

But from the perspective of finding a solution, you clearly have one. A mathematician can't argue with your defined solution set O, but they can argue with your proof...

Answer 12

Yes I know, describing something as intuitively obvious is always a no no in math. I will try to write up a rigorous proof.

Answer 13

Actually, you don't need a specific case...you can have a set of cases.

Suppose n=n1 provides a cover for X. Then any rational in the interval (sqrt(1/2)-(1/n), sqrt(1/2)) would not be included. And there are infinitely many rationals in this interval.

Answer 14

Basically, you're *saying* "intuitively" without using the actual word :)

Answer 15

Heh