Constructing an Open Cover for the Rational Numbers in the Interval [0, 1]

[Bachelier]

Need an open cover with NO FINITE subcovers

{x ∈ Q : 0 ≤ x ≤ 2}

I was thinking: {A_n} n = 1 to infinity s.t.

A_n = (1/(sqrt 2^n) - 1/(2^10) , 3)

 

[Bachelier]

The following answer was given, I think it's wrong, for one, x=0 is not covered, and x= 2 is not neither. I think a way to correct it would be to choose A_n = (-1, X_n) and instead of √2 we should choose 2√2. What do you think?

The answer:
Because the rationals are dense in R we can construct an infinite collection of rational points that
get close to √2, but never reach it. Let x1 = 1/2 . Since x1 and √2 are real numbers, there is a rational number between them, call it x2. Now since x2 ∈ R, there is some rational number between x2 and √2, call it x3. Continue to get x1 < x2 < x3 < · · · < xn < · · · < √2. Now let An = (1, xn). Then J = {An} is an open cover of our set, but it has no finite subcover.

 

[AxiomOfChoice]

{x ∈ Q : 0 ≤ x ≤ 2}

I was thinking: {A_n} n = 1 to infinity s.t.

A_n = (1/(sqrt 2^n) - 1/(2^10) , 3)

Hmmm...well, $\mathbb R$ is an open cover of this set! After all, an open cover of $S$ is just a collection $U_\alpha$ of open sets such that $S\subset \bigcup_\alpha U_\alpha$. Are you looking for an open cover with no finite subcover?

 

[Bachelier]

Hmmm...well, $\mathbb R$ is an open cover of this set! After all, an open cover of $S$ is just a collection $U_\alpha$ of open sets such that $S\subset \bigcup_\alpha U_\alpha$. Are you looking for an open cover with no finite subcover?

exactly. With no finite subcovers

 

[disregardthat]

Try covering it with disjoint open intervals. E.g. ( [0,r_1) U (r_1,r_2) U (r_2,r_3) U ... ) U (r,2].

$$r_n = \sum^{n}_{k=1}sqrt(2)/2^k, \ \ r = \lim r_n$$

 

[AxiomOfChoice]

Try covering it with disjoint open intervals. E.g. ( [0,r_1) U (r_1,r_2) U (r_2,r_3) U ... ) U (r,2].

$$r_n = \sum^{n}_{k=1}sqrt(2)/2^k, \ \ r = \lim r_n$$

How does that work? $[0,r_1)$ and $(r,2]$ are not open sets. (By the way...what's your $r$ there?)

 

[disregardthat]

How does that work? $[0,r_1)$ and $(r,2]$ are not open sets. (By the way...what's your $r$ there?)

I assume he is giving [0,2] cap Q the subspace topology of R. In which case e.g. [0,r_1) = (-1,r_1) cap ([0,2] cap Q), where (-1,r_1) is an open interval in R. Or just: if [0,2] cap Q has the order topology, then [0,a) is open since 0 is the least element.

I defined r as the limit of the r_n's above.

 

[Bachelier]

I assume he is giving [0,2] cap Q the subspace topology of R. In which case e.g. [0,r_1) = (-1,r_1) cap ([0,2] cap Q), where (-1,r_1) is an open interval in R. Or just: if [0,2] cap Q has the order topology, then [0,a) is open since 0 is the least element.

I defined r as the limit of the r_n's above.

So the limit r in this case will approach sqrt 2, right?

 

[Bachelier]

Is there an intuition behind using the Sqrt 2?

 

[Bachelier]

What would you recommend for {x ∈ Q : 0 ≤ x ≤ 1}?

with respect to another summation series

 

[Bachelier]

$$r_n = \sum^{n}_{k=1}sqrt(2)/2^k * 1/1.4141..., \ \ r = \lim r_n$$

to cover interval [0,1]

 

[disregardthat]

Is there an intuition behind using the Sqrt 2?

The r_n's and r are irrational, so they are not in your set. If they were, it wouldn't be a cover. And yes, r = sqrt(2).

For your other set you could just divide the r_n's by 2...

$$r_n = \sum^{n}_{k=1}sqrt(2)/2^k * 1/1.4141..., \ \ r = \lim r_n$$

to cover interval [0,1]

I don't know what you're trying to say here.

 

[Bachelier]

The r_n's and r are irrational, so they are not in your set. If they were, it wouldn't be a cover. And yes, r = sqrt(2).

For your other set you could just divide the r_n's by 2...I don't know what you're trying to say here.

Thanks Jarle. great work.

I think I found another cover for {x ∈ Q : 0 ≤ x ≤ 1}

Basically the union of R_n U R_0 such that R_n = (-1 , 1/sqrt(2) - 1/n ) and R_0 = ( 1/sqrt(2) , 2)