Is the Open Cover of the Interval [0,1) with Infinite Subcovers Effective?

[Bachelier]

Given the interval [0,1)

is this a good Open cover with infinite subcovers

{(An)} such that An =(-1/n, n) with n \in lN

Is there any reason we should stay to the boundaries of the set we're trying to cover?

I'm thinking that even (-n, n) should work.

Am I wrong?

 

[Bachelier]

I think I just answered my own question.

I think to make sure that the portion that covers the interval is infinite.

Hence (-1, 1 - (1/n)) would work.

 

[AxiomOfChoice]

Given the interval [0,1)

is this a good Open cover with infinite subcovers

{(An)} such that An =(-1/n, n) with n \in lN

Is there any reason we should stay to the boundaries of the set we're trying to cover?

I'm thinking that even (-n, n) should work.

Am I wrong?

Neither of these is an example of a cover without a finite subcover. In the first case, (-1,1) \supset [0,1); same in the second case.

 

[Bachelier]

Neither of these is an example of a cover without a finite subcover. In the first case, (-1,1) \supset [0,1); same in the second case.

Agreed, but (-1,1-1/n)
is.

 

[AxiomOfChoice]

I think I just answered my own question.

I think to make sure that the portion that covers the interval is infinite.

Hence (-1, 1 - (1/n)) would work.

Yep. No finite subcovers here!