Show that B is not a topology on R

[Tomath]

Homework Statement

Let B be the family of subsets of \mathbb{R} consisting of \mathbb{R} and the subsets [n,a) := {r \in \mathbb{R} : n \leq r < a} with n \in \mathbb{Z}, a \in \mathbb{R} Show that B is not a topology on \mathbb{R}

Homework Equations

The Attempt at a Solution

If B were a topology then we would need:
\emptysetand \mathbb{R} \in B (1), the arbitrary union of any opens in B to be in B (2) and any finite union of opens in B to be in B (3). Now the first two conditions (1), (2), seem to be valid so if B is not a topology on \mathbb{R} then certainly condition (3) would have to fail. My question is, does condition (3) indeed fail and if it does, how can I show this?

Thanks in advance

 

[Dick]

Homework Statement

Let B be the family of subsets of \mathbb{R} consisting of \mathbb{R} and the subsets [n,a) := {r \in \mathbb{R} : n \leq r < a} with n \in \mathbb{Z}, a \in \mathbb{R} Show that B is not a topology on \mathbb{R}

Homework Equations

The Attempt at a Solution

If B were a topology then we would need:
\emptysetand \mathbb{R} \in B (1), the arbitrary union of any opens in B to be in B (2) and any finite union of opens in B to be in B (3). Now the first two conditions (1), (2), seem to be valid so if B is not a topology on \mathbb{R} then certainly condition (3) would have to fail. My question is, does condition (3) indeed fail and if it does, how can I show this?

Thanks in advance

What is the union of all of the [n,a)?

 

[Tomath]

What is the union of all of the [n,a)?

If I am not mistaken the union of all of the [n,a) is \mathbb{R}

 

[micromass]

If I am not mistaken the union of all of the [n,a) is \mathbb{R}

Yes. What if you take a little bit less of the $[n,a)$?? Can you form some half-open interval?

 

[Dick]

If I am not mistaken the union of all of the [n,a) is \mathbb{R}

No, I don't think it's all of R. a or a+1 isn't in it. Oh, and your definition of topology is a little off. You want finite intersections to be in the topology. Specifying finite unions after you already said arbitrary union would be a little redundant.