Is the set of all the finite half-open intervals closed under finite intersections and unions?

Question

Let $\mathcal{J} = \{ [a,b) | a,b \in \mathbb{R}, a \leq b \}$ is $\mathcal{J}$ closed under finite intersections and unions?

• $\mathcal{J}$ is closed under finite intersections Let $J_1 =\{ x \in \mathbb{R} | a_1 \leq x <b_1 \}$ Let $J_2 =\{ x \in \mathbb{R} | a_2 \leq x <b_2 \}$ where $a_1 \leq a_2 < b_2 < b_1$

Let $J=J_1 \cap J_2 \implies J=\{z \in \mathbb{R} | (a_2 \leq z <b_2) \text{ and } (a_1 \leq z<b_1) \}$ Let $z_1 < z_2$ where $z_1, z_2 \in J$

Now, choose any $\epsilon >0$ and let $t=z_1 + \dfrac{z_2 - z_1}{\epsilon}$

Then $t \leq z_2 < b_2$ and $a_2 \leq z_1 \leq t$

$ \implies z_1 \leq t \leq z_2$ and $t \in J$

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$J_1 \cup J_2$ is not a half-open interval but how to show that? also is the previous proof correct?

Answer 1

$[a,b) \cap [c,d)$ is empty if $b \le c$ (so equal to $[a.a)$ in $\mathcal{J}$ in that case), and if $b > c$ and $b <d$, the intersection is $[c,b)$, if $b > c$ and $b \ge d$, the intersection is $[c,d)$, all in $\mathcal{J}$.

$[0,1) \cup [2,3)$ is not in $\mathcal{J}$, obviously (it's not order convex). That one example suffices.

Answer 2

It is not closed under union. In fact

$[0,1)\cup [2,3)\not \in J$

In any case $J$ is a base for a Topology on $\mathbb{R}$. This Topology is important because it is an example of a Space which it is slitgly different from the Euclidean Topology but that is not separable.