Proff of half infinite intervals through set theory

[congtongsat]

Problem:
We define half infinite intervals as follows:
(a, \infty) = {x\in R | x>a};
[a, \infty) = {x\in R | x\geqa};

Prove that:
(i) (a, \infty) \subseteq [b, \infty) \Leftrightarrow a\geqb,
(ii) [a, \infty) \subseteq (b, \infty) \Leftrightarrow a>b.

I've got pretty much no idea how to do this. Then again I've been struggling at this for a couple hours and my mind doesn't work particularly well at 2:25 am PST. Help would be greatly appreciated on this.

Thanks.

 

[HallsofIvy]

Problem:
We define half infinite intervals as follows:
(a, \infty) = {x\in R | x>a};
[a, \infty) = {x\in R | x\geqa};

Prove that:
(i) (a, \infty) \subseteq [b, \infty) \Leftrightarrow a\geqb,
(ii) [a, \infty) \subseteq (b, \infty) \Leftrightarrow a>b.

I've got pretty much no idea how to do this. Then again I've been struggling at this for a couple hours and my mind doesn't work particularly well at 2:25 am PST. Help would be greatly appreciated on this.

Thanks.

You prove that A\subseteq B by proving that any member of A is a member of B.
If a\geq b, then for any member, x, of (a, \infty), x> a\ge b so x> b and therefore x\in [b, \infty).

The converse is a litlle harder because we are looking at (a, \infty) rather than [a,\infty), for all \epsilon> 0, a+ \epsilon> aso a+ \epsilon\in (a, \infty). If (a,\infty)\subseteq [b, \infty) and so a+\epsilon\ge b. Taking the limit as \epsilon goes to 0, a\le b.<br /> <br /> (ii) is actually simpler.

 

[congtongsat]

much appreciated. cleared things up for me.