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$$\geq$$a};

Prove that:
(i) (a, $$\infty$$) $$\subseteq$$ [b, $$\infty$$) $$\Leftrightarrow$$ a$$\geq$$b,
(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$$\geq$$a};

Prove that:
(i) (a, $$\infty$$) $$\subseteq$$ [b, $$\infty$$) $$\Leftrightarrow$$ a$$\geq$$b,
(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 $\epsilon> 0$, [itex]a+ \epsilon> a[itex]so $a+ \epsilon\in (a, \infty)$. 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.[/itex][/itex]

 

[congtongsat]

much appreciated. cleared things up for me.