1. closure of semi-infinite intervals

consider $[a, \infty)$

Is this set open or closed or neither?

It's pretty clearly not open as any epsilon neighborhood about $a$ will contain points both in and not in the interval.

Is it closed?

One definition I've read is that closed intervals contain all their endpoints and as there is no endpoint on the right to be contained this interval is closed.

Another definition I read here is that closed intervals will contain all their limit points. I believe $\infty$ is a limit point to this interval yet it is not contained within it.

Therefore the interval is not closed.

Is there some fundamental disagreement here or is it simply the choice of definitions?

Is there some definite assertion you can make about the closure of $[a, \infty)$ or $(\infty, a]$ ?

2. Re: closure of semi-infinite intervals

Originally Posted by romsek
consider $[a, \infty)$
Is this set open or closed or neither?
It's pretty clearly not open as any epsilon neighborhood about $a$ will contain points both in and not in the interval.
1) $\infty$ is not ever anything more than a concept. So it cannot be a point nor limit point.

2) The point $a$ is a boundary point of the set $[a,\infty)$ in fact in $\mathbb{R}^1$ it is the only boundary point.

3) The complement of $[a,\infty)$ is $(-\infty,a)$ which is open therefore $[a,\infty)$ is closed