Why is $(0, \infty)$ an open set and $ [0, \infty)$ a closed set?

Question

Specifically, I don't understand how to think about the infinity part. Closed and open seem to have pretty intuitive definitions when not considering infinity.

Answer 1

One definition of a "closed" set is a set that contains all if its boundary points. And a definition of an "open" set is a set that contains none of its boundary points. And a "boundary point" is a point such that every open neighborhood contains at least one point in the set and at least one point that is not in the set. Given either of these two sets, the only boundary point is $0$. No negative number, $x$, is a boundary point because $(x-1, x/2)$ is an open neighborhood that contains no member of the set (negative numbers are "exterior points" of the set). No positive number, $y$, is a boundary point because $(x/2, x+ 1)$ is an open neighborhood that contains only members of the set (positive numbers are "interior points" of the set). But $(-1, 1)$ is an open neighborhood of $0$ that contains both negative numbers (so not in the set) and positive numbers (so in the set). "Infinity" is, of course, not a point in the real number line at all.

So the only boundary point of $[0,\infty)$ and $(0, \infty)$ is $0$ itself. It is in $[0, \infty)$, so that set is closed. It is not in $(0, \infty)$, so that set is open.

(Edited thanks to J W Tanner's comment.)

Answer 2

Let's play a game.

Let's use, the notation of $(a,b)$ to mean an interval with $a $ at the low end and $b $ at the high end. If an end point is not part of the interval, we write a round bracket. If an end is included then we write a square bracket.

So far that's the usual notation.

Now for a different rule. If there isn't a low end or a high end we simply don't write any number and we have a reverse bracket. For example $(a, ($ means the interval has $a $ on the low end and doesn't have a high end; it has all numbers greater than $a $.

Is it clear $(0, ($ is open? Around every $x\in (0, ($ we can find a small interval $(x-e,x+e) $ so it is open.

And is it clear $[0, ($ is closed. It has all it's limit points, especially $0$. So it is closed.

But what about its "infinity part"? Um... there is no "infinity part". "Infinity" has nothing to do with the intervals. "Infinity" is not an endpoint of the intervals. There's nothing to "think about" about the "infinity part".

Note: the notation $(0,\infty)$, for better or worse, has nothing to do with infinity. Infinity is not an end point. The infinity sign only means there is no finite upper end point. But there isn't an infinite end point either.

(In case it's not clear; $(a, ($ in my made up pretend notation, is exactly the same thing as $(a,\infty) $ in conventional notation.)

Answer 3

$(0, \infty)=\{ x\in \mathbb{R}: x > 0\} = \bigcup \{(0,x): x >0\}$ and if we know all $(0,x)$ are open, then also their union is open. A similar reasoning we can holds for $(-\infty, 0) = \{x \in \mathbb{R}: x < 0\} = \bigcup\{(x,0): x < 0\}$ which is thus also open and has $[0,\infty)$ as its complement (and so that latter set is closed).