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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.

J. W. Tanner
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user3180
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  • Use the formal definition. – anomaly Feb 08 '19 at 01:31
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    Comment: $(- \infty,0)$ is open, so its complement is closed – J. W. Tanner Feb 08 '19 at 01:31
  • What are the definitions of "open set" and "closed set"? Which of those two sets satisfies the definition of "open set"? Which of them satisfies the definition of "closed set"? Would it make you feel better if the sets were described without the $\infty$ symbol, as ${x:x\gt0}$ and ${x:x\ge0}$? – bof Feb 08 '19 at 01:31
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    In $\mathbb{R}$ there's no $\infty$ for a sequence to try to converge to, so $[0,\infty)$ is closed because sequences that "go to infinity" just aren't convergent. There's a thing called the extended real line that contains the elements $-\infty,\infty$; in the topology of this object, $[0,\infty)$ is not closed because there are sequences that bona fide converge to $\infty$. – Ian Feb 08 '19 at 01:33
  • I am using this page for the definition of open and closed set, but actually the definition seems incomplete (it only describes open interval in some detail): http://planning.cs.uiuc.edu/node125.html – user3180 Feb 08 '19 at 01:39
  • For example, the three axioms given for what constitutes an 'open set' do not seem to allow me to reason about infinity – user3180 Feb 08 '19 at 01:39
  • The infinity part isn't relevent. And there is nothing to "think about". The "infinity" is not in any way part of the set and infinity certainly is not an element of the set. The notation of the infinity sign means it extends forever and has no upper bound. – fleablood Feb 08 '19 at 01:42
  • Do you know what the notation $(0,\infty)$ means? –  Feb 08 '19 at 01:48

3 Answers3

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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.)

J. W. Tanner
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user247327
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    I think you meant "and at least one point that is not in the set". – J. W. Tanner Feb 08 '19 at 01:49
  • How is neighborhood defined? – user3180 Feb 08 '19 at 01:49
  • This is a good answer, although the intuitive geometric idea of 'closed' would semantically seem to be incompatible with the idea of unboundedness, but at least the concept of boundary points clears up the mathematical definition – user3180 Feb 08 '19 at 01:58
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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.)

fleablood
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    This is the intuitive answer. The symbol $\infty$ doesn't represent an end point; it represents the fact that everything in that direction is inside the set. Infinite implies "you can't get out that way." – David K Feb 08 '19 at 13:19
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$(0, \infty)=\{ x\in \mathbb{R}: x > 0\} = \bigcup \{(0,x): x >0\}$ 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).

Henno Brandsma
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