Under this definition, 'closed' makes sense in the larger context, since in mathematics, if an operation is 'closed' on a set, that means that applying the operation to elements of the set always yields another element of the same set.

Open is a bit more awkward. A set O is 'open' if and only if the complement of O (i.e. the set of all points not in O) is closed. So open is kind of the opposite of closed, hence the convention. Of course, it isn't really the opposite of closed, since there are sets which are neither closed nor open.

tt;dr (too technical; didn't read) in maths, 'closed' usually means 'I can do stuff inside this set without falling out of it'. In the case of intervals, the 'stuff' in question is taking a limit of a convergent sequence.

This set is not closed there are non-rational numbers in that interval which are limit points of sequences that consist only of rationals. For example, any of the algebraic numbers. I think that all real numbers are limits of such sequences, but I might be mis-remembering some subtlety of Dedekind Cuts (one method for constructing the Reals).

This set is not open because any rational is the limit of a sequence of non-rational reals. This probably makes intuitive sense, but just for the sake of formality: To construct such a sequence for any rational r, start with the number x_1 = 1/pi, and approach by a factor of 1/pi at each step, i.e. x_n+1 = x_n + (r-x_n)/pi . x_n is irrational because pi is transcendental.

Any simple interval in R will be either closed or open on each end (but it could be closed on one and open on the other). It's more illustrative to create a set with a non-compact interior. In higher dimensions it's possible to have more exotic borders on an interval, but I think that border will just end up being isomorphic to a non-compact set in a lower dimension.

Proof that it's not closed: the sequence (1, 1/2, 1/4, 1/8, ...) is entirely inside the interval, but converges on 0, which is outside the interval, therefore etc.

Proof that it's not open: the sequence (2, 3/2, 5/4, 9/8, ...) is entirely outside the interval (i.e. inside the complement), but it converges on 1, which is inside the interval (i.e. outside the complement). Thus the complement of the interval is not closed, therefore etc.

http://en.m.wikipedia.org/wiki/Clopen_set

    a[..]b

With a closed interval it eventually return a bound instead of getting smaller/larger - continuing.

This no doubt sounds strange, but it surely comes down to the slightly odd behavior of real numbers. In an open interval like (0, 1), there is no single real number which is less than one but is greater than all other real numbers less than one. (Proof by Cantor's diagonal: write out a list of all such numbers and you can always construct another which wasn't on the list, even when the list is infinite.) In contrast, the closed interval [0, 1] has a definite maximum element, the number 1.

This seems like a good example of poor UX in mathematics; the terms are too similar and can only be learned by memorization, the notations likewise. Good luck changing it though; Wikipedia informs me that there are no less than two ISO standards codifying it.

There is a perfectly clear notation in the Wikipedia page, however: (a,b) = {x∈ℝ|a<x<b}; [a,b] = {x∈ℝ|a≤x≤b}

The rational numbers are like this too.

All the rational numbers between 0 and 1 also increase arbitrarily close to 1 with no largest element. You can prove this by simple contradiction: For every rational number x < 1, there is another rational number y = (x+1)/2 with x < y < 1. So the set of rational numbers less than 1 also has no largest element.

Actually real numbers can be defined (axiomatized) in terms of the least upper bound property -- every set of real numbers that's bounded above has a least upper bound. So you could actually have a set S of rational numbers that gets arbitrarily close to something like sqrt(2) from below. S then has no rational least upper bound -- for every rational number x greater than or equal to every element of S, there is another rational number y greater than or equal to every element of S but smaller than x.

Note that a set of real numbers need not contain its upper bound -- as you noted, a bounded open interval doesn't contain its upper bound.

Also, any notation will have to be learned. Concise ones a bit more, but there is nothing one can do about that.

And that perfectly clear notation requires you to learn quite a bit by memorization, such as the meaning of those {} brackets, the |, and the symbols ∈ and ℝ.

Finally, you don't need Cantor's diagonalization for a proof. It is easy(1) to show that, for any real x<1, x+(1-x)/2 is real, less than one, and strictly greater than x.

Alternatively, assuming 0<x<1, write x in decimal, and increase the first digit less than nine by one to get a larger real less than one. There is such a digit because 0.9999999… is not less than one.

(1) depending on how deep you want to descend into the foundations of mathematics.

To take your thinking a bit further the interval is "shut" precisely because it's closed exactly at that point. And the interval is thrown "open" by not including the point.

BTW; Rudin's classic text (1) covers the fascinating topic of neighborhoods which builds on the concept of intervals. It took me a while to completely understand the concept but after that understanding limits was relatively easier; even otherwise "neighborhood" as a concept is interesting in itself.

[1] http://www.math.boun.edu.tr/instructors/ozturk/eskiders/guz1...

Thanks for everyone's input. I think I have a clear understanding of the open/closed paradigm now.