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I'm having a little trouble grasping the definition of a topological space in terms of the open-sets. Suppose that we're defining a topology on the real line, in theory could we arbitrarily select the elements of the open-sets, so long as the set of all open sets satisfies the properties of a topological space?

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So, you have two open sets which consist each of two points, right? Then you can generate a discrete topology with this. That is, if you add a number of extra axioms like translating an open set is still giving an open set, but that is quite natural to do on the reals. – Raskolnikov Dec 20 '10 at 21:18
Sorry, I just deleted that to make the question more clear. What I'm current confused about is whether an arbitrary set can be defined as "open" without satisfying other properties such as convergence? – niale Dec 20 '10 at 21:21
Yes, you can define any set to be open, provided when you take the collection of sets that you defined to be open and you make unions and finite intersections, these sets also be open. But of course, not every choice will lead to interesting topologies. The discrete and the trivial topology are not really interesting from the viewpoint of analysis. – Raskolnikov Dec 20 '10 at 21:26
I'm not sure what you're asking, but if you're asking whether a set of subsets of the real line defines a topology if and only if it satisfies the conditions in the definition of a topology, then the answer is yes. – Jonas Meyer Dec 20 '10 at 21:27
Yes arbitrary set can be defined as open, but the catch is, that you need to define all open sets. The convergence is then determined by the open sets you have. If you have non-interesting open sets (trivial topology for example), your convergence can be non-interesting. For trivial topology for example any element in topological space is a limit for any sequence. – mpiktas Dec 20 '10 at 21:38

The answer to your specific question is yes: if you are given a set $X$, then (even in practice!) you may choose any collection $\{U_i\}_{i \in I}$ of subsets of $X$ which (i) contains the empty set and $X$ itself, (ii) is closed under the operation of finite intersections and (iii) is closed under the operation of arbitrary unions, and then this family will be the open sets for a unique topology on $X$.

In fact, all of the above is nothing else than the definition of a topology on a set $X$. But let's be honest: this is quite a profound and general definition, especially at first.

You can absolutely define "strange" topologies on the real line. For instance, a famous one is the lower limit topology in which the open sets are decreed to be unions of half-open intervals of the form $[a,b)$. It is easy to see that this gives a strictly larger family of open sets than the usual "Euclidean" topology on $\mathbb{R}$, i.e., every open interval $(c,d)$ is an infinite union of half-open intervals, but no $[a,b)$ is an infinite union of open intervals! This lower limit topology has all sorts of bizarre properties, many of which are detailed in the wikipedia article linked to above. In fact it is primarily of interest as a "counterexample", i.e., an example of a topological space that does not have many of the properties of more familiar spaces -- like metrizable spaces -- do.

A brief discussion of the family of all topologies on a fixed set $X$ can be found here (although I do not necessarily recommend these notes for a true beginner; they are somewhat more sophisticated than is necessary for a first pass). In particular, right at the beginning I have left as a (not that easy) exercise that on any infinite set $X$, there are $2^{2^{\# X}}$ different topologies on $X$. When $X = \mathbb{R}$, this is a really, really large infinite number, larger than one usually encounters outside of set-theoretic considerations. One is certainly not going to meet most of these topologies personally.

Note though that if you change the topology on "the real numbers", it loses much of the structure that made it the real numbers in the first place.

On a first pass through topological spaces, you should definitely keep your eye on the metric topologies: i.e., the topology associated to a metric space $(X,d)$ by decreeing a set to be open if it contains an $\epsilon$-ball around each of its points. For these types of topologies one can use the spatial and analytic intuition developed in one's previous studies in a useful way.

One may well ask: why do we speak of topological spaces rather than just metric spaces? There are (at least) two good reasons for this:

1) Even if the spaces that we want to work with are metrizable spaces -- i.e., the topology is derived from a metric -- often it turns out to be the case that the properties of the space that one wants to study depend only on the topology induced by the metric. For instance, continuous functions and convergent sequences -- two basic concepts in elementary analysis -- are purely topological in this way. (Arguably the only important concept of basic analysis which is not purely topological is that of Cauchy sequence and the related notion of completeness. Even these depend only a structure intermediate between the metric and the induced topology, namely the uniform structure of the space.) If one is really working with topological notions like convergence, after a certain point it becomes simpler not to carry around the metric structure. The standard example I like to give of this is that of product spaces: if $(X_1,d_1),\ldots,(X_n,d_n)$ are finitely many metric spaces, one would like to discuss concepts like convergence on the product set $X = X_1 \times \ldots \times X_n$. Experience shows that we want e.g. a sequence in the product to converge iff each of its components converges. There is one natural topology on the product which gives this convergence, the product topology. The product topology on a finite product of metric spaces is a metric topology, but not in a canonical way: even to put a metric on $\mathbb{R}^N$ there are plenty of choices, e.g. the $L^p$ norm $||(x_1,\ldots,x_n)||_p := (|x_1|^p + \ldots + |x_n|^p)^{\frac{1}{p}}$ for every $1 \leq p < \infty$ and also the infinity norm $||(x_1,\ldots,x_n)||_{\infty} = \max_i |x_i|$. (Kaplansky in his book Set Theory and Metric Spaces describes this situation as an "embarrassment of riches".) When one wants the product of a countably infinite family of metric spaces, there is again an obvious topology which turns out to be induced by a metric, but now it becomes slightly tricky to write down a single metric. But the product topology exists -- and is very natural and well-behaved -- for any family of topological spaces.

2) Remarkably -- even spookily -- in many parts of mathematics topological spaces show up which are far from being metrizable. For instance, non-Hausdorff topologies are useful in functional analysis, algebraic geometry and order theory. A few examples:

a) As above, any family $\{X_i\}_{i \in I}$ of (say nonempty) topological spaces admits a product space $X = \prod_i X_i$. Even if all of the spaces $X_i$ are metrizable, then as long as uncountably many of them consist of more than a single point, the product topology will not even be first countable, let alone metrizable. But still any product of Hausdorff spaces is Hausdorff and any product of compact spaces is compact: these are immensely useful facts. Taking "big" products of spaces -- and in particular of the interval $[0,1]$ -- comes up much more often in topology and related branches of mathematics than one might at first suspect.

b) A space is Boolean if it is compact, Hausdorff and totally disconnected. From the perspective of classical geometric topology, such spaces are of rather limited interest. Indeed if such a space has no isolated points and is metrizable, then it must be homeomorphic to the classical Cantor set. But M. Stone proved that to every Boolean space one can associate a Boolean ring and conversely: there is a categorical (anti-)equivalence between Boolean spaces and Boolean rings. (More generally, there is an equivalence between locally compact Hausdorff totally disconnected spaces and Boolean rings-without-multiplicative-identity.)

c) A space is Alexandroff if the intersection of any collection of open sets is again open. The obvious examples are discrete spaces and finite spaces. Indeed it is not hard to show that a Hausdorff space is Alexandroff iff it is discrete, so in studying interesting Alexandroff spaces we are leaving Hausdorff spaces far behind. Every $T_0$-Alexandroff space determines a partial ordering on the underlying set, and conversely given a partially ordered set one can define an Alexandroff topology. In this way one gets an equivalence between $T_0$-Alexandroff spaces and partially ordered sets. (More generally, there is an equivalence between Alexandroff spaces and quasi-ordered sets, i.e., with the axiom of anti-symmetry removed.)

d) There is up to homeomorphism a unique 2-point space in which exactly one of the points is closed. In general topology this is sometimes called the Sierpinski space. It is arguably the single most important topological space in arithmetic geometry, since it is the spectrum (in the sense of Zariski) of any discrete valuation ring. The germ of the idea of specialization in algebraic geometry is already contained in this two-point non-Hausdorff space!

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The main drive in mathematics is usually to look at some known object and to generalize the notion. Once you defined "open intervals" and saw that the union of them remains open (in the sense that each point has some interval around it) and that a finite intersection is also open, and so on...

Once you have, say, a list of properties that some space you know has, and you want things to look more or less like your known space (in this case $\mathbb{R}$). So you start looking for a "right" generalization. In topology it was found out that it is enough to request that the open sets have these three properties, namely - any union, and a finite intersection of open sets is still open, the empty set and the entire space are open as well. Once you have these properties you can talk about convergence and continuity.

How did we define convergence of sequences on the real line? $a_n \to a$ if and only if, for an interval around $a$ which is arbitrarily small we have only finitely many elements outside the interval. So we do the same with open sets. Eventually it was apparent that sequences are not always enough, and you need something stronger (namely nets). The main idea here, that you took something that well-defined to begin with and sought a way to generalize it so it would be both consistent and useful. And when groups, rings and things like function spaces (that is all the functions from some set $X$ to $\mathbb{R}$ or $\mathbb{C}$) are useful to you, in one way or another - you might want to define some general notion that would capture the essence in all the spaces that you encounter, granted that they have certain properties.

About continuity, it turns out that one can simply require that the preimage of an open set is an open set, and that is enough for a function to be continuous. Combine this with convergence and you have that by defining the open sets you already know a lot about the space at hand (with the given topology), because you know to say when something converges and when functions are continuous, and that is quite a lot.

To your question, you can define several different topologies on the real line. The standard one is generated by open intervals, while others might have different properties and would be generated by other kind of sets (e.g. sets of the form $[a,b)$) or maybe stated explicitly (e.g. "Every subset is open"), but different topologies might have different characteristics.

If you have two topologies, and you have a basis for each one, if you can show that each open set in one topology contains an open set from the other - you can say that the topologies are essentially the same, and would have the same properties. A good example is when you take $\mathbb{R}^2$ and you take the topology generated by open balls (that is given a point and a radius under the familiar Euclidean distance) and by squares, for example. These two families would generate the same topology. That is, both the topologies will have the same open sets and the same properties.

The main issue, when you approach a new space and try to define topology on it, is just one: You want the topology not to be "too loose" (i.e. not enough open sets, so nothing really converges, and nothing is really continuous) and not "too tight" (i.e. too many open sets, so not enough functions are continuous and not enough sequences/nets would converge.)
This is not always an easy task, although many times there is a natural way to define things, so not all is lost.

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As others have commented and answered, the answer is, roughly, "yes". I might add only that to see examples of 'other' topologies, ones in which open sets are not what you may be used to, check out (a) the problems in a point-set topology book or (b) the book Counterexamples in Topology.

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