What does a topology do, and what makes a particular topology the 'right' one?

What does a topology do, and what makes a particular topology the 'right' one?

A topology lets you prove certain things; it's the right one if it helps you with your present problem; it's the natural one if it helps you 99 times out of 100.

To illustrate this, I'll give an example of a problem you'd never think is solvable with topology, but it absolutely is. This is Fürstenberg's proof there are infinitely many primes (with some steps skipped you can try filling in yourself):

We can verify there is a topology on $\Bbb Z$ whose open sets are the unions of two-sided arithmetic progressions in $\Bbb Z$. In this topology, each open set is also closed, and is either infinite or empty. If there are finitely many primes, the set of integers not divisible by any prime number is clopen, and hence infinite or empty. But this set is $\{-1,\,1\}$, giving a contradiction.

This example inspired this question; all its answers are worth reading for other "creative" topology examples. (My own answer there discusses another example that proves any Jordan curve passes through some rectangle's vertices.) But that question is about unexpected uses of topology. The proofs' authors always picked the right topology, but typically not a natural one.

Topologies are intimately related to metrics, norms etc., the choice of which may hint at what is natural. For example, our options on $\Bbb Q$ are limited, and the Euclidean norm gives birth to $\Bbb R$ and $\Bbb C$ in a way that non-trivial alternatives give us the $p$-adics, which continue to have limited intuition and utility.

Let me start with the sentence

The same set can have different topologies. For instance, the real line, the complex plane, and the Cantor set can be thought of as the same set with different topologies.

This is totally correct and, at the same time, useless. To make sense of this statement, one observes that all the three sets have the same cardinality. Hence, one takes, for instance, a bijection $f: {\mathbb R}\to {\mathbb C}$ and takes the image under $f$ of the standard topology ${\mathcal T}_1$ on ${\mathbb R}$. The result is a topology ${\mathcal T}_2$ on ${\mathbb C}$ such that $({\mathbb C},{\mathcal T}_2)$ is homeomorphic to $({\mathbb R},{\mathcal T}_1)$. However, given the fact that our map $f$ was an arbitrary bijection, we did not gain any insight into topological structure of either one of these spaces.

Next, to the

The Euclidean topology is the natural topology for ${\mathbb R}^n$.

the answer is "it depends". First of all you have to decide what ${\mathbb R}^n$ is. I assume this means the $n$-dimensional real vector space (with a particular choice of a basis). If you are doing, say, analysis or PDEs or differential geometry or geometric topology on this space then indeed, the Euclidean topology is the natural choice. For instance, from the analysis viewpoint, this topology is natural since it agrees with the notion of limits, continuity etc. that one uses in analysis on ${\mathbb R}^n$. In fact, (real) analysis was one of the original sources of topology which arouse out of needs of several branches of mathematics and analysis (real and complex) was one of these.

At the same time, if you are doing (real) algebraic geometry, you quickly realize that besides the Euclidean topology, there is another natural topology on ${\mathbb R}^n$, namely, the Zariski topology (which is non-Hausdorff as long as $n>0$). This topology is strictly weaker than the Euclidean topology but captures solutions of algebraic equations better than the Euclidean topology. The same (even more so) applies to ${\mathbb C}^n$. Moreover, in algebraic geometry one frequently uses both topologies and compares the answers. Furthermore, the Zariski topology on ${\mathbb C}^n$ has a mild generalization, called the etale topology (which is, strictly speaking not a topology on on ${\mathbb C}^n$) which is also quite useful. So, from the algebraic viewpoint, say, Zariski topology is more natural than the Euclidean topology.

One way to think of this is that Zariski closed sets are zero sets of polynomial (vector) functions while subsets of ${\mathbb C}^n$ closed in the classical topology are zero level sets of smooth functions.

Dealing with topology such as the Zariski topology (on general algebraic varieties/schemes) forces one to reconsider the standard topological notions one is used to in analysis, such as compactness and connectivity. One replaces them with appropriate "algebraic" counterparts (completeness and irreducibility).

Thirdly,

What, if anything, makes a specific topology the right one for a given context - that is, given an arbitrary set, and perhaps some additional information, which topology best describes that set, and why?

does not have a universal answer. As you learn more mathematics, you will discover this from reading and trying to solve problems. If your math specialty is, say, PDEs, then you use Euclidean topology on ${\mathbb R}^n$ and "whatever works" on various functional spaces. For instance, you try to gain some form of compactness (if possible), which requires "fewer" open sets and continuity of some functionals (which requires more open sets). There is a clear tension between the two, which makes things interesting.

Edit. A classical example (going back to the origins of topology) is the Dirichlet problem. The idea (due to Riemann) is to solve the problem by minimizing a suitable (nonlinear!) functional (the energy) on a certain space of functions. Thus, the topology one looks for is the one for which the energy functional is continuous and its sublevel sets are compact. Riemann's original expectation was that even in this infinite-dimensional setting, a continuous functional on a closed bounded subset will attain its minimum (the "Dirichlet principle"). Hilbert observed that this is utterly wrong in full generality, but can be made work in the special setting of the Dirichlet problem by choosing carefully the functional space and its topology.

One more example, this time from algebra: Topology on groups. In general, in algebra it is a good idea to use topology “consistent with the algebraic structure as much as possible”. For instance, for the group $G=SL(n, {\mathbb C})$ the Euclidean (also called “classical”) topology will equip $G$ with the structure of a topological group (multiplication $G\times G\to G$ and the inversion $G\to G$ are continuous) and even of a Lie group (the group operations are differentiable with respect to the natural smooth manifold structure). However, in many cases, it is also useful to equip the same group $G$ with the Zariski topology. And with this topology, $G$ is only a paratopological group: The multiplication $G\times G\to G$ is continuous with respect to each “variable” but is not jointly continuous. But looking at the situation a bit closer, the source of the problem is that we are using the "wrong topology" on the product group $G\times G$: Instead of the product topology one should use the Zariski topology on this product.

In this example you see how algebra (both group theory and algebraic geometry) governs the choice of topology (on both $G$ and $G\times G$).

The group $G$ with Zariski topology has much fewer closed subgroups which makes it possible to analyze them. Furthermore, they are frequently “nicer” than closed subgroups in the classical topology. The example that I like is taking the Zariski closure of an infinite subgroup $\Gamma$ which is discrete (and, hence, closed) in the classical topology. The resulting subgroup of $G$ is again a Lie group but now it has only finitely many connected components (in the classical topology!). Hence, this Zariski closure can be further analyzed by means of the Lie theory (say, looking at its Lie algebra). So, in this example, we go from Euclidean topology to Zariski topology and then back to Euclidean topology.