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In the beginning chapters of real analysis we learn the algebraic properties of the real no.s (i.e., operations defined on $\mathbb{R}$, order properties and completeness properties of $\mathbb{R}$) as an algebraic structure imposed over $\mathbb{R}$. And then suddenly we are introduced to the topology of real no.s where we learn about neighbourhood, open set, limit points etc. But doesn't the algebraic properties of real no.s like the total order property, play any role in the topology of real no.s? I feel like there's an abrupt gap between the bridging of algebra and topology of $\mathbb{R}$ and I want to know if there's any source which can explain to me how topology of real no.s (or in that sense general topology) flows out naturally from it's algebraic structure.

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    $\begingroup$ Here are two properties of a field $F$ you want to have with respect to any "reasonable" topology: The product and the sum should define continuous maps $F\times F\to F$ and also the additive and multiplicative inversions should be continuous as well. $\endgroup$ – Moishe Kohan Jan 25 at 20:12
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In general, a mathematical object $\mathfrak{O}$ can have both algebraic and topological structure which are unrelated. However, in practice there are usually some connections. One of the standard ones is that the algebraic aspects of $\mathfrak{O}$ should be "topologically simple," e.g., all the relevant functions on $\mathfrak{O}$ should be continuous with respect to the relevant topology. See e.g. the notion of topological groups.

This happens with for example the fields of real and complex numbers: in each case there is a natural topology, and the field operations are continuous with respect to that topology (and the appropriate product topologies it generates).

In the case of $\mathbb{R}$ in fact there's an even tighter connection: addition and multiplication of real numbers alone determine the usual topology on $\mathbb{R}$. Specifically, the topology on $\mathbb{R}$ is the order topology coming from the usual ordering on $\mathbb{R}$, and that ordering is definable from addition and multiplication alone as $$x\le y\quad\iff\quad \mbox{ for some $z$ we have }x+(z\cdot z)=y.$$ But this is rather special to $\mathbb{R}$ - the usual topology on $\mathbb{C}$, for example, cannot be defined from addition and multiplication of complex numbers alone. (HINT: $\mathbb{C}$ has lots of discontinuous field automorphisms.)

Interestingly, the topology on $\mathbb{Q}$ is definable from the algebraic structure alone. Like $\mathbb{R}$ this goes by defining the order on $\mathbb{Q}$ from the algebraic operations, but things are much harder now since we can't take square roots. One way to do it (very briefly) is to first define $\mathbb{Z}$ in $\mathbb{Q}$, and building off that use the four squares theorem to define $\mathbb{Z}_{>0}$ in $\mathbb{Q}$. A rational is positive iff it is a ratio of two positive rationals, so this gets us $\mathbb{Q}_{>0}$, and this in turn gets us the ordering on $\mathbb{Q}$ since $x\le y$ iff $y-x$ is either $0$ or in $\mathbb{Q}_{>0}$. But there's a lot of work there (and there is probably a much simpler approach I'm missing here).


For a deeper connection between algebraic and topological structure, you may be interested in how topology can be applied to prove the fundamental theorem of algebra, which is of course a topology-free result. More generally, there is a whole subject of algebraic topology.

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If the algebraic structure is unrelated to the topological structure no, but if they are linked, then absolutely yes.

For example, a topological vector spaces become completely regular when they are only Hausdorff because the vector space structure with continuous operations makes it a uniform space.


This isn't quite what you asked for but I think it is also interesting: when you have a smooth manifold, the differentiable structure inherited by the cotangent bundle gives rise to a special algebraic operation on the algebra of smooth functions on the cotangent bundle: a Poisson bracket.

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