Topological Spaces: The Standard Topology on R^n

Today, we construct the standard topology, which gives us the way we usually think about R^n.

[angelmendez-rivera]

Actually, what gives the field of real numbers the geometric structure of a line with no gaps is not its topology, but its total ordering. The unique axiomatic characterization of the real numbers is that it is the totally ordered field such that it is a Dedekind complete lattice. A totally ordered field is densely ordered, and on top of being densely ordered, it is Dedekind complete, and both of these properties in combination make the field of real numbers a continuum. This is what gives the real numbers the structure of a line. The standard topology of the field of real numbers is its order topology, the topology induced by its total ordering, and since the field forms a continuum with respect to its ordering, this is what makes the order topology so desirable as the standard topology.

Also, the topology on R^n does not require assuming the Euclidean metric on R^n, it just requires using the product topology, assuming that you have already endowed R with its order topology. Essentially, you want the Cartesian products of the open sets in the standard topology of R to generate the topology in R^n.

[wuyqrbt]

I was waiting for it.
Thank you so much

[thepirateage]

very intuitively explained ! im keepin up with this series

π and 3 are definitely the same element... but on a more serious note, I love this series! amazing explanations, please make more videos on this in the future :)

[narfwhals]

Thank you for this. At 15:02 you say we take for granted that we know how to do basic arithmetic in R (which I think means we assume R is a vector space?) But doesn't defining addition and such already make some assumptions about the topology?
I don't quite know how to properly phase this rigorously but I _feel_ like to do standard math we assume that our set has no "holes".