For many infinite-dimensional vector spaces of interest we don't care about describing a basis anyway; they often come with a topology and we can therefore get a lot out of studying dense subspaces, …

6 Show that if a $\sigma$-algebra is infinite, that it contains a countably infinite collection of disjoint subsets. An immediate consequence is that the $\sigma$-algebra is uncountable.

Oct 3, 2015 · A pedantic point: is a complex number with a 0 imaginary part the same as a real number?

The reason for this has to to with power series, because the Taylor series is a power series, as well as our approximations. See, if we were to carry out our approximation over and over (in infinite amount …

Sep 10, 2019 · Your further question in the comments, whether a vector space over $\mathbb {Q}$ is finite dimensional if and only if the set of vectors is countable, has a negative answer. If the vector …

Apr 24, 2013 · What might be interesting to notice (although it might be a little too advanced, if this is your first course in topology) that the family of sets in your example is locally finite. Union of a locally …

You are right to be suspicious. We usually define an infinite sum by taking the limit of the partial sums. So $$1+2+3+4+5+\dots $$ would be what we get as the limit of the partial sums $$1$$ $$1+2$$ …

I am looking for a small treatment of basic theorems about infinite products ; surprisingly enough they are nowhere to be found after googling a little. The reason for this is that I am beginning t...

57 Why are box topology and product topology different on infinite products of topological spaces ? I'm reading Munkres's topology. He mentioned that fact but I can't see why it's true that they are different …

Just out of curiosity, I was wondering if anybody knows any methods (other than the integral test) of proving the infinite series where the nth term is given by $\frac {1} {n^2}$ converges.