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 …

May 12, 2024 · Except for $0$ every element in this sequence has both a next and previous element. However, we have an infinite amount of elements between $0$ and $\omega$, which …

Why is the infinite sphere contractible? I know a proof from Hatcher p. 88, but I don't understand how this is possible. I really understand the statement and the proof, but in my imagination this...

Infinite decimals are introduced very loosely in secondary education and the subtleties are not always fully grasped until arriving at university. By the way, there is a group of very strict …

What do finite, infinite, countable, not countable, countably infinite mean? [duplicate] Ask Question Asked 13 years, 2 months ago Modified 13 years, 2 months ago

Dec 3, 2020 · However, while Dedekind-infinite implies your notion even without the Axiom of Choice, your definition does not imply Dedekind-infinite if we do not have the Axiom of Choice …

The dual space of an infinite-dimensional vector space is always strictly larger than the original space, so no to both questions. This was discussed on MO but I can't find the thread.

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Jun 6, 2020 · The reason being, especially in the non-standard analysis case, that "infinite number" is sort of awkward and can make people think about ∞ ∞ or infinite cardinals …

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.