Recent content by nuuskur

Oh it gets much worse than that. Consider the statement "there is a cardinal number strictly between ##|\mathbb N|## and ##|\mathbb R|##. ZFC alone can neither prove nor disprove this (Gödel & Cohen). A basis for that sequence space exists, is uncountable and good luck determining anything else...

I don't think it does, Gödel's theorem is about consistency and completeness, roughly stating that you can't have both if your theory is "rich" enough (ZF is rich enough). Regardless, if there was some way to demonstrably show in ZF alone that R over Q (or some other weird vector space) has no...

Yes, a statement is either true or false. The best we have in this instance is that it is consistent with ZF that R over Q does not have a basis. In similar vein, it is consistent with ZF that every subset of R is Lebesgue measurable (as a consequence of R being a countable union of countable...

If AC is false, then there's no guarantee this is one of those vector spaces without a basis. Of course it's clear, a generating subset has to be uncountable, because countable subsets can only ever span countable subsets over ##\mathbb Q##. But now you're in the uncountable territory, it's very...

The argument is if AC is false, then the statement "every vector space contains a minimal generating subset" is also false. Therefore, there exists (at least one) a vector space such that none of its linearly independent subsets are generating. .. but can we find an explicit example of one?

It is known that the axiom of choice (in its full power) is equivalent to the statement that any vector space has a basis. For finite dimensional spaces, we don't need choice to be able to produce a basis. Without choice, there is no hope for a basis in ##\mathbb R_\mathbb Q## for example. Even...

I missed a memo or something. E=mgh , ## E = mgh ##, E = mgh In some of my past posts the displaymode tex environment doesn't even include the tex brackets at all, they're replaced by some br /br thingamajig. As a result, the tex doesn't render. Is the ## still functional ? edit: and again...

The column of zeros would make the set of column vectors linearly dependent. But that does not imply the set of row vectors would be linearly dependent. There is no designation like vector x is linearly dependent on vectors y,z,w. A system of vectors is either linearly dependent or linearly...

This cannot be a definition, there will be only the trivial solution only if the rank of the system is equal to the number of variables. Undetermined? Otherwise correct, any subset that contains the zero is linearly dependent. Curious, which textbook that is. What does undetermined mean? Ah...

Specifically this part. What I understand is we consider the set of integer points bounded by the ellipse ## \frac{x^2}{R^2} + \frac{y^2}{\left(\frac{R}{\sqrt{2}} \right)^2} = 1 ## Are you saying that ##N(R)## is equivalent to ##cR^2 + \mbox{Error}(R)## as ##R\to\infty##? Why is that true and...

it always be floridians o0)

A point to consider is why the function is well defined, i.e, why does the minimum exist for ##x>0##. (If it doesn't then change minimum to infimum) Assuming the function is well defined and the limit exists, then the limit must be zero, because we'll just pick a sequence ##x_n =...

A textrobot compiles a list of data points and generates a (convex) combination of them based on your query. It is incapable of generating new data points. But humans (and other sentients) have this ability.

Turn the volume knob up gradually when holding the C key.