I wanted to better understand dfa. I wanted to build upon a previous question:
Creating a DFA that only accepts number of a's that are multiples of 3
But I wanted to go a bit further. Is there any way we can have a DFA that accepts number of a's that are multiples of 3 but does NOT have the sub...
Let $X$ be a measurable space and $Y$ a topological space. I am trying to show that if $f_n : X \to Y$ is measurable for each $n$, and the pointwise limit of $\{f_n\}$ exists, then $f(x) = \lim_{n \to \infty} f_n(x)$ is a measurable function. Let $V$ be some open set in $Y$. I was able to show th...
I was wondering If it is easier to factor in a non-ufd then it is to factor in a ufd.
I can come up with arguments for that , but I also have arguments in the opposite direction.
For instance : It should be easier to factor When there are more possibilities ( multiple factorizations in a non-ufd...
Consider a non-UFD that only has 2 units ( $-1,1$ ) and the min difference between 2 elements is $1$. Also there are only a finite amount of elements for any given fixed norm. ( Maybe that follows from the other 2 conditions ? )
I wonder about counting the irreducible elements bounded by a lower...
@Balarka so it seems the plan is to give something closer to a sketch of the quasifibration part of Dold-Thom and focus instead on the resulting homology bit
How would you make a regex for this? L = {w $\in$ {0, 1}* : w is 0-alternating}, where 0-alternating is either all the symbols in odd positions within w are 0's, or all the symbols in even positions within w are 0's, or both.
I want to construct a nfa from this, but I'm struggling with the regex part
A precise statement found via google: “The moduli space of holomorphic quadratic differentials is isomorphic to the cotangent bundle over the moduli space of complex structures.“
"o" <-- this isn't a circle, not because it's imperfect: even if I manage to arrange the atoms such that it's perfectly continuous and perfectly circular, it still isn't a circle
The definition I know of general polynomials involves the extension $F(x_1, \dots, x_n)$ over the field of symmetric rational functions over $F$ in $x_1, \dots x_n$
this will always have Galois group $S_n$, regardless of the field $F$
the general polynomial of degree $n$ is then just $(T-x_1) \dots (T-x_n)$
which has symmetric polynomials as coefficients
because the symmetric rational functions are those rational functions fixed by the action of $S_n$ permuting the variables, this has Galois group $S_n$