A bottom element is not an atom.

]]>I’m not so sure TVTropes is an appropriate link here (the previous edit made “trope namer” link there), so I’ve rolled back to #15.

]]>added link

Anonymous

]]>Add structure and a paragraph beginning ’This should be distinguished from barring the trivial object entirely.’.

]]>I don’t follow your conclusion. Yes, any space $X$ has a finite subset $\emptyset$, so the “unbiased” definition on the page says that for it to be path-connected there must exist a path in $X$ that passes through every point of $\emptyset$. But this is true as soon as $X$ is nonempty, since *any* path in $X$ (such as a constant path) vacuously passes through every point of $\emptyset$.

A space is path connected if for each (Kuratowski)-finite subset there is a path passing through every point of that subset.

Then the empty space is not path connected because it has no paths at all and hence no path through the empty subset.

This argument leads me to the conclusion that no space is path-connected, because every space has the empty set as

a finite subset. The traditional definition of a path-connected space is one that admits a path between any pair of its

elements. ]]>

Rephrased the field example; ’except’ may be taken to imply 0 is an exception, i.e. the full non-naïve definition is assumed already.

Anonymous

]]>Nothing wrong with Oscar’s definition of path-connected.

]]>The definition at integral domain allowed for the trivial ring, but I fixed that (and expanded some other stuff).

]]>I'm also inclined to take the zero ideal to be principal, for the reason that Mike gives. And I'm the most constructivist person in this conversation, I think. But constructivism isn't giving me any insights here. A principal ideal is still just an ideal that is generated by a singleton subset. (Constructivism also gives us a principal antiideal in a ring-with-apartness, as one anti-generated by a singleton subset, but that doesn't help any.)

]]>On prime numbers: I don't want to say that $0$ is a prime number. Saying that $1$ is not a prime number is a valid correction of the classical definition, which did not quite get at what it was trying to get at. But saying that $0$ is a prime number is going beyond that. (One way to see this is that general ring theory has a concept that generalizes that of prime number; it's just that this concept is called ’irreducible element’ rather than ’prime’.) I don't know why the meaning of the word ’prime’ changed in the transition from elementary number theory to ring theory, but it did.

]]>My initial inclination is also that the zero ideal is principal, because we generally talk about generators for ideals rather than “bases”. Sometimes the nullary cases are clarified by looking at constructive mathematics: what’s the right notion of “principal ideal” there?

]]>I’m conflicted on this one. You’re right that it’s generated by no elements. But it’s also generated by $0$ (there’s no law that says the generators have to be linearly independent). It would take a little rewriting over at ideal, which covers cases besides rings, to make it come out the way you want. (Maybe not much rewriting, but I’m not convinced it would be a good idea in the end.)

Somehow the argument about free (commutative) monoids doesn’t sway me. I think I’m happy referring to the monoid of nonzero principal ideals.

Just by-the-bye: one nice characterization of principal ideal domains is that they’re commutative rings where all ideals are free modules.

]]>I think I found another example: We should say that the zero ideal isn’t principal. This is because it’s generated by *no* elements of the ring rather than just one. This has the advantage of making the statement of unique factorization easier: “The monoid of principal ideals is isomorphic to a free monoid”. It might seem that it makes the definition of principal ideal ring a bit less nice, namely “the minimal number of generators of any ideal is 1 or less”, but this actually makes the definition analogous to that for a Noetherian ring “the minimal number of generators of any ideal is finite”.

(Rod, I haven’t forgotten your question, but I’d like to think more before replying. If Lawvere were here, he’d have a great deal to say, I’m sure.)

]]>Todd - don’t non 2 biased (finite) Boolean lattices, (“Oolean”), correspond to number divisor lattices?

$2^n$, a 2oolean, is the divisor lattice of any product of n different primes - $div( \prod_{i=1}^n p_i)$

$3^n$, a 3oolean, is the divisor lattice of any product of n different squared primes - $div( \prod_{i=1}^n p{_i}^2)$

$k^n$, a koolean, is the divisor lattice of any product of n different $k-1$ powers of primes - $div( \prod_{i=1}^n p{_i}^{k-1})$

$1^n = 1^m = div(1) = 1$ is not a special “too simple” case.

and a (potentially) mixed bias Oolean lattice is the divisor lattice of any arbitrary number. Does your approach also apply to the mixed bias case? Is $div(0)$ an exceptional case for the “too simple” number $0$?

EDITED: mainly to change some wrong “some”s to “any”s.

]]>Does it look okay now?

]]>Yes, of course. Do you still think there’s a problem?

]]>There is a tacit universal quantifier. To be pedantic, a subset S is linearly independent if, for each finite subset T of S and each $a \in K^{T}$, if $\sum_{v\in S'} a_v v = 0$, then every $a_v = 0$.

]]>Colin, I don’t see a problem. At linearly independent set it says

The subset $S$ is

$\sum_{i=1}^n a_i v_i = 0_V ;$linearly independentif, conversely, every $a_i = 0_K$ whenever the sumotherwise, $S$ is

linearly dependent.

(Here the $v_i$ are tacitly distinct elements of $S$.) From the statement, it is vacuously true that the empty set is a linearly independent set. For it is vacuously true that every scalar $a_i = 0$, since there are no scalars to speak of in an empty sum of scalar multiples on the left side of the displayed equation.

]]>