@SteamyRoot yes, a bijection from an empty set to an empty set is the same point for me from the beginning of the study. A subset of the cartesian product of nothing with nothing is nothing. I don't like "limit"-cases.
(The following comment may not make sense due to poor terminology) It is kinda interesting how when thinking about $0^0$ in the more global context of the function $z=x^y$, that the combinitorics convention works seemed quite tempting to suggest that all of combinitorics took place along the line $y=x$
A function $f$ from $A$ to $B$ is a subset of $A \times B$ such that $\forall a \in A: \exists b \in B: (a,b) \in f$ and $(a,b_1), (a,b_2) \in f \implies b_1 = b_2$
You have $0^1=0$, $1^1=1$, and $1^0=1$. Nothing about real numbers.
@Kirill if you define $x^y$ as the number of functions from $y$ to $x$, then indeed $0^0=1$ makes sense. But, this definition would only work for integers $x$ and $y$.
(cc @Secret ^ the motivation of defining $0^0=1$ in combinatorics)
@SteamyRoot yes, a bijection from an empty set to an empty set is the same point for me from the beginning of the study. A subset of the cartesian product of nothing with nothing is nothing. I don't like "limit"-cases.
Hmm, I need to think about how to take integer sequence limits. I can easily find a topology on $\Bbb{R}^2$ that will do the job for me, but I am not too familar to write that in notation $\lim_{(x,y) \in \Bbb{N}^2 \to (0,0)}x^y$...?
@LeakyNun So we cannot use topology to define a notion of continuity on the integer lattice such that we can take paths of the form {5,4,3,2,1,0} to approach a point of interest?
Basically, I want a "curve" that has values when the parameter is an integer, in addition to the smooth ones. But I am not very sure the machinary I need. I think they might be just sequences in $\Bbb{R}^2$
(Also I am talking about generic f(x,y) now, moved on from the x^y discussion)
@LeakyNun $\lim_{(x,y)\to (a,b)} f(x,y)$ is computed by taking all possible paths on $\Bbb{R}^2$ to $(a,b)$ and if the result is path independent, then the limit exists. I want those paths which are sequences on $\Bbb{R}^2$, but I am not sure how to write that out as the subscript notation in the limit sign
the real-life logic would be this: a subset of a box is formed by taking out some elements of the box. For example, from the box $\{1,2,3\}$, take out $1$ to form the subset $\{2,3\}$. Now, take everything out, and you have proved that the empty set is the subset of any set.
[Linearly ordered $\aleph_2$ set construction] Hmm, given the set of all indicator functions of the positive reals $\{0,1\}^{\Bbb{R}}$, it might be possible to construct a linear order by imposing a lexicographic ordering onto this set. But then because the domain is the reals, this means any attempt to define an open interval on this set is going to contain the rational indicator functions and since there are only countably many rationals so I expect some contradiction may happen, hmm...
That will depend on what position the $+1$ corresponds...
if the $+1$ is at an even position, then both will be the same, if the $+1$ is at an odd position then $\chi_{2\Bbb{Z}+1} > \chi_{\Bbb{Z}}$. Hmm, case 1 breaks totality in the definition of linear ordering, thus I can only have a partial ordering
Wait, totallity means $x \leq y$ or $y \leq x$, but I guess our problem here is the ambiguity of the set $2\Bbb{Z}+1$
even if we consider the set $\{0,1\}^{\Bbb{R}^+}$, the ambigurity remains
We might... be able to get around this ambiguirity by having $2\Bbb{Z}+1$ to denote the equivalence class of all sets of the form $2\Bbb{Z}+\{x\},x\in \Bbb{R}$ and thus the set which the indicator functions is 1 will need to be wrote out explicitly. Alternately, we can take $2\Bbb{Z}+1$ to mean $2\Bbb{Z}+\{1\}$, that will remove the ambiguity also
@LeakyNun O, sorry now's its more clear. In that case, $\chi_{2\Bbb{Z}}$ and $\chi_{2\Bbb{Z}+1}$ is still ambiguous (because there is no smallest element to begin the lexicographic ordering that can distinguish between them), we can take $\chi_{2\Bbb{Z}}=\chi_{2\Bbb{Z}+1}$ but then the problems is then we have uncountably many things that are equal with each other
Let $a,b$ be the elements of a commutative ring $R$ that contains $\mathbb{Q}$. Then $$F_a=\sum_{k=0}^{\infty}\binom{a}{k}x^k \in R[[x]] $$ is called the binomial series. It is $F_a \cdot F_b = F_{a+b}, F_{a}^{-1}=F_{-a}, F_{a}^{n}=F_{na}$ for $n \in \mathbb{N}$. @LeakyNun
@LeakyNun interesting is, why it is necessary that $\mathbb{Q}$ is in there.
@LeakyNun Hmm this gives {-ve stuff, 1,1/3,1/5,1/7,1/9,...} and {-ve stuff, 1/2,1/4,1/6,1/8,1/10}. Then clearly the former is bigger because the former has position 1 being nonzero, whereas for the latter all possitions that are nonzero are < 1
@LeakyNun In that case it is even easier, we then have {1,1/3,1/5,1/7,1/9,...} and {1/2,1/4,1/6,1/8,1/10,...}. Then the former is bigger since for the latter all positions with 1s are < 1
The lexicographical order is defined with the same ordering of $\Bbb{R}^+$ thus analogous to decimal expansions, if the positions (which are indiced by $\forall x,y \in \Bbb{R}^+$, then if $x < y$ and $\chi(x), \chi(y) = 1$ and $0$ otherwise ,then $\chi(x) < \chi(y)$
basically, the indicator functions allows a uncountable generalilsation of a binary sequence (at least that's what I think so far)
So, $\chi_{\dfrac1{2\Bbb N+1}}$ corresponds to the indicator function where $\chi(\{1,1/3,1/5,1/7,1/9,...\})=\{1,1,1,1,1,...\}$ and zero otherwise
hmmm... I think my set might be isomorphic to some subset of the surreals, since there is clearly a maximum, the indicator function $\chi : x \mapsto \{1\}$ (or in plot on a graph, $y=1$)
The set $\{0,1\}^{\Bbb{R}^+}$ contains elements $x$ of the form:
$$x = \prod_{i \in R^+} \chi_x x _i$$
Lexicographical ordering is imposed by: Let $\chi_y,\chi_x$ be the indicator distributions/functions which is nonzero at y an x respectively. Given for all $x,y \in \Bbb{R}^+$, and for all $\chi_y,\chi_x \in \{0,1\}^{\Bbb{R}^+}$ such that $\chi(x)=\chi(y)=1$, if $x \leq y$, then $\chi_x < \chi_y$
so e.g. $\chi_{\pi} < \chi_{4}$, $0<1$, $\chi_{\dfrac1{2\Bbb N+2}} < \chi_{\dfrac1{2\Bbb N+1}}$
$\chi_A$ because $12>11$ (and all the rest are easy to see.
Hmm, something's interesting, it seems only the maximal and minimal elements that are different and form the domain of the indicator function matters...
no, we need the smallest pair of $x \in B$ and $y \in A$ such that $y > x$ (and because the reals are in a linear order, that guarentee uniqueness, which means $y \not \in B$ and $x \not \in A$
suffice to say the answer to the ordering of a given $\chi_A, \chi_B$ is uncomputable without oracles if $A$ or $B$ are uncountable subsets of $\{0,1\}^{\Bbb{R}^+}$ and there exists no formula that uniquely enumerate all elements of the sets
but most transcendentals are uncomputable too, so that should not be a big problem
Does there exist a totally ordered set $S$ with cardinality greater than that of the real numbers? Sequences are continuous functions with domain $\mathbb{N}$ and paths are continuous functions with domain $\mathbb{R}$; both of them are very important "traversals" of the points inside a space. Wo...
smallest pair is guarenteed since the reals satisfy the upper bound property hence an infimum always exists
Actually, if the lexicographical order relies only on infimums, then the set could be isomorphic to some subset of the surreals since we can come up with sets $A$ and $B$ which has the same max/min hence inf/sup, and there exists one subset within that shares the same inf/sup. There will then be uncountably many such $\chi$ that are equivalent to each other thus making them effectively infintesimal elements when an addition is defined by e.g. $\chi_A + \chi_B = \chi_{A \cup B}$
I might try to construct a proof of today's results later, now to finish that python program
If successful, then I have some rather intuitive $\aleph_2$ dense linear orders to play with
that's how the ordering is determined. The worse case scenario is e.g. $\chi_{[0,...,2]}$ and $\chi_{[0,...,2]'}$ where all subsets share the same inf/sup and where in the absence of a formulae, the only thing we can write is if $\chi_{[0,...,2]} > \chi_{[0,...,2]'}$, then $\exists x \in [0,...,2], y \in [0,...,2]'$ s.t. $x > y$
So things like addition, sine, exponentiation etcetera, although not all of them are written as a word followed by argument-filled brackets, are functions.
And they can be reversed if you give all but one argument etcetera.
buuuuuuuuuut...
What about set theory's operators? The 1-input complement operator seems to work just like... say, sin(x)
but a set can't contain sets
So my curiosity asks what kind of "function" intersection, complement etcetera is
when if a function is the relationship between multiple sets to make a function that deals with sets you would need to have the sets in a set!! so that's where i'm curious
But I am trying to construct explicitly $\aleph_{\geq 2}$dense linear order when GCH holds. That MSE link said it is possible. The above discussion then suggest $\{0,1\}^{\Bbb{R}^+}$ under the lexicographical order can be a good candidate
also interestingly en.wikipedia.org/wiki/Function_(mathematics) says that a function is a [definition of a functions] with the property that each input is related to exactly one output... but i always thought that radicals like the square root of... say, 16, had two answers: 4 and -4. and radical is a function (base and... thing under the line comes out with exponent.) What is there to know about multiple-answer mathematical expressions?
i like to frolic amongst the numbers, bro. i'm learning this 'cause i find it pretty. i'll probably be asking a lot of questions so i'd not talk to me if you've got anything better to do :P
and aha oho!
I understand
also
is the inverse of a function a function or some kind of extension of the function its inversing, if you capice
I cannot guarentee I can answer all questions. The only part of maths where I had sufficient strength is linear algebra, anything else I might need other mather's help
@Danu This looks like a suspicious question. You mean if $M$ and $N$ are equidimensional homotopy equivalent manifolds, does $M$ being spin also mean $N$ is spin?
$M$ being spin means the map $M \to BSO$ classifying the tangent bundle of $M$ lifts to $M \to BSpin$ if I recall correctly. That should not have anything much to do with homotopy equivalence, because if you pull that map back by a homotopy equivalence $N \to M$ that need not classify the tangent bundle of $N$.
Homotopy equivalences do not have anything at all to do with the tangent bundle of the corresponding manifolds. So my money is on no.
It is well known that Stiefel-Whitney classes are homotopy invariant for closed smooth manifolds. But in the case of open manifolds even $w_1$ is not a homotopy invariant (take just open cylinder and open Mobius strip).
Therefore, the following question naturally arises.
Conjecture 1. Suppose t...
Is the set $S = \{ f : \mathbb R \to \mathbb R \}$ of functions $f$ from the reals to the reals a totally ordered set ?
I think the question is clear. Note that there is a bijection to $g(x)$ gives $0$ or $1$ with that choice depending on the real $x$. The cardinality of this set is $P(R)$ wher...
but then that raises a question, stuff from axiom of choice is not supposed to be constructible, so what on earth is that I just talked about a few hours ago?
Right, so the issue is that once any subset is allowed to have a sup/inf that does not coincide with a min/max, and when the topology is not discrete, then in general we cannot assign $>$ or $<$ on uncountably many of them. If these elements are considered equal under the ordering then it becomes a partial order
Now to see how the lower cardinals escape from that, this will give us more idea on how the high cardinal sets are quite different from the lower ones (at least under their natural topologies)
Finite base systems: {0,1,2,...,n} (clearly no issue in assigning lexicographical ordering)
Countable systems: $\Bbb{N}$ Discrete topology 0,1,2,3,4,... Each element is uniquely defined since they have finite support (in particular, $\chi_x = \delta(y-x), \forall y \in \Bbb{N}$
$\Bbb{Z}$, discrete topology Basically two copies of $\Bbb{N}$, thus ok
$\Bbb{Q}$, subspace topology of reals
$\frac{n}{m}, n,m \in \Bbb{Z}$
This is fine since each element is not defined by a sequence, but by two countable sets, hence countable
It is well known that Stiefel-Whitney classes are homotopy invariant for closed smooth manifolds. But in the case of open manifolds even $w_1$ is not a homotopy invariant (take just open cylinder and open Mobius strip).
Therefore, the following question naturally arises.
Conjecture 1. Suppose t...
Is the set $S = \{ f : \mathbb R \to \mathbb R \}$ of functions $f$ from the reals to the reals a totally ordered set ?
I think the question is clear. Note that there is a bijection to $g(x)$ gives $0$ or $1$ with that choice depending on the real $x$. The cardinality of this set is $P(R)$ wher...
