hey chat, if someone of you knows a bit homogeneous space and want to earn a bounty, please give a look at this question, it would help me alot https://math.stackexchange.com/questions/3234925/understanding-projection-of-vector-field-in-homgeneous-spaces
@anakhro: Generally lack of responsiveness, lack of interaction. With too small a class, people are quieter, in general.
Hmm...
I want to create a mathematical object that behave somewhat like this
However, it is not clear what context needs to be abstractised to produce said object
A "void lecture" ,assuming Ted does capture most of the gist, is one where you transmit information into it, and getting only total silence as a response
Thus one way to abstract this is the zero map, which is trivial
What if, you can produce "a zero with intricate structure", abstracted from how in most practical case, what seemed silent is full of implied information in nonverbal form
and said information may be recovered given the correct tools
One of the easiest nontrivial way to imbue a structure on an element that otherwise behave like zero is to have a semigroup where there is a subsemilattice. Then you have elements that behave like an identity except for a few elements.
Likewise, nth order absorbers are also straightforward to implement
So, an "absorber" is defined an element $a$ of a magma $M$ satisfying that $am=ma=a$ for all $m\in M$. Obviously, you can only have one absorber per operation on the object, but is there a concept of a "second order absorber?" Ie an element $b$ satisfying that $bm=mb=b$ for all $m\neq a$?
In addition, suppose we have a strange zero function $\mathscr{o}$ such that:
$$\int_a^b \mathscr{o} dx > 0$$
but $\mathscr{o} + f = f$ for some functions $f$, then we will often end up with divergence issues e.g.:
$$\int_a^b f dx = \int_a^b \mathscr{o} + \mathscr{o} + \cdots + f dx$$
Hey folks, how do you remember that $f(A\cap B) \subseteq f(A)\cap f(B)$ is the only operation where in general we don't have equality? I always know there's one thing prohibiting the image from being a lattice homomorphism, but I keep forgetting which one.
@Luke: one way to think of the equality is as a generalization of the definition of injective functions. Indeed, the definition of injective functions is literally the same as $f(\{a\}\cap\{b\})=f(\{a\})\cap f(\{b\})$.
@user170039 The problem is not „understanding it on a deeper level“, I just needed a minimal example or mnemonic so I get the correct fact with minimal effort / distraction when proving other things.
Three girls $G_1,G_2,G_3$ and three boys $B_1,B_2,B_3$ are made to sit in a row randomly. The probability that at least two girls are together is ....
My try: Probability of no girl together: B_B_B here in dashed places girls be seated so probability is $\dfrac{4*3*2*3*2}{6!}=\frac15$ and hence probability of at least two girls are together is $1-\frac15=\frac45$. Am I correct?
Hey everyone, I am going over some information theory stuff, and I came across the sentence "Let $\mu_n$ and $\mu'_n$ be two probability distributions on the state space of a Markov chain at time $n$, and let $\mu_{n + 1}$ and $\mu'_{n + 1}$ be the corresponding distributions at time $n + 1$". I am kind of confused at what this means exactly. How can you have two different distributions on the same markov chain?
@Ultradark Maybe they aren't meant to be stationary, and that is what I am misunderstanding. I think what i am confused by is this: given a distribution on the state space of a Markov Chain, how exactly does that relate to the actual probability of an event in the Markov Chain?
Baire's theorem: Let $X$ be a complete metric space then for every countable collection of nowhere dense sets $A_n$ then $X\setminus \bigcup A_n$ is dense in X
obviously $(-\infty,0] \cup [1,\infty )$ is not dense in $\mathbb{R}$
Assume by contradiction $(-\infty,0] \cup [1,\infty)$ is first category. That is, there exists $(A_n)_{n=1}^\infty$ nowhere dense sets such that $(-\infty,0] \cup [1,\infty) = \bigcup_{n\in \mathbb{N}} A_n $, By Baire's category theorem we get that $\mathbb{R} \setminus (-\infty,0] \cup [1,\infty)$ is dense in $\mathbb{R}$, but (0,1) is not dense in R
In topology, the long line (or Alexandroff line) is a topological space somewhat similar to the real line, but in a certain way "longer". It behaves locally just like the real line, but has different large-scale properties (e.g., it is neither Lindelöf nor separable). Therefore, it serves as one of the basic counterexamples of topology. Intuitively, the usual real-number line consists of a countable number of line segments [0, 1) laid end-to-end, whereas the long line is constructed from an uncountable number of such segments.
== Definition ==
The closed long ray L is defined as the cartesian...
Why would gluing together more than ω1 copies of [0,1) make the space no longer locally homeomorphic to R?
@AlessandroCodenotti Let $X,Z$ be metric spaces such that $X$ is complete, let $f:X \rightarrow Z$ s.t the set of continuity points of $f$ is dense in $X$, How can I prove that the set of discontinuities is of first category?
Let's look at $\omega_2\times [0,1)$ for concreteness
With the order topology induced by the lexicographic order
That is $(\alpha,x)<(\beta,y)$ iff $\alpha<\beta$ or $\alpha=\beta$ and $x<y$, ok?
Where $\alpha$ and $\beta$ are compared with the usual order on ordinals and $x,y$ with the usual order on $\Bbb R$
And a basis of the topology is given by the intervals $(a,b)=\{x\mid a<x<b\}$ for $a,b\in\omega_2\times[0,1)$
Now let's look at the point $(\omega_1,0)\in\omega_2\times[0,1)$. We want to find an open set containing it, so in particular it must contain an interval around this point
In particular this interval must contain some point smaller than $(\omega_1,0)$, but those must be of the form $(\alpha,x)$ for some $\alpha<\omega_1$ and $x\in[0,1)$ (They can't be of the form $(\omega_1,x)$ because $0$ is the minimum of $[0,1)$)
But there's uncountably many ordinals between $\alpha$ and $\omega_1$ for every $\alpha<\omega_1$, so in particular every nbhd of $(\omega_1,0)$ must contain uncountably many copies of $[0,1)$, hence it cannot be locally euclidean