what is the $\pi_1$ of this space?: $X$ is a sphere with three disks removed (so three boundary components) and each boundary is identified with $S^1$ by degree $n,m,k$ maps
I think it should be like $\langle a,b,c\mid a^nb^m = c^k\rangle$
Something I'm confusing now: For $S^1\times I$, if I do identification only on one boundary component $S^1\times\{0\}$ with other space $X$, then still I can just deformation retract $S^1\times I/\sim$ to $S^1\times\{0\}$ but if I do identification on both boundary components, then I cannot just deformation retract as before right? @Thorgott
In particular, attaching two $2$-cells on $S^1$ by degree $n$ and $m$ maps is different from attaching two $2$-cells on $S^1\times I$ on each boundary by degree $n$ and $m$ maps are different
this is funny: I went to my probability professor's office to ask some questions and he was doing a weekly meeting with his student. One of his advice to his student was "don't take my class. it's a waste of time. Just google it if you forgot some concepts".
trying to determine if you fix a unit hypercube and inscribe spheres in them for each dimension if the sequence of the volumes of the spheres strictly increases, strictly decreases or stays constant
Revised problem: (Optimization) Consider a set of maximal (volume) codimension one hyper-surfaces of revolution $S_n$ with constant positive curvature and embeddings $e_n:S_n \hookrightarrow X^n$ for $X^n=[0,1]^n$ with conjugate points $p,q$ anchored on $\partial X^n$ where $\partial X^n=X^n-(0,1)^n$ for $\mathrm{dist}(p,q)=\sqrt{n}$. Is the sequence $\lbrace \mathrm{Vol}(S_n) \rbrace_{n\in \Bbb N}$ strictly increasing, strictly decreasing or constant for all $n$?
@JohnZimmerman yes, i saw some interviews of him. he says we look at our consciousness and study the question "what physical system could have this feature?" and he comes up with integrated information theory
i think he is hypothesizing a correlation between the integrated information and consciousness
In my lecture notes there are the following two sentences:
> The space $\mathbb{R}^{n}$ is a locally compact topological (abelian) group with respect to translation, which is a continuous operation. More generally, there exists a (left or right) translation-invariant measure, called Haar measure, on any locally compact topological group; this measure is unique up to a scalar factor.
When the author says $\mathbb{R}^n$ is a topological group with respect to translation, does this make sense? Maybe they mean addition, although translation is closely associated with that, but I wouldn't say they are the same thing, since translation is map from $\mathbb{R}^{n}\to\mathbb{R}^{n}$ whereas addition is a map from $\mathbb{R}^{n}\times \mathbb{R}^{n}\to\mathbb{R}^{n}$.
If someone could guide me with this I'll be very happy - I'm trying to prove, in a way that involves using the integral test the following inequality: for each $n \in \mathbb{N}$: $\sum_{k=1}^{n} \frac{1}{\sqrt{k}} \leq 2\sqrt{n} - 1$. However I struggle to bound it tightly enough
yet we also get two more generator from picking a path in the sphere part between the basepoints of the 3 circles that get identified
in particular, the existence of an open cover by two path-connected sets whose intersection has 3 components implies that $\pi_1$ needs to have a retract that is free on two generators
@Thorgott If you glue a cylinder to a circle by a degree $k$ map, the belt curve $a$ now winds $k$ times around the circle curve $d$ that it is attached to. So $a^k = d$.
You could prove it like this: $$\sum_{k=1}^n \frac{1}{\sqrt{k}} \leq 1+ \sum_{k=2}^n \int_{k-1}^k \frac{1}{\sqrt{x}}dx = 1+\int_1^n \frac{1}{\sqrt{x}}dx = 2\sqrt{n}-1$$
I think the result be something like $\langle\alpha,\beta,\gamma\vert\alpha^n=\beta\alpha^m\beta^{-1}\gamma\alpha^k\gamma^{-1}\rangle$
where $\alpha$ is the loop in the $S^1$, $\beta$ a path from the circle $a$ to $b$ that becomes a loop in the resulting quotient and $\gamma$ similarly a path from circle $a$ to $c$
@Jakobian that's the definition of a translation. You wouldn't say that's a binary operation, would you? The binary operation in my view here is addition. If that is defined, then we can speak of translation.
psie generally speaking if you are reading a textbook and your internal type checker starts beeping at something that can be resolved by interpreting a statement another way, you can just interpret the statement the other way (i.e. the author isn't trying to send you secret messages that you need to decode)
psie: i.e. while i don't know what maps an author might specifically have in mind when they say "translation" (note that a group operation can often be encoded in all sorts of equivalent ways), i think you're right that they mean the usual addition there
if you need separate affirmation that in general a map from R^n to itself is a different thing from a binary operation on R^n, i concur
right, when you say some space is a group with respect to some operation, then that operation would probably be the group operation, which is a binary operation, which translation isn't
@RyderRude it follows from yoneda lemma ... We follow the usual argument of inverting the colour spectrum ... That is red and yellow are interchanged .... Now can we reach a consensus different people will reach the consensus "some swapping has happened" ... No ... This is the classic philosopher conundrum .... What about if I invert the sound spectrum? Do we find the samd music theory? Nope ... Music theory changes and it is detectable ... Hence by yoneda lemma there is a mapping
Let $A = \mathbb{R}^\mathbb{N}$ with the product topology. Let $B = \{ (x_n)_n \in A \mid \forall n : |x_n| \leq 1\} = [-1, 1]^\mathbb{N}$ be the $\infty$-dimensional cube. Let $C = \{ (x_n)_n \in A \mid \sum_{n=1}^\infty |x_n| \leq 1 \}$ be the $\infty$-dimensional cross-polytope. Let $\operator...
a map S x S -> S being regarded as the same data as a map S -> (S -> S) pops up so often enough that many people don't even have a name for it, that might be what is going on in the example above
the real expository sin is immediately following a use of 'translation' to mean a specific operation on R^n, for purposes of explaining how R^n can be a topological group, with a use of 'translation' to mean 'whatever the operation is in a topological group.' if you aren't already familiar with the first example, you might not understand the second
i.e. the second use makes "X is a topological group under translation" sound sufficiently tautological to naive ears that they may not "get it" when you say, somewhat sloppily, that "R^n is a topological group under translation"
psie what is the personal email of whoever wrote these notes, i want to spend the rest of my day bothering them
@leslietownes Any infinite dimensional convex compact subset of $\ell^2$ is homeomorphic to the Hilbert cube. I suspect you can insert $\{x : \sum |x_i|\leq 1\}$ into $\ell^2$ as such subspace
at least, until the day comes when text editors all come with an AI assistant named Jakobian who points out when you're about to do something like what that author did
i'd really like to answer that question with another question, which is, what possessed them to call something an "infinite dimensional cross-polytope"
@Thorgott imagine living in a world where a map A -> (B -> C) does not canonically give rise to a map A x B -> C, but does so only upto contractible choice
(Though I think that $\mathrm{d}C$ looks better than either of the alternatives.)
In other news, the weather just cannot make up its c*rr**ng mind today. When I got in this morning, it was clear. An hour later, it was pounding down snow. 45 minutes after that, it was bright and sunny, and all the snow had melted. An hour ago, it was pounding down snow again. And now it is sunny again!
if you go into the "settings" menu you can change the speed at which in-game weather changes. i think you'd need to install a mod if you want to fully get snow out
Opinion: Is an unnumbered subsubsection appropriate to set up a lengthy paragraph's worth of notation before I state and prove a theorem? Or should I just manually write \textit{Notation.} blah blah? Or should I write nothing at all and just get on with the paragraph?
this seems like it might be an organizational issue whose optimal solution would go deeper than whether/how you label something, but "long paragraph immediately before the theorem explaining any notations" is a common enough way of handling this that a reader probably wouldn't be offended by that choice
As @leslietownes, this sounds like a deeper organizational / structural issue. Hard to give advice without seeing exactly what you are doing. And it all comes down to taste at the end of the day.
the deeper issue maybe being, if you really need the notation to understand what the theorem is saying (e.g. for purposes of applying it like a black box in some context outside of your paper), this might militate in favor of working the notation (however long it is) into the statement of the theorem itself, or into definitions that precede the theorem
if you feel like you're going to be making explicit reference to something like "the definitional material immediately preceding Theorem X" a lot, such that it might be handy to have a number, it probably does need something that makes it easier to refer to
but as far as quick fixes go, i'm happy as long as everything i need to make sense of a theorem is right before the theorem, if it isn't in numbered items of its own
There are two instances I had in mind when I asked the above. One of them has a short paragraph preceding the theorem with some standard discussion on notation. I have kept it as is, an isolated paragraph with a \medskip on the final line before it begins -- for better visibility of it as an isolated paragraph discussing some notation in the theorem that follows
i once handled this by putting, at the end of the statement of the theorem, "where here and elsewhere X and Y are defined as in (3.1) and (3.2)" where 3.1 and 3.2 were numbered equations expressing most but not all of definitions of X and Y given right before the theorem. in retrospect, the "and elsewhere" feels particularly slimy, but i had the benefit of knowing nobody would ever read what i wrote
When using LaTeX, you should focus on syntactic markup, and let the typesetting engine handle the precise spacing. That is the advantage of LaTeX over TeX.
@BalarkaSen Basically, I would prefer that you drown some children and eat some puppies rather than use \medskip in a LaTeX document. At least puppy-eating can be justified in some circumstances.
i dont want to things to look like this (space space) anyway, let us move on with the discussion and talk about Leslie Lamport's interesting work on LaTeX
hello. I am trying to prove that the Cantor set with subspace topology induced by the usual topology on $\mathbb{R}$ is not discrete. In particular, I am constructing a subsett of the Cantor set that is nonempty, not the whole thing, and is not open.
Let $C$ be the Cantor set, for which $0 \in C$ and let $d_2$ be the usual metric. so I have that the sequence $(1/3^n)_{n \in \mathbb{N}}$ trivially converges to $0$, as there are no elements of the convenient base $\{U = V \cap C \ \lvert \ V \in \tau_{d_2} \}$ containing $0$.
Hence, $\{0\} \cup \{1/3^n \ \lvert \ n \in \mathbb{N} \}$ is not closed. Hence, its complement is not open. Its complement is certainly not empty because the Cantor set is uncountably infinite and it is certainly not the whole Cantor set because the aforementioned set is nonempty.
If $x = \sum a_i/3^n$ is some element of Cantor set where $a_i\in \{0, 2\}$ and $U$ is an open set containing $x$, then we can simply keep terms up to some points the same, say $b_i = a_i$ for $1\leq i\leq N$, where $N$ is big enough, but let $b_i = 2-a_i$ for $i > N$
@SillyGoose While the argument that @Jakobian gives is correct, your comment indicates to me that you perhaps are not taking a course in topology, and are not expected to use those tools. What tools are you expected to use?
this is for a first course in undergrad topology, but my prof has a very analytic slant in the sense that I think he wants us to use convergence arguments and so on a lot
so we talk abt limits a lot but not limit points i guess :P
Right, so all you have to do is show that there exists a closed singleton, which is much easier than showing that there exists a non-trivial non-open set.
But that is immediate, since singletons are closed in the ambient topology, yes?
@SillyGoose They boil down to the same thing. My observation that the set of endpoints of removed intervals is countable and dense means that if you pick any point of the Cantor set that is not in that set, then you can find a sequence in the set which converges to that other point.
That is, take $F$ to be the set of endpoints of removed intervals, choose any other point $a$ of the Cantor set, and use density to get a sequence $F \ni x_n \to a$.
