i am looking at the Hilbert cube $\Pi_{n \in \mathbb{N} -\frac{1}{n}, \frac{1}{n} \subseteq \mathbb{R}$ w.r.t. the topological space $(\mathbb{R}, \tau_\Pi)$ (using the product topology)
sorry, I am looking at the Hilbert cube $H_C = \Pi_{n \in \mathbb{N}} [-\frac{1}{n}, \frac{1}{n}] \subseteq \mathbb{R}$ w.r.t. the topological space $(\mathbb{R}, \tau_\Pi)$
I am given that this subset should be closed. However, I am getting that it seems to be neither closed nor open. I make the observation that $p_n(H_C) \neq \mathbb{R}$ for all $n \in \mathbb{N}$ where $p_n$ projects onto the $n$th product factor. Hence, no union of $U \in B_\Pi$ can construct this set.
If $x\in \prod_i \overline{A}_i$, and $V = \{y : y_{i_j}\in U_{i_j}, j = 1, ..., n\}$ is a neighbourhood of $x$, then taking $y_{i_j}\in A_{i_j}\cap U_{i_j}$ and $y_i$ for other $i$ to be any element of $A_i$ (assuming $A_i$ is non-empty), then $y\in V\cap \prod_i A_i$
and this shows that $x\in \overline{\prod_i A_i}$
The other direction is easier, since the set $B_j = \{y : y_j\in \overline{A_j}\}$ is a closed set containing $\prod_i A_i$
(since complement of $B_j$ is open by definiton)
but $\bigcap_j B_j = \prod_i \overline{A_i}$
hence $\overline{\prod_i A_i}\subseteq \prod_i \overline{A_i}$ from definition of the closure
I'm not sure why authors choose to use $\prod_n [-1/n, 1/n]$ instead of $[0, 1]^\mathbb{N}$, since the two are homeomorphic
this only really makes sense if you were to consider $H_C$ as subset of $\ell^2$, in which case you need this modification for it to be homeomorphic to the Hilbert cube
i think we are meant to prove this later on we are just getting into continuous functions at this point in the course :P -- it seems a little bit late, but I think at my uni there are several related topology courses and this one is not supposed to focus on homeomorphisms or something like that
but this is "the first course in topology" course
well i do not know what a "usual" intro topology course is like. but this course seems unusually focussed towards building up to functional analysis and ignoring the "classification" sort of mathematics, i.e., defining a notion of isomorphism and then talking about invariants of classes of spaces.
silly yeah i dunno if there is a 'usual' intro to topology class. it is frequently taught as an adjunct to analysis (some departments will not even have regularly taught undergrad-level topology classes, and teach it if at all within an analysis class). it is also often taught without analysis (or indeed much of anything else) as a prerequisite, which severely restricts what a first course can cover in limited time.
someone mentioned armstrong earlier, if i recall correctly it gets roughly as far as defining the fundamental group (which does indeed distinguish at least some spaces, although it is not super useful in general as an invariant), and the classification of surfaces (which could easily be "the" focus of the course, but some people who teach out of the book skip it entirely for lack of time).
so, it isn't surprising if your course offering is kind of 'weird,' it occupies a somewhat unsettled place in many curricula
there are some tough tradeoffs involved. roughly, assume nothing and maybe not get very far (and bore students who 'know' the subject from prior exposure in other classes), or actually use a bunch of prerequisites, and maybe not have a lot of people take the class :D
departments that have grad programs often deal with this by just having sufficiently motivated undergrads take a graduate level class, and not offer a specifically undergrad version
and yeah at least in the US a typical math department might have many/most of its majors be interested in k-12 math education, where just getting sequences down well enough to understand "calculus with proofs" is often the ceiling of what you can expect people to care about
Let $f:A\to \Bbb R$ where $A\subseteq \Bbb R$. We say that, $f$ is uniformly continuous on $A$ if for any $\epsilon\gt 0$ there exists $\delta(\epsilon)=\delta\gt 0$ such that for any $x_1,x_2\in A$ and satisfying $|x_2-x_1|\lt \delta$ we have, $|f(x_2)-f(x_1)|\lt\epsilon.$
Now, my question is: Say, for a particular $\epsilon_0\gt 0$ there exists a $\delta\gt 0$ such that for any $x_1,x_2\in A$ and satisfying $|x_2-x_1|\lt \delta$ we have, $|f(x_2)-f(x_1)|\lt\epsilon_0.$ But, what if, no two distinct points in the domain of $f$ say, $A$ has a distance of $\delta$ or, in other words, what if every pair of distinct points in $A$ has a distance strictly greater than $\delta$ ? Will $f$ be still uniformly continuous?
My answer is "yes". This is because, the definition of uniform continuity says, that if any two points say, $x_1,x_2$ have the distance between them the required $\delta$ or even less than $\delta$ (, for some choice of $\epsilon$) then $|f(x_2)-f(x_1)|\lt \epsilon$ must hold, but NEVER in the definition of uniform continuity it assumes that there must exist two distinct points in the domain of the function ,$A$ such that the distance between them is at most the required $\delta.$
thomas: yes, if A has that property (there is d > 0 with the property that any pair of distinct points in A is at least d apart), then any function from A to R will be uniformly continuous on A. as you point out (or equivalently to what you point out) the definition of uniform continuity does not require anything of f in this case, other than that it be a function from A to R
Both traveler's Alice ($A$) and Bill ($B$) start at node 0. There are a total of six nodes or places to go. The chords represent alternative routes but to take a chord is the same thing as taking the two series hexagon sides the chord is parallel with. Time passes at the same rate for both trav...
@BalarkaSen That is actually not my question, that is my friend's question (he's in differential geometry, more specifically Chern-Simons theory) lol
well I don't know much about fiber bundles (except when I studied higher homotopy group in Hatcher several years ago) so I don't have much to say.
But I recently studied a bit of flat plane or circle bundles and their relation to euler classes so.... But btw, Euler class completely classifies the isomorphism type of the circle bundle over a manifold in particular a surface. Is there anything similar in the case of 2-orbifolds so that it covers Seifert fibered spaces?
apparently it's somewhat standard to use arrows only in handwriting and boldface for printing.
I definitely don't like typing \vec{} all the time. Also, it should be clear from the context what's a vector and what isn't. If we're using $\times$ then obviously they're vectors. It also doesn't matter for inner product so yea
@Obliv In a context where some things are scalars and some things are vectors (or whatever), I would prefer for the notation to distinguish between them. Either by changing the symbol (making it bold, putting a line over it, whatever), or (and this is my preference) just by choice of names, e.g. $f,g,h$ for vectors, $\alpha, \beta, \gamma$ for scalars.
the point is that every such bundle can be "linearized", i.e. is the sphere bundle of a complex line bundle and the latter are classified by their Euler classes (=first Chern class)
I know that \ell gives me this nice mathematical "calligraphic" letter l, which I want to use throughout my document. However, it is also a common typographic convention that vectors are typeset in boldface as opposed to having a little arrow on top of them.
Hence: How do I get a boldface \ell?...
Idk what I'm doing. I got a circle in the x-y plane concentric on the z-axis, I want a vector from a point on the circle to a point on the z-axis. I'm given $R$ is the radius of the circle. So clearly I have $\sqrt{z^2+R^2} = ||\mathbf{r}||$ right
I'd honestly prefer polar coordinates since I'm going to be taking an integral from $0$ to $2\pi$ around the circle anyway
Or, oh jeez! Did you lose your toes in some kind of accident? If so, I'm sorry for being insensitive. Do you have a friend who has toes? Maybe you can borrow theirs?
koro: degree is 'multiplicative', isn't it? so might you not get something like det(that matrix) [maybe up to sign] when computing degree, based on considering more specific examples of maps of that form
Note that the $2$-torus $T^2$ can be seen as a quotient space $\Bbb R^2/\Bbb Z^2$ of $\Bbb R^2$. Then any $2\times 2$ integer matrix $A=(\begin{smallmatrix}
a & b\\
c & d
\end{smallmatrix})$ gives a well-defined map $A:T^2\to T^2$. On the other hand, we have $H_1(T^2)=\Bbb Z^2$ and $H_2(T^2)=...
all I'm doing is shortcutting by relying on the observation that a degree $d$ covering map of compact manifolds induces multiplication by $\pm d$ on (co-)homology, sign depending on orientation
Given a map $f:T^2 \rightarrow T^2$, the degree of $f$ is given by the induced homomorphism $f^*:H_2(T^2) \rightarrow H_2(T^2)$.
If I know the induced homomorphism $f^{**}:H_1(T^2) \rightarrow H_1(T^2)$ is $f(x+y) = mx + ny$ (since $H_1(T^2) = \mathbb{Z} + \mathbb{Z}$) can I calculate the degree...
I'm trying to get my head around covering maps.
I've seen a question that asks to show there are covering maps $p$ from the torus $T$ onto $T$ with degree equal to any positive integer.
I'm not sure how to show this. I know there are covering maps from $S^1$ to $S^1$ with degree equal to any ...
koro: so with just using multiplicativity of degree and something like smith normal form on the matrices, it feels like at least if you ignore sign, you will get the determinant, as long as you can check that this works for diagonal A with one entry equal to 1, e.g. the maps (z,w) -> (z^n, w) for some n
math being a seamless web, it should be possible to work your way from worked out answers that use stuff you "can't use" to things that you can
koro if you look at the comment on that second answer, the answerer is also talking about local degrees and preimages of points. this is not a completely different world just because it says "universal cover"
this is what I was getting at: just because you're trying to solve homework that tells you to "not use <x>" doesn't mean you should ignore every instance of "<x>" in trying to understand the problem
doing so is just bad pedagogy
if you understand that the map is a normal degree $d$ cover, then it becomes clear how to do the "by hand" proof: any value is a regular value with exactly $d$ preimages and the local orientation at any of these is the same cause the deck transformation group acts transitively on the fibers, so the degree of the map is $\pm d$
you can also make this very explicit if you follow leslie's suggestion and first bring the matrix into upper-triangular form
@Thorgott I thought Koro said that the rule to "use only things in the lecture" is an unspoken one, so its not like its clear that you can't use them, you probably can, why not? If the lecturer is understanding
"the" is used to point forward to a following qualifying or defining clause or phrase, or denoting one or more people or things already mentioned or assumed to be common knowledge. @thorgott yw
@Thorgott sadly, I've never studied deck transformations :(
here is what I was thinking prior to this: locally a torus is a square. Assume that (p,p), where p=(1,0) is regular value. So we can apply local degree theorem here. Take a chart (t,s)|--->(exp(it), exp(is)) Calculations would give restriction on a,b,c,d. More precisely, a^2+b^2=1, c^2+d^2=1. So Jac determinant of the composition chart \circ f\circ chart inverse is +1 or -1 depending upon the cases. So degree (f) should be 1 or -1 depending upong a,b,c,d where a,b,c,d lie in {-1,0,1}. but this is wrong because a,b,c,d are not restricted.
The context in which I'm asking is if you have two smooth manifolds $M\times N$ and you pushforward a vector $(v,w)\in TM\oplus TN$ in the tangent space of the product space
The context is also physics so forgive my notation/precision or lack of
koro: what do you mean by 'calculations would give restriction on a,b,c,d. More precisely, a^2 + b^2 = 1, c^2 + d^2 = 1.' how do these constraints arise?
Let's try to find the inverse image of the point (p,p) where p =(1,0) is on S^1.We have a exp(it)+ b exp(is) =p, c exp(it)+ d exp(is)= p. This gives a cos t +b cos s =1, a sint +b sin s =0 hence a^2+b^2 =1.
But this is not correct because a,b,c,d are not restricted.
it's incorrect. Even with this, I should not get a^2+b^2=1. There is an error in it.
I'll put this question on hold till I do a revision of Poincare duality. I forgot what it was.
I've felt that there's a sort of distinction of two ways of doing math, but haven't seen it be discussed anywhere. The first way is like you find a problem you find interesting, and then you develop tools to try to solve it. The second is you look at what tools you have, and ask yourself which problems they can solve, and go after solving them.
I'm reading the following paragraph in a handout on ordinals and transfinite induction:
> Let $S$ be an arbitrary set. Then it may be indexed as $$S=\{s_\alpha:\alpha\in A\}$$ where $A$ is an ordinal. Moreover, we may assume $A$ is minimal among all ordinals of cardinality $|A|$; otherwise we may simply re-index suitably. (Note that the ordinals $\alpha\in A$ such that $|\alpha|\leq A+1$ gives a non-empty subset of $A$; so by well-ordering, it has a least element. This least element can replace $A$ as an index set for $S$.)
What confuses me is that the author has used the notation $|A|\leq|B|$ to mean there exists an injection from $A$ to $B$. In the paragraph above, the author writes $|\alpha|\leq A+1$. What does this mean? Moreover, why does it give a non-empty subset? I do not really understand why we can simply replace $A$ by doing what is suggested.
Anyway it should be easier than whatever the author is doing, the set of ordinals that are in bijection with $S$ is nonempty, so it has a least element
And the bijection from that least element to $S$ is exactly the indexing we're looking for
What you said is true, but I think one wants to take ordinals $\alpha$ in which $S$ injects into, since we don't know if there exists a set bijective to an ordinal a priori
@Jakobian depending on how we phrased the well-ordering principle
@AlessandroCodenotti but that gives us a surjection $A\to S$ so we need to invoke axiom of choice to find an injection $S\to A$, and then find least such ordinal.
and then prove that this ordinal is in bijection with $S$. Seems like its still a lot of work
why do something exist? Outside of the fact that I exist in the first place, how is it possible that I can see, I can observe, that I am here
I suspect this is a problem of a question that can't be answered from within itself
> Philosopher Stephen Law has said the question may not need answering, as it is attempting to answer a question that is outside a spatio-temporal setting, from within a spatio-temporal setting.
Just like how ZFC cannot prove that its consistent, we cannot give an argument for why we exist
@AlessandroCodenotti ok, I agree. It's taken me a while to grok this, but the least element of $\{ \alpha \in A : |\alpha| \leq A + 1 \}=A$ would indeed just be $0$, which is the empty set, so this can't be right (though this is how it's written in the notes). Could you elaborate why the set of ordinals that are in bijection with $S$ is non-empty?
yeah. your question is an extreme edge case. but still, no answer to questions like this. i think we can only answer questions about the universe in terms of other things about the universe that we know. so only these questions have meaning
maybe we are just capable of asking meaningless questions...
these questions certainly seem out of the domain of what questions are supposed to be
i bought the Cartesian argument that the very fact that im questioning it would mean i would have to exist, because i would be the one doing the questioning
so for "me being uncertain" to be well defined, "me" needs to exist
Both traveler's Alice ($A$) and Bill ($B$) start at node 0. There are a total of six nodes or places to go. The chords represent alternative routes but to take a chord is the same thing as taking the two series hexagon sides the chord is parallel with. Time passes at the same rate for both trav...
Note that the $2$-torus $T^2$ can be seen as a quotient space $\Bbb R^2/\Bbb Z^2$ of $\Bbb R^2$. Then any $2\times 2$ integer matrix $A=(\begin{smallmatrix}
a & b\\
c & d
\end{smallmatrix})$ gives a well-defined map $A:T^2\to T^2$. On the other hand, we have $H_1(T^2)=\Bbb Z^2$ and $H_2(T^2)=...
I'm trying to get my head around covering maps.
I've seen a question that asks to show there are covering maps $p$ from the torus $T$ onto $T$ with degree equal to any positive integer.
I'm not sure how to show this. I know there are covering maps from $S^1$ to $S^1$ with degree equal to any ...
