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 $(\mathbb{R}, \tau_\Pi)$

$\mathbb{R}^\mathbb{N}$ you mean

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.

Yes sorry you are right

No, its closed

If $A_i\subseteq X_i$ are closed, then so is $\prod_{i\in I} A_i \subseteq \prod_{i\in I} X_i$

this is a general fact

@SillyGoose this only means its not open

right but $p_n(\mathbb{R}^N \backslash H_C) \neq \mathbb{R}$ for all $n \in \mathbb{N}$ as well

which would imply that $H_C$ is not closed, but i must be doing something incorrect

in fact, if $A_i$ are arbitrary then $\overline{\prod_i A_i} = \prod_i \overline{A_i}$

@SillyGoose thats not true

Complement of $H_C$ is the union of sets $U_n$ such that $x\in U_n$ when $x_n\notin [-1/n, 1/n]$

oh i see i have been taking the complement incorrectly

and as such, $p_n(\mathbb{R}^\mathbb{N}\setminus H_C) = \mathbb{R}$ for all $n$

hm okay so the constraint is on the vector $\vec{x}$, but the individual "coordinates" $x_n$ are actually free to take on any value in isolation

for instance, $\vec{x} = (0, 1, 1, 1, ...) \not\in H_C$ even though $x_1 \in [-1, 1]$.

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

hm i see okay i guess this way is better. i think the way i wanted to solve the problem actually just doesn't work at all

since $\Pi_{n \in \mathbb{N}} p_n(X) \neq X$ in general

The two are definitely very non-equal in general

bleb okay well this was a good problem

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.

@SoumikMukherjee Hi. Did you solve the exercise?

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

i see. i wonder why topology has been placed in limbo as the subject always half-taught in every course along a math curriculum :P

i have personal found it quite a nice language to know...but i guess other people disagree :P

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

Some people don't go outside of the world of sequences

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 $\delta.$

Is my reasoning correct?

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

@leslietownes thanks for the clarification

@Jakobian Hi, no not yet, I will tell you once I manage to solve it.

Thanks for asking:)

@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?