I am currently reading in Hatcher's book at page 522 about the construction of open sets in a CW complex. They start with an arbitrary set $A \subset X$ and want to construct an open neighborhood $N_{\varepsilon}(A)$.

You might also look at page 4 in this reference top of page 4

The construction is inductive over the skeleta $X^n$. Hence, he constructs a sequence of $N_{\varepsilon}^n(A)$ sets, where $N_{\varepsilon}^n(A)$ is an open neighbourhood of $A \cap X^n$.

He starts by saying: Take $N_{\varepsilon}^0 = A \cap X^0$. I don't see why this should be open. This is an arbitrary set of discrete points, why should this be open?

I also don't see why this induction works: I mean, there does not need to be an intersection between $A$ and a particular $n+1$ cell, so how do you construct this open $\varepsilon_{\alpha}$ neighborhood. The only possible explanation to me is: If the intersection is trivial, then you are allowed to take any epsilon ball and otherwise you take an open covering of $A$ where each ball has a distance less than $\epsilon_{\alpha}$ away from $A$, correct?

And maybe you could comment on this spherical coordinates thing. What is a Cartesian product with respect to spherical coordinates? Does this mean that you the radius is taken between $(1-\varepsilon_{alpha},1]$ and the angle such that the second product factor is completely covered by all combination $(r,\theta)$. Maybe anybody here could add a few more details on this?

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    $\begingroup$ Well, step one is that the $N_{\epsilon}^{n}(A)$ are open in $X^{n}$, not necessarily in all of $X$. But when you get to the top dimension, you'll get an open set. $\endgroup$ – coolpapa Jun 1 '14 at 23:57
  • $\begingroup$ but why is $N_{\varepsilon}^0 $ open in $X^0$? $\endgroup$ – user66906 Jun 1 '14 at 23:58
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    $\begingroup$ Well, as you said, $X^{0}$ is a disjoint collection of points. So it carries the discrete topology - EVERY subset of it is open in $X^{0}$. $\endgroup$ – coolpapa Jun 1 '14 at 23:58
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    $\begingroup$ Your comment in the third paragraph seems on track - that's the point of choosing a neighborhood of $\phi^{-1}_{\alpha}(A)$ that misses the boundary - if $A$ intersects the interior of the cell, you need to choose an open set around it, but you can choose that pretty arbitrarily. Then the spherical coordinates are just expressing the idea that you want to choose an open set of the cell that includes the entire boundary. When gluing in a disc (2-cell), that describes a small annulus containing the boundary circle. For a 3-cell, that's...whatever a spherical annulus is. $\endgroup$ – coolpapa Jun 2 '14 at 0:13
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    $\begingroup$ Generally, spherical coordinates means that in dimension $n$, you can choose a point of the $n$-ball by choosing a point on the $n$-sphere and then choosing a distance $r$ from 0 to 1 (with, naturally, identification of all the points with $r=0$.) So that set is all the points in the $n$-ball that are a distance of at least $1-\epsilon_{\alpha}$ away from the center. $\endgroup$ – coolpapa Jun 2 '14 at 0:15

Spherical coordiantes are a way to express a point on the $n$-sphere via angular coordinates $$\phi, \theta_1,...,\theta_{n-1} \qquad \text{ where }\; 0\le\phi\le2\pi,\;\; 0\le\theta_i\le\pi$$ Then if you want to define in point in the $n$-ball, you have to add a radius $0\le r\le 1$. The topology on $D^n$ goes well with this description, in particular, there is a quotient map $[0,1]\times[0,2π]×[0,π]×...×[0,π]\to D^n$ $$ (r,\phi,\theta_1,\dots,\theta_{n-1}) \mapsto \pmatrix{ r \sin\theta_1 \dots \sin\theta_{n-1} \sin\phi \\ r \sin\theta_1 \dots \sin\theta_{n-1} \cos\phi \\ r \sin\theta_1 \dots \cos\theta_{n-1} \\ \vdots \\ r \sin\theta_1 \sin\theta_2 \cos\theta_3 \\ r \sin\theta_1 \cos\theta_2 \\ r \cos \theta_1 \\ } $$ so if your set $N^{n-1}_ε(A)$ is already open in the sphere $\partial D^n_\alpha$, then adding an open range of $r\in(1-ε_\alpha,\ 1]$ gives an open set in $D_\alpha^n$. Also see this wiki article.

In dimension zero any set in $X^0$ is open (and closed) as $X^0$ has the discrete topology.

If $\Phi^{-1}_\alpha(N^{n-1}_ε(A))$ is empty in $D^n_\alpha$, then you could simply take the empty set in that ball.

  • $\begingroup$ I don't think Hatcher idea was to use that complicated map you defined there. I think the right map is $r:[0,1]\times S^{n-1}\to D^n$ defined as $r(t ,s)=ts$ (someone actually posted this in the comments of this question). This also makes more sense since we are using a sphere ($S^{n-1}$) in my definition, where is the sphere in your map?. I also put a complete answer in my linked question, It'd be nice if you check it and tell me what you think. $\endgroup$ – Zero Oct 3 '15 at 14:02

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