Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

currently taking my first class in homotopy theory and have just seen the definition of cw complex for the first time. im still a bit confused about how the construction works when there are infinitely many cells of the same dimension. suppose $n$ is fixed and $B_\epsilon(a) \subset \mathbb{R}^n$ is the ball about $a$ of the radius $\epsilon$. can anyone explain how the space $\cup_k B_{2^{-k}}((2^{-k},0,\cdots,0))$ is formally constructed as a cw complex?

share|improve this question
Let $a_k = (2^{-k}, 0, \ldots, 0)$, then $|a_k - a_{k+1}| = 2^{-(k+1)}$. So if $|y-a_{k+1}|\le 2^{-(k+1)}$ then, $|y-a_k| \le |y-a_{k+1}| + |a_k - a_{k+1}| \le 2^{-(k+1)} + 2^{-(k+1)} = 2^{-k}$. In other words, your space is just $B_{2^{-k}}(a_k)$ for the smallest value of $k$ in the union, which is trivially a CW complex. Maybe you meant the sphere? In that case you have a generalized Hawaiian earring which is a classic example of a non-CW-complex. You can see the Hawaiian earring is not a CW complex by proving it is not locally contractible. –  Justin Young Apr 19 '12 at 12:13
add comment

1 Answer

The space you have described: $$\bigcup_kB^n_{2^{-k}}(2^{-k},0,\ldots,0)$$ assuming that $k$ indexes over $\Bbb{N}$, is simply $B^n_{\frac 12}(\frac 12,0,\ldots,0)$ because each subsequent element in the union is just a subset of the first. Since this is just $D^n$ it is obviously a CW complex.

I'll assume that you are referring to the sphere ,$S^{n-1}$ with center $2^{-k}$ and radius the same. The "usual" an equivalent way to describe this space is with radius and center $\frac 1n$ for $n\in \Bbb{N}$, rather than $2^{-k}$, so we will use the usual definition. We'll call this space in the $n$th dimension $X^n$. This is a far more interesting space than the open ball. This is not, however, a CW complex. For $n=2$ this is called the Havaiian earing and looks like this:


This is not the wedge sum of an infinite number of $S^{n-1}$, although at first it looks as if it is. We'll first look at $X^2$ which has been extensively studied. The wedge sum of a countable number of $S^1$, that is the CW constructed with one $0$-cell and a countably infinite number of $1$-cells attached to this cell, is $\bigvee_{\Bbb{N}}S^1$. This has a relatively simple fundamental group. The wedge point has a contractible neighborhood, so we can apply Van Kampen's theorem and get that $\pi_1(\bigvee S^1) $ is the free group with countably many generators. $X^2$ is far more bizarre. Every neighborhood of the intersection point at $(0,0)$ contains all but a finite number of the circles, and is not contractible. It turns out that the fundamental group of this space is actually uncountable. We see therefore that the difference between these spaces is the topologies they have. $X^n$ inherits its topology from $\Bbb{R}^n$ and therefore has no contractible neighborhood of the intersection point, while the wedge sum of $S^n$ does, although for $n>3$ the fundamental groups are the same and trivial.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.