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In my Algebraic Topology course we are studying CW complexes and at some point my teacher gave us a non-example I do not totally understand, let me explain.

Consider the space $X = \bigcup_n X_n$ where $X_n$ is the circle in $\mathbb R^2$ of radius $1/n$ centered at $(1/n, 0)$, which should look like something like this:

enter image description here

He told us this was not a CW complex but did not tell us why. Clearly it may be constructed by attaching a countable number of circles to a point. The only axiom in the definition of CW complex that is likely to be problematic in this case is (in my opinion)

A subset $A \subset X$ is closed in $X$ if and only if $A \cap X^n$ is closed in $X^n$ for all $n$.

But still I am unable to construct an closed set in $X$ that is not closed in $X^n$, I am not totally confortable with the topology on CW complexes. Could one of you help me with this ?

Falcon
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    If $X$ were a CW-complex under the subspace topology inherited from $\Bbb R^2$, then $A:=\left{\left(\frac2n,0\right):n\in \Bbb N\right}$ would be closed under the CW-complex topology, but $A$ is not closed subset of $X$ under subspace topology of $X$ inherited from $\Bbb R^2$. – Sumanta Aug 01 '22 at 10:06
  • There is another way of showing the same thing: At first show every CW-complex is semi-locally simply connected, hence has universal cover. Though Hawaiian earring not semilocally simply connected, see here https://math.stackexchange.com/questions/1104275/why-is-hawaiian-earring-not-semilocally-simply-connected – Sumanta Aug 01 '22 at 10:15
  • The "$X^n$" in the definition of the CW complex are the skeleta. These cannot be the same as the $X_n$ in the Hawaiian earring (they are not an increasing sequence). So, the first question you ought to contemplate (especially since you say you aren't fully comfortable) is what potential CW structure on $X$ you are envisioning (CW structures are not unique) and what its skeleta are.

    For an alternative way, you could also use the fact that CW complexes are locally contractible.

     – Thorgott Aug 01 '22 at 12:01
  • You can argue it by using the first homology of the space. Look at this https://math.stackexchange.com/questions/3860464/showing-hawaiian-earrings-are-not-cw-complexes – Yeipi Aug 01 '22 at 17:10
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    Also see https://math.stackexchange.com/q/3860464/349785 – Paul Frost Aug 02 '22 at 06:04

2 Answers2

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It's the right-to-left direction that causes the problem (a closed subset of $X$ is automatically a closed subset in any subspace of $X$ that happens to contain it). To see what goes wrong in the left-to-right direction, look at the subset $A = \{ (2/n, 0) : n = 1, 2, \ldots\}$. It isn't closed in $X$, because its limit point $(0, 0)$ is not in $A$, but its intersection with any $X^n$ is finite and hence closed.

Rob Arthan
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An extended comment: the Hawaiian earring is not obtained by attaching a countable number of circles to a point, which is the whole point of this construction; that space would be a CW complex. You can see the difference by examining neighborhoods of the wedge point. In the Hawaiian earring any neighborhood of the wedge point contains all but finitely many of the circles; this is not true for the wedge of countably many circles, where a neighborhood of the wedge point would just consist of a union of neighborhoods of the wedge point in each circle. (Roughly speaking the true wedge of countably many circles is not "shrinking"; each of the circles has equal status, which is not true here.)

Qiaochu Yuan
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