Shrinking wedge of circles is not a CW complex 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:

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 ?
 A: 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 $A = \{ (2/n, 0)\}$. 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.
A: 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.)
