TITLE: Counterexample for associativity of smash product
QUESTION [15 upvotes]: In Section 1.7 of Parametrized Homotopy Theory by May and Sigurdsson it is stated that the smash product of pointed topological spaces is not associative (which is just another hint that $\mathrm{Top}$ is the "wrong category"). Specifically, they claim that $\mathbb{N} \wedge (\mathbb{Q} \wedge \mathbb{Q})$ is not isomorphic to $(\mathbb{N} \wedge \mathbb{Q}) \wedge \mathbb{Q}$ in $\mathrm{Top}_*$. But actually  they just prove that the canonical bijection $\mathbb{N} \wedge (\mathbb{Q} \wedge \mathbb{Q}) \to (\mathbb{N} \wedge \mathbb{Q}) \wedge \mathbb{Q}$ is not an isomorphism. Equivalently, there is no isomorphism in the slice category $(\mathbb{N} \times \mathbb{Q} \times \mathbb{Q}) / \mathrm{Top}_*$. Therefore, my question is as follows:

How to prove that there is no isomorphism between $\mathbb{N} \wedge (\mathbb{Q} \wedge \mathbb{Q})$ and $(\mathbb{N} \wedge \mathbb{Q}) \wedge \mathbb{Q}$ in $\mathrm{Top}_*$?

I had asked the same question on math.SE.

REPLY [9 votes]: Building on Fernando's answer, here is a proof that they are not homeomorphic.  By Fernando's answer, it suffices to show that if a sequence $(x_k,y_k,n_k)$ converges to the basepoint in $\mathbb{Q}\wedge(\mathbb{Q}\wedge\mathbb{N})$, it also converges to the basepoint in $(\mathbb{Q}\wedge\mathbb{Q})\wedge\mathbb{N}$.  Let $(x_k,y_k,n_k)$ be such a sequence.  We may assume that the points $(x_k,y_k,n_k)$ are all distinct from the basepoint.  I claim $\{n_k\}\subseteq\mathbb{N}$ must be a finite set; the result then follows easily.
Suppose for a contradiction that $\{n_k\}$ is infinite.  Passing to a subsequence, we may assume that for each $n\in\mathbb{N}$, there are only finitely many $k$ such that $n_k=n$.  For definiteness of notation, let us say that $0$ is our chosen basepoint in both $\mathbb{Q}$ and $\mathbb{N}$.  For each $n>0$, let $\epsilon_n>0$ be such that $\epsilon_n<|x_k|$ and $\epsilon_n<|y_k|$ for all $k$ such that $n_k=n$.  Now let $$U=\{(x,y,n):|x|<\epsilon_n\text{ or }|y|<\epsilon_n\}\subset \mathbb{Q}\wedge(\mathbb{Q}\wedge\mathbb{N}).$$
Then $U$ contains the basepoint and is disjoint from our sequence.  Furthermore, the pullback of $U$ to $\mathbb{Q}\times(\mathbb{Q}\wedge\mathbb{N})$ is open: it is the union of the open sets $$V=\mathbb{Q}\times\{(y,n):|y|<\epsilon_n\}$$ and $$W_n=\{x:|x|<\epsilon_n\}\times \{(y,n):y\neq0\},$$
the latter being a separate open set for each fixed $n>0$.  Thus $U$ is open, contradicting the assumption that our sequence converges to the basepoint.