Counterexample for associativity of smash product 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.
 A: This is not a real answer since there is a gap, a claim I'm not currently able to prove, but it is too long for a comment and I hope somebody can fix the gap. Consider the topological property of being sequential. The idea is that $(\mathbb Q\wedge \mathbb Q)\wedge\mathbb N$ is sequential but $\mathbb Q\wedge (\mathbb Q\wedge\mathbb N)$ should not. 
First countable spaces, such as $\mathbb Q\times \mathbb Q$, are sequential. Sequential spaces are closed under quotients and disjoint unions, hence $\mathbb Q\wedge \mathbb Q$, $(\mathbb Q\wedge \mathbb Q)\times\mathbb N$ and $(\mathbb Q\wedge \mathbb Q)\wedge\mathbb N$ are sequential.
We can identify the underlying sets of $(\mathbb Q\wedge \mathbb Q)\wedge\mathbb N$ and $\mathbb Q\wedge (\mathbb Q\wedge\mathbb N)$ in the obvious way, so we're really speaking about different topologies on the same set. May and Sigurdsson 
show that the former has strictly more open sets than the latter. I think both of them have the same convergent sequences, but I haven't been able to prove it (the only problem is when the limit is the base point). If this is true, any open set in $(\mathbb Q\wedge \mathbb Q)\wedge\mathbb N$ would be sequentially open in $\mathbb Q\wedge (\mathbb Q\wedge\mathbb N)$, so the latter cannot be sequential, otherwise the two topologies would be the same.
A: 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.
