In Section 1.7 of [Parametrized Homotopy Theory](http://www.math.uchicago.edu/~may/EXTHEORY/MaySig.pdf) 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:



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> 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 $\mathrm{Top}\_\*$?
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I had asked the same question on [math.SE](https://math.stackexchange.com/questions/755070/smash-products-of-pointed-spaces-is-really-not-associative).



Counterexample for associativity of smash product
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.


