Timeline for Counterexample for associativity of smash product
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 9, 2023 at 19:32 | comment | added | FShrike | @EricWofsey Oh, in my though process I was blindly assuming this "contain an entire rectangle" thing should hold true (neighbourhood of $x_0\times Y$ was getting jumbled with sets of form $U\times Y$, but as you say this isn't true in general). Well, that clears everything up, thank you. I apologise for the elementary mistake | |
| Feb 9, 2023 at 19:19 | comment | added | Eric Wofsey | On the other hand, this is true if you replace $\mathbb{N}$ by a finite discrete space $F$ (or more generally, a compact space), since a neighborhood of the basepoint in $X\wedge F$ must contain an entire rectangle $U\times F$ where $U\subseteq X$ is a neighborhood of the basepoint. | |
| Feb 9, 2023 at 19:11 | comment | added | Eric Wofsey | @FShrike: It is not true that $(x_k,y_k,n_k)$ converges to the basepoint in $(\mathbb{Q}\wedge\mathbb{Q})\wedge\mathbb{N}$ iff $(x_k,y_k)$ converges to the basepoint. Indeed, my answer shows that $(x_k,y_k,n_k)$ cannot converge to the basepoint unless $n_k$ takes only finitely many different values. | |
| Feb 9, 2023 at 17:26 | comment | added | FShrike | So the sequence must also converge in $(\Bbb Q\wedge\Bbb Q)\wedge\Bbb N$. I feel like this is simpler. Would you mind elaborating how you intended to use the fact that $\{n_k:k\}$ is finite? | |
| Feb 9, 2023 at 17:24 | comment | added | FShrike | Let every $[x_k,y_k,n_k]$ be distinct from the basepoint. This converges in $(\Bbb Q\wedge\Bbb Q)\wedge\Bbb N$ iff. $(x_k,y_k)$ converges to the basepoint in $\Bbb Q\wedge\Bbb Q$. If that is not true, then it must be that for some $\epsilon>0$, "$|x_k|<\epsilon$ or $|y_k|<\epsilon$ is eventually true" is false. The convergence in $\Bbb Q\wedge(\Bbb Q\wedge\Bbb N)$ then entails that $(y_k,n_k)$ converges to the basepoint in $\Bbb Q\wedge\Bbb N$. Since $\Bbb Q\times\{0\}$ is open, it is clear that $y_k\to0$ has to hold in $\Bbb Q$. But that is a contradiction. (cont.) | |
| Feb 12, 2015 at 23:31 | vote | accept | Martin Brandenburg | ||
| Feb 12, 2015 at 14:39 | comment | added | Eric Wofsey | The complement of the basepoint has the same topology in both spaces, which is just the subspace topology from $\mathbb{Q}\times\mathbb{Q}\times\mathbb{N}$. I'm not sure what you mean by your second question. | |
| Feb 12, 2015 at 10:10 | comment | added | Martin Brandenburg | Thank you. Why does it suffice to look at sequences converging to the base point? What about $(0,y,n)$ for instance? | |
| Feb 10, 2015 at 3:30 | history | answered | Eric Wofsey | CC BY-SA 3.0 |