Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am asked to show that $(X_{1}\times X_{2}\times \cdots\times X_{n-1})\times X_{n}$ is homeomorphic to $X_{1}\times X_{2}\times \cdots \times X_{n}$. My guess is that the Identity map would work but I am not quite sure. I am also wondering if I could treat the the set $(X_{1}\times X_{2}\times \cdots\times X_{n-1})\times X_{n}$ as the product of two sets $X_{1}\times X_{2}\times \cdots\times X_{n-1}$ and $X_{n}$ so that I could use the projection maps but again I am not sure exactly how to go about this. Can anyone help me?

share|improve this question
1  
The identity map is a homeomorphism iff the topologies are the same. Just check that the topology you define in both ways is the same. –  Ilya Oct 26 '11 at 15:54
    
@Gortaur. Yes you are right about the identity map. So are you then claiming that the topologies are the same? How do I show this if that were the case? –  johnny Oct 26 '11 at 15:58
1  
You cannot use the identity map here. The underlying sets are different. –  Asaf Karagila Oct 26 '11 at 16:09
1  
$(X_1\times\cdots\times X_{n-1})\times X_n$ is a set of things shaped like $((x_1,\ldots,x_{n-1}),x_n)$, whereas $X_1\times\cdots\times X_n$ is a set of things shaped like $(x_1,\ldots,x_n)$. So they are not formally the same thing (and so are not connected by an identity map), but it's trivial to write down the natural correspondence that is just as good. What you need to show is that this is a homeomorphism -- just work through the elements of the relevant definitions one by one. –  Henning Makholm Oct 26 '11 at 16:09
1  
This sort of problem is all about the definitions, so we need to know how $X_1 \times ... \times X_{n}$ is defined in the context of this problem. –  Thomas Andrews Oct 26 '11 at 16:24
show 2 more comments

1 Answer

up vote 1 down vote accepted

Let us denote $A = X_1\times \cdots \times X_{n-1}$ and $X = X_{1}\times \cdots\times X_{n-1}\times X_n$. The box topology $\tau_A$ on $A$ is defined by the basis of open product sets: $$ \mathcal B(A) = \{B_1\times\cdots \times B_{n-1}:B_i \text{ is open in } X_i,1\leq i\leq n-1\}. $$ The box topology $\tau_X$ on $X$ is defined by the basis: $$ \mathcal B(X) = \{B_1\times\cdots\times B_{n}:B_i \text{ is open in } X_i,1\leq i\leq n\}. $$ Let us follow Henning and put $f:A\times X_n\to X$ as $$f((x_1,\ldots,x_{n-1}),x_n) = (x_1,\ldots,x_n)$$ so $$ f^{-1}(x_1,\ldots,x_n) = ((x_1,\ldots,x_{n-1}),x_n). $$ Clearly, it is a bijection. Then we should check that $B\in\tau'$ iff $B\in \tau_X$.

Let us check it:

  1. if $B\in\tau_X$ then $$ f^{-1}(B) = \bigcup\limits_{\alpha}(B_{1,\alpha}\times\cdots\times B_{n-1,\alpha})\times B_{n,\alpha}\in \tau'$$ since $B_{1,\alpha}\times\cdots\times B_{n-1,\alpha}\in \tau_A$.

  2. if $B\in \tau'$ then $$ B = \bigcup\limits_\alpha C_\alpha \times B_{n,\alpha} $$ where $C_\alpha \in \tau(A)$. But we know the basis for the latter topology, so $$ C_\alpha = \bigcup\limits_\beta C_{1,\alpha,\beta}\times\cdots\times C_{n-1,\alpha,\beta} $$ where $C_{i,\alpha,\beta}$ are open in $X_i$, here $1\leq i\leq n-1$. Finally we substitute these expressions and get $$ f(B) = \bigcup\limits_{\alpha}B_{1,\alpha}\times\cdots\times B_{n-1,\alpha}\times B_{n,\alpha}\in \tau_X $$ where we denote $$ B_{i,\alpha} = \bigcup\limits_{\beta}C_{i,\alpha,\beta}\text{ - open in }X_i. $$ Note that we also implicitly interchanged unions w.r.t. $\alpha$ and $\beta$.

share|improve this answer
    
I think that is a genius's way of solving this question. Thank you. –  johnny Oct 26 '11 at 16:20
    
Isn't that the box topology rather than the product topology? That could be a problem if we don't already know that they are the same for finite products. –  Henning Makholm Oct 26 '11 at 16:21
    
Thanks, but I think now that's the wrong way of solving it. To be correct, I should consider $B$ and $f(B)$ where $f$ is a map @Henning has provided. –  Ilya Oct 26 '11 at 16:22
    
@HenningMakholm: you're right, I assumed it to be known. Fixed it, as well as your name mentioned in my answer with a typo. –  Ilya Oct 26 '11 at 16:24
    
@HenningMakholm. I do not think it makes any difference. Because for finte products the two topologies are the same. And by the way which map have you provided? –  johnny Oct 26 '11 at 16:30
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.