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?
Homeomorphism between two spaces
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1This 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. – 2011-10-26
1 Answers
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:
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 $B_{1,\alpha}\times\cdots\times B_{n-1,\alpha}\in \tau_A$.
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 $\beta$.
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0@HenningMakholm: In Lee: 'Introduction to Topological Manifolds' 2nd ed. the product topology for the finite product is given by the basis of open product sets. The box topology is only defined for infinite products – 2011-10-27