Exercise 19.4

Proof. First, we note that since we are dealing with finite products, the box and product topologies are the same; we shall find it most convenient to use the box topology definition. Also, as there are no intervals involved here, we use the traditional tuple notation using parentheses. So define $f:X_1\times \cdots \times X_n\to (X_1\times \cdots \times X_{n-1})\times X_n$ by

It is obvious that this is a bijection, and it is trivial to prove. Also obvious and trivial to prove based on the definition of $f$ is that $f(A_1\times \cdots \times A_n)=(A_1\times \cdots \times A_{n-1})\times A_n$ when each $A_k\subset X_k$.

First, we show that $f$ is continuous by showing that the inverse image of every basis element in $(X_1\times \cdots \times X_{n-1})\times X_n$ is open in $X_1\times \cdots \times X_n$. So consider any basis element $C$ of $(X_1\times \cdots \times X_{n-1})\times X_n$ and let $U=f^{-1}(C)$ so that of course $f(U)=C$ and $U\subset X_1\times \cdots \times X_n$. We then have that $C=V^{\prime }\times V_n$ where $V^{\prime }$ is open in $X_1\times \cdots \times X_{n-1}$ and $V_n$ is open in $X_n$ by the definition of the box/product topology. Now consider any $x\in U$ so that $x=\left(x_1,\dots ,x_n\right)$ and we have that $f(x)=(\left(x_1,\dots ,x_{n-1}\right),x_n)\in f(U)=C$. Hence $x^{\prime }=\left(x_1,\dots ,x_{n-1}\right)\in V^{\prime }$ and $x_n\in V_n$. Since $V^{\prime }$ is open in $X_1\times \cdots \times X_{n-1}$ there is a basis element $C^{\prime }$ containing $x^{\prime }$ that is a subset of $V^{\prime }$. By the definition of the box topology, we then have that $C^{\prime }=V_1\times \cdots \times V_{n-1}$ where each $V_k$ is open in $X_k$.

We then have that $B=V_1\times \cdots \times V_n$ is a basis element of $X_1\times \cdots \times X_n$ and also clearly $B$ contains $x$ since $\left(x_1,\dots ,x_{n-1}\right)=x^{\prime }\in C^{\prime }=V_1\times \cdots \times V_{n-1}$ and $x_n\in V_n$. Now suppose that $y=\left(y_1,\dots ,y_n\right)\in B$ so that each $y_k\in V_k$. Then we have that $y^{\prime }=\left(y_1,\dots ,y_{n-1}\right)\in C^{\prime }$ so that also $y^{\prime }\in V^{\prime }$ since $C^{\prime }\subset V^{\prime }$. Since also of course $y_n\in V_n$, we have that $(y^{\prime },y_n)\in V^{\prime }\times V_n=C$. Also clearly $f(y)=(y^{\prime },y_n)\in C=f(U)$ so that $y\in U$. Since $y$ was arbitrary this shows that $B\subset U$, which suffices to show that $U$ is open since $x$ was arbitrary. This completes the proof that $f$ is continuous.

Next, we show that $f^{-1}$ is continuous, which is a little simpler. Let $B$ be any basis element of $X_1\times \cdots \times X_n$ so that $B=U_1\times \cdots \times U_n$ where each $U_k$ is open in $X_k$ by the definition of the box topology. Then we have that $f(B)=(U_1\times \cdots \times U_{n-1})\times U_n$. By the definition of the box topology, we then have that $U^{\prime }=U_1\times \cdots \times U_{n-1}$ is a basis element of $X_1\times \cdots \times X_{n-1}$ and is therefore open. Since $U_n$ is also open, we have that $f(B)=U^{\prime }\times U_n$ is a basis element of $(X_1\times \cdots \times X_{n-1})\times X_n$ by the definition of the box/product topology, and is therefore open. Since $f(B)={(f^{-1})}^{-1}(B)$ is the inverse image of $B$ under $f^{-1}$, this shows that $f^{-1}$ is also continuous.

We have shown that both $f$ and $f^{-1}$ are continuous, which proves that $f$ is a homeomorphism by definition. □