# Homeomorphism in product and box topology

Given sequences $(a_1,a_2,...)$ and $(b_1,b_2,...)$ of real numbers with $a_i\gt0$ for all $i$

Define $h:\Bbb R^\omega\to\Bbb R^\omega$ by the equation $h((x_1,x_2,...))=(a_1x_1+b_1,a_2x_2+b_2,...)$

Now if $\Bbb R^\omega$ is given by the product topology the it is fairly east to show that $h$ is a homeomorphism

I am testing whether $h$ is homeomorphic if $\Bbb R^\omega$ is given in box topology

My Try:

Given fuction can be broken down into 2 functions $h_1$ and $h_2$ defined as

$h_1((x_1,x_2,...))=(a_1x_1,a_2x_2,...)$

$h_2((x_1,x_2,...))=(x_1+b_1,x_2+b_2,...)$

Now the original function $h$ is a composition of these functions i.e. $h=h_2\circ h_1$

Clearly both $h_1$ and $h_2$ are one-one and onto.

Now we show $h_1$ is continuous.

Consider the coordinate maps $h_{1_i}:\Bbb R\to \Bbb R$ for each $i\in \Bbb Z^+$ defined by $h_{1_i}(x)=a_ix$

Its clear that $h_{1_i}$ is continuous for each $i$

Now for each $i$ consider the open set $V_i$ then $h_{1_i}^{-1}(V_i)$ is open

$\prod_{i=1}^{\infty}V_i$ is a typical basis element for box topology

Also $h_1^{-1}(\prod_{i=1}^{\infty}V_i)=\prod_{i=1}^{\infty}h_{1_i}^{-1}(V_i)$

where the RHS is a basis element of box-topology hence it is open

Therefore $h_1$ is continuous, Similarly continuity can be established for $h_1^{-1},h_2,h_2^{-1}$

Thus $h_1,h_2$ are homoemorphic and hence their composition namely $h$ is also homeomorphic.

Have I gone wrong somewhere?

In the question they have taken $a_i>0$ for all $i$. If $a_i=0$ then there will be a problem in the inverse map. But what if $a_i<0$, I guess even in this case the the above solution (if at all it is correct) must hold. So can we relax the condition on $a_i$'s to just being nonzero?

The argument is fine, and yes, it works so long as every $a_n$ is non-zero; they need not be positive. In fact, we have the following general result.
Proposition. Let $X=\prod_{n\in\Bbb N}X_n$ have the box topology. Suppose that $h_n:X_n\to X_n$ is a homeomorphism for each $n\in\Bbb N$, and let $$h:X\to X:\langle x_n:n\in\Bbb N\rangle\mapsto\langle h_n(x_n):n\in\Bbb N\rangle\;;$$ then $h$ is a homeomorphism.
Your proof can easily be adapted to this more general setting. And since the map $$h:\Bbb R\to\Bbb R:x\mapsto ax+b$$ is a homeomorphism whenever $a\ne 0$, in your result we need only that each $a_n$ be non-zero.
• @Abhishek: Example $\mathbf{2.3.10}$ in Ryszard Engelking’s General Topology discusses a more general form of the result for the ordinary product; the proof for the box product is essentially the same. Both are straightforward applications of the fact that if $B_i\subseteq Y_i$ for each $i\in I$, and $f_i:X_i\to Y_i$ for each $i\in I$, then $$f^{-1}\left[\prod_{i\in I}B_i\right]=\prod_{i\in I}f^{-1}[B_i]\;.$$ – Brian M. Scott Aug 14 '16 at 15:57
You are in right direction, my suggestion is you can do it directly without using composition of two function. I believe you can make the $a_i$ nonzero for each i .
• Is it justified to say $h_1^{-1}(\prod_{i=1}^{\infty}V_i)=\prod_{i=1}^{\infty}h_{1_i}^{-1}(V_i)$ ? – Naive Aug 9 '16 at 17:08