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



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.

  • $\begingroup$ I had not come across the above proposition you quoted. Hence I was a little skeptical about coordinate maps I had defined. Moreover, since the above proposition holds for Box topology can we say that the above proposition holds for product topology as well since Box is finer than product topology? Also could you please take the trouble of sending me a pdf link or name a book which has this proposition. Just for my future references. Thank You. $\endgroup$ – Naive Aug 14 '16 at 8:45
  • $\begingroup$ @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 $$f^{-1}\left[\prod_{i\in I}B_i\right]=\prod_{i\in I}f^{-1}[B_i]\;.$$ $\endgroup$ – 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 .

  • $\begingroup$ Is it justified to say $h_1^{-1}(\prod_{i=1}^{\infty}V_i)=\prod_{i=1}^{\infty}h_{1_i}^{-1}(V_i)$ ? $\endgroup$ – Naive Aug 9 '16 at 17:08

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