Solving the Homeomorphism Problem Using Real Sequences and Topology Theorems

[radou]

Homework Statement

Here's another problem from Munkres.

Let (a1, a2, ...) and (b1, b2, ...) be sequences of real numbers, with ai > 0, for every i. Define h : Rω --> Rω with h((x1, x2, ...)) = (a1x1 + b1, a2x2 + b2, ...). Show that if Rω is given the product topology, h is a homeomorphism.

Homework Equations

I used a theorem which states that if f : A --> ∏Xj is given by the equation f(a) = (fj(a)) (j is from some indexing set J), where fj : A --> Xj, for each j, then f is continuous if and only if fj is continuous, for each j.

I'm not sure if I can use this theorem here, since there's no information about what the set A is.

The Attempt at a Solution

First I tried to prove that h is continuous using the theorem up there.

h(x) can be represented with (h1(x), h2(x), ...), where hj(x) = aj∏j(x) + bj. Here ∏j(x) denotes the projection mapping onto the j-th coordinate of x. Since hj(x) is continuous for every j (the projection mapping is continuous, addition and multiplication are, too), we conclude that h is continuous.

Proving one to one and onto is easy:

Let h(x) = h(y), then for every j, ajxj + bj = ajyj + bj, hence xj = yj, for every j. Let c = (c1, c2, ...) be in the image set of h. Then, for every j, (cj - bj)/aj = xj, and hence (x1, x2, ...) maps to c.

The j-th component of the inverse of h is given with (hj(x) - bj)/aj, and the inverse is continuous too, since it's continuous for every index j.

All this is well defined, since aj > 0, for every j.

Hence, h is a homeomorphism from R∞ to R∞.

 

[radou]

Here's an alternative proof for continuity, which seems even easier.

Let x be a point in Rω, and Uω a neighborhood of h(x). Every component of h(x) lies in a neighborhood <a, b>, i.e. a < aj xj + bj < b. It follows that <(a-bj)/aj, (b-bj)/aj> is a neighborhood of xj whose image is in <a, b>. The product of all such neighborhoods in Rω is a neighborhood of x whose image under h lies in Uω.