3
$\begingroup$

Assume we are given a space $A$ with a metric $d$. Assume $A = A_1 \times A_2 \times A_3 \cdots$, ie. $A$ is a Cartesian product of spaces $A_i$, where $i \in I$. $I$ is countable or countably infinite. Assume also that we know that each $A_i$ is homeomorphic to a space $B_i$.

Is the Cartesian product $A = \prod A_i$ homeomorphic to Cartesian product $B = \prod B_i$?

Because each $A_i$ is homeomorphic to each $B_i$, let the homeomorphism be $h_i: A_i \rightarrow B_i.$ Can I apply these in some way to an element $x \in A$ to get to $y \in B$? Like "apply $h_i$ to the $i$th element of $x$"? I dread to use projections as a projection isn't a homeomorphism.

$\endgroup$
6
$\begingroup$

You're quite right. We need not restrict ourselves to metric spaces or countable or finite products:

Let $A_i, B_i, i \in I$ be a family of topological spaces such that $h_i: A_i \rightarrow B_i$ is a homeomorphism for every $i$. Then $h: A = \prod_{i \in I} A_i \rightarrow B = \prod_{i \in I} B_i$ defined by $h( (x_i) ) = h(x_i)_i $ is a homeomorphism as well.

The fact that $h$ is a bijection is simple set theory. Any map $f: X \rightarrow \prod_{i \in I} X_i$ is continuous iff for every $i$, $\pi_i \circ f$ is continuous between $X$ and $X_i$, where $\pi_i$ is the projection onto the $i$'th coordinate. This is the universal property for the product topology.

From the universal property $h$ is continuous as by construction $\pi_i \circ h = h_i \circ \pi_i$, which is continuous, as a composition of continuous functions. The product of the inverses of the $h_i$ is the required continuous inverse, using the universal property for the $\prod_i A_i$ instead.

$\endgroup$
  • $\begingroup$ Shouldn't it be $\pi_i\circ h=h\circ\pi'_i$ or something like that. (The map $h_i$ goes from $A_i$ to $B_i$, so it cannot be equal to the map $\pi\circ h$ going from $A=\prod A_i$ to $B_i$. Unless I have misunderstood your notation or overlooked something.) $\endgroup$ – Martin Sleziak Oct 2 '13 at 12:39
  • $\begingroup$ You're right, of course. This is still continuous, so the proof still works. Edited. $\endgroup$ – Henno Brandsma Oct 2 '13 at 19:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.