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.


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.

  • $\begingroup$ 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

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