Are $\prod_{i\in I,j\in J}X_{ij}$ and $\prod_{i\in I}\prod_{j\in J}X_{ij}$ homeomorphic?

Question

For each $i\in I, j\in J$, $X_{ij}$ is a topological space. Are

$$\prod_{i\in I,j\in J}X_{ij}$$ and $$\prod_{i\in I}\prod_{j\in J}X_{ij}$$ homeomorphic? What's the homeomorphism?

Answer 1

Set $X = \cup\left\{X_{ij}: i \in I, j \in J \right\}$. Then first set in your question equals (as a set, by the definition of Cartesian products):

$$ \prod_{i \in I,j \in J} X_{ij} = \left\{ f: I \times J \rightarrow X: \forall_{i \in I}\,\forall_{j \in J}\, f(i,j) \in X_{ij} \right\}\mbox{,}$$ while if we set (for $i \in I$):

$$X_i = \prod_{j \in J} X_{ij} = \left\{ f: J \rightarrow \cup_{i \in I} X_{ij} : \forall_{j \in J}\,f(j) \in X_{ij} \right\}\mbox{,}$$ the second set equals

$$\left\{f: I \rightarrow \cup_{i \in I} X_i : \forall_{i \in I}\,f(i) \in X_i \right\}\mbox{.}$$

Now, in the last definition we thus have that each $f(i)$ is itself a function from $J$ to $\cup_{i \in I} X_{ij}$ such that $f(i)(j) \in X_{ij}$, for all pairs $(i,j)$ from $I \times J$. So if we have a member $f$ of the first set, set $H(f)$ in the other set as the function defined by $(H(f)(i))(j) = f(i,j)$: $H(f)$ is a function defined on $I$, so we have to define $H(f)(i)$ for all $i$, but these themselves are functions from $J$, so we need to define all $(H(f)(i))(j)$ values. One checks easily that the functions have the right co-domains and satisfy the conditions to be in $X_i$ and the second set.

This is the required bijection (the inverse is similarly defined, I leave that to the reader).

Now use that function into a product is continuous iff all compositions with the projections of the product are continuous. This allows one to easily show that $H$ and its inverse are both continuous.