# Equality of product topologies on $X_1\times X_2\times X_3$

I am reading "Introduction to Topological Manifolds" by John M. Lee and encountered the following problem.
I have to show that the product topology is "associative", i.e. that the three product topologies $$X_1\times X_2\times X_3, (X_1 \times X_2)\times X_3$$ and $$X_1\times (X_2 \times X_3)$$ on the set $$X_1\times X_2 \times X_3$$ are equal.
The product topology $$X_1\times X_2\times X_3$$ is spanned by $$\mathcal{B}_1=\{U_1\times U_2\times U_3\,\vert\, U_i\, \text{open in } X_i, i=1,2,3\}$$.
But it is not clear to me how the other two are defined on $$\mathbf{X_1\times X_2\times X_3}$$.

Should I consider the topologies spanned by $$\mathcal{B}_2 = \{V\times U\,\vert\,V \text{ open in }X_1\times X_2, U\,\text{open in } X_3 \}$$ and $$\mathcal{B}_3 = \{U\times V\,\vert\, U \text{ open in } X_1, V \text{ open in } X_2 \times X_3 \}$$ or $$\tilde{\mathcal{B}}_2= \{ (U_1\times U_2)\times U_3\,\vert\, U_i \text{ open in } X_i, i = 1,2,3 \}$$ and $$\tilde{\mathcal{B}}_3= \{ U_1\times (U_2\times U_3)\,\vert\, U_i \text{ open in } X_i, i = 1,2,3 \}$$? Or something else?

It is immediately clear to me, that we can find the homeomorphisms for $$\tilde{\mathcal{B}}_2$$ and $$\tilde{\mathcal{B}}_3$$, so they are equal in the homeomorphic sense.
However, I wonder if equality and topological equivalence is the same notion, or if equality means that they are the same?
In the second case I don't see how they could be the same, since we deal with different sets entirely.

• This can easily be seen from the universal property of product topology. The product of two spaces has a universal property, and the product of three spaces has a smililar universal property. It is then a tautological checking that the universal properties of $(X_1 \times X_2)\times X_3$, $X_1\times(X_2\times X_3)$ and $X_1 \times X_2 \times X_3$ are all equivalent. Commented Nov 12, 2021 at 21:48
• @WhatsUp Agreed, assuming the OP has seen some category theory before. Commented Nov 12, 2021 at 23:05
• The sets are different formally but they have a canonical bijection between them and as this respects projections it's a homeomorphism too. Commented Nov 12, 2021 at 23:06

If $$\mathcal B_i$$ is a basis for $$Y_i,$$ $$i=1,2$$ then the set $$\{U_1\times U_2\mid U_i\in \mathcal B_i\}$$ is a basis for the product topology on $$Y_1\times Y_2.$$
Then for each $$X_i$$ take the whole topology as it’s basis.
Then use this to get a basis for $$(X_1\times X_2)\times X_3$$ which is $$\{(U_1\times U_2)\times U_3\mid U_i \text{ open in }X_i\}.$$