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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.

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    $\begingroup$ 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. $\endgroup$
    – WhatsUp
    Commented Nov 12, 2021 at 21:48
  • $\begingroup$ @WhatsUp Agreed, assuming the OP has seen some category theory before. $\endgroup$ Commented Nov 12, 2021 at 23:05
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    $\begingroup$ The sets are different formally but they have a canonical bijection between them and as this respects projections it's a homeomorphism too. $\endgroup$ Commented Nov 12, 2021 at 23:06

1 Answer 1

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Hint: Show:

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 $Y_1\times Y_2.$

This is relatively simple to prove.

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 $$\{(U_1\times U_2)\times U_3\mid U_i \text{ open in }X_i\}.$$

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