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The PRODUCT topology is the coarsest (smallest, weakest) topology you can define such that the projections are continuous. (it is apparently the unique topology with this property)

The BOX topology is just this similarly defined thing.

However I've been informed by the book the product topology is unique, this means the box topology (if it is distinct) will not have continuous projections.

What does this actually mean (please may I have an example?)

Questions: Can somebody prove that the product topology is the unique topology such that the projections are continuous? The book annoyingly leaves it as an exercise.

What have I done? Read Continuity of product of fuctions w.r.t. product and box topology and got an example

found Why is the box topology finer than the product topology? and Examples on product topology $ \gg $ box topology?

Clemens Bartholdy
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Alec Teal
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  • In that example http://math.stackexchange.com/questions/72781/product-and-box-topology I'm not sure what $f$ is – Alec Teal Feb 27 '15 at 19:54
  • Because surely $f_\beta^{-1}(U_\alpha){\alpha\in I}$ is $(b\alpha){\alpha\in I}$ where $b\alpha=X_\alpha$ (for $\alpha\ne\beta$) and $b_\beta=U_\beta$ – Alec Teal Feb 27 '15 at 19:57
  • I'm not sure what you mean by "The box topology is just this similarly defined thing." The box topology is certainly not the coarsest topology such that the projections are continuous. It's true that the projections are continuous with respect to the box topology, but in a product of infinitely many spaces with nontrivial topologies, the box topology is strictly finer than the product topology. For example, any product of infinitely many proper nonempty open subsets is open in the box topology, but not in the product topology. – Jack Lee Feb 27 '15 at 20:03
  • Sorry @JackLee the product topology was defined (before I've seen this "fundamental property" form) as products of open sets. The books/notes were not wrong because it was a finite case. – Alec Teal Feb 27 '15 at 20:05

2 Answers2

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As far as I understand the question, you ask why the product topology is unique. If $\tau_1$ and $\tau_2$ are two topologies that are both "coarsest" (that is, minimal elements wrt. inclusion) such that the projections $p_{\alpha}: \Pi_\beta X_\beta\to X_\alpha$ are continuous, then they contain all sets $p_\alpha^{-1}(U)$ for all $U$ open in $X_\alpha$; that is, both topologies contain all sets of the type $(\Pi_{\beta\neq\alpha} X_\beta)\times U$ where $U$ is open in $X_\alpha$. Further, both $\tau_1$ and $\tau_2$ contain all unions and finite intersections of such sets.

But neither $\tau_1$ nor $\tau_2$ cannot contain anything else. So they are equal.

In the box topology, open sets are generated by "infinite products of open sets" whereas in the product topology, they are generated by only finite product of open sets and "full $X_\beta$'s".

Be careful, the projection are continuous also in the box topology. The projections go from the products.. so having more open sets in the domain doesn't contradict continuity.

Peter Franek
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It is quite possible for a function into a product under the box topology to be continuous in each coordinate and not be continuous. This does not happen in the the product topology. For example, if you map the real numbers into the box topology equivalent of Hilbert space by f(x)=(x,x,x,x,x,x,x,...), you'll find the image is discrete.

Syd Henderson
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