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I'm studying the topological product using Schubert's book. He starts by defining the product topology as follows:

DEFINITION. Let $\{X_{\lambda}\}_{\lambda\in\Lambda}$ be a (non-empty) family of topological spaces. The coarsest topology on the set $X=\underset{\lambda\in\Lambda}{\prod}X_{\lambda}$ , for which all projections $p_{\lambda}:X\rightarrow X_{\lambda}$ are continuous, is called the product topology on X. If $X$ is considered as having this topology, then $X$ is called the topological product of the spaces $X_\lambda$.

He then states the following theorem:

THEOREM: Sets of the form $\prod Q_{\lambda}$, where $Q_{\lambda}$ is open in $X_{\lambda}$ and $Q_{\lambda}=X_{\lambda}$ with a finite number of exceptions, form a basis for the product topology.

Which the usual definition of the product topology.

I tried to prove this theorem from the given definition. I'm still very shaky in this topic (topology), and I found this proof particularly tricky, so I would like you to check it.

My attempt: I know that the coarsest topology on $X$ containing all the sets of a certain family $A$ of subsets of $X$ is the topology generated by $A$ ($A$ thus being a subbasis for this topology).

Hence, if we let $A$ be the set of all subsets of $X$ containing all the sets of the form $p_{\lambda}^{-1}(Q_{\lambda})$ (with $Q_\lambda$ an open set of $X_\lambda$), $A$ is a subbasis of the product topology (because all projections $p_{\lambda}:X\rightarrow X_{\lambda}$ are continuous precisely when the sets of the form $p_{\lambda}^{-1}(Q_{\lambda})$ (with $Q_\lambda$ an open set of $X_\lambda$) are open in $X$).

I also know that, if $Q_{\mu}$ is a set of $X=\underset{\lambda\in\Lambda}\prod X_{\lambda}$, then $p_{\mu}^{-1}(Q_{\mu})=X_{1}\times X_{2}\times...\times Q_{\mu}\times...$

Now, the set of all finite intersections of the elements of $A$ is a basis of $X$. The finite intersections of the $p_{\mu}^{-1}(Q_{\mu})=X_{1}\times X_{2}\times...\times Q_{\mu}\times...$ are precisely of the form described in the theorem.

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    $\begingroup$ Yes this looks exactly right to me. By the way I really like the question. Also you might find this interesting: en.wikipedia.org/wiki/Product_topology#Properties If you ever learn category theory, the reason to use the product topology (instead of the seemingly more straightforward box topology) is that it is a categorical product which is equivalent to the universal property definition Schubert gives above (to check that a function into a product is continuous, we just verify that each coordinate function is continuous). $\endgroup$ – Chill2Macht Jan 15 '17 at 22:04
  • $\begingroup$ Yep, that's how it's done imho. $\endgroup$ – user159517 Jan 15 '17 at 22:26
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What your attempt shows is that A is a sub-base for a topology $T$ on $X$ that's a subset of the product topology given by the "DEFINITION". What is missing is the converse. Your 2nd-last sentence needs to proven. That is, show that every $p_{\lambda}$ is continuous with the topology $T$ on $X.$ To show this, let $S$ an be open subset of $X_{\lambda}.$ Then $p_{\lambda}^{-1}S\in A\in T,$ so $p_{\lambda}^{-1}$ is $T$-continuous. So by the DEFINITION, $\;T$ must have the product topology as a subset.

This is often called the Tychonoff product topology,or the topology of point-wise convergence, to distinguish it from the box product topology, which is stronger, and, for some types of $X_{\lambda},$ there is also the uniform product topology (also called the topology of uniform convergence.) The Tychonoff product has been found to be an extremely useful tool.

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