Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A short question: Why does not the Tychonoff theorem (the arbitrary product of compact spaces is compact) hold in the box topology? I don't know how to show that there is no finite sub-cover of any open cover of the product space! Can anyone please give me some hint? Thank you!

share|improve this question
    
Are you looking for a counterexample or for an intuitive explanation for why the theorem fails to hold? –  Srivatsan Dec 9 '11 at 1:56
    
Hi, I'd like to see an explanation for it. However, a counterexample should also work. Thanks. –  Feri Dec 9 '11 at 2:01
3  
For an explanation, the intuition is that there are too many open sets in the box topology, so being compact is more difficult –  M Turgeon Dec 9 '11 at 2:03

2 Answers 2

up vote 10 down vote accepted

Consider the space $2^{\mathbb{N}}$ in the box topology, where $2=\{0,1\}$ is the two-point discrete space. So each factor is a finite space, and clearly compact. In the box topology, however, the product space $2^{\mathbb{N}}$ is discrete, since every individual sequence $x\in 2^{\mathbb{N}}$ is the unique member of the corresponding open set determined by its coordinates. So every point is isolated and the space is discrete. But no infinite discrete space is compact, so Tychonoff fails for the box topology.

share|improve this answer
1  
The same argument shows more generally that the box product of discrete spaces is discrete, and this violates Tychonoff when the factors are finite (hence compact), since the product is infinite. –  JDH Dec 9 '11 at 2:09
    
Thank you for your answer. To make it concrete to myself, should we consider the set $2=\{0,1\}$ in the discrete topology? So, then every one point set is open. Right? Thanks again. –  Feri Dec 9 '11 at 2:47
    
Yes, consider $\{0,1\}$ to be discrete, so that every set is open on each factor. Thus, if $x\in \{0,1\}^{\mathbb{N}}$, then it is the unique element of $\Pi_n \{x(n)\}$, which is open in the box topology. –  JDH Dec 9 '11 at 2:54
    
I see. Great! Thanks. –  Feri Dec 9 '11 at 3:01

Here are the cold hard facts. Let $(X_i)_{i \in I}$ be a nonempty family of nonempty topological spaces. Let $X = \prod_{i \in I} X_i$. Let $P$ be $X$ in the product topology. Let $B$ be $X$ in the box topology.

$$ B \text{ is Hausdorff} \Leftrightarrow P \text{ is Hausdorff } \Leftrightarrow X_i \text{ is Hausdorff for each } i \in I$$

$$ B \text{ is compact } \Rightarrow P \text{ is compact } \Leftrightarrow X_i \text{ is compact for each } i \in I$$

The above can be got using Tychonoff's theorem and some elementary arguments. A few remarks:

  • the box topology is finer than the product topology (refinements of Hausdorff topologies are Hausdorff, and vice versa for compact topologies)
  • the projections are continuous for $B$ and $P$ (and continuous images of compact spaces are compact)
  • by fixing a base point in X, we can embed the $X_i$ into $B$ or $P$ (note subspaces of Hausdorff spaces are Hausdorff).

The point of the above is that, if $B$ is a compact Hausdorff space, then so is each $X_i$ which means $P$ is a compact Hausdorf space. But any two compact Hausdorff topologies which are comparable must coincide, so this implies $P = B$.

Moral: if the box topology turns out to be compact Hausdorff, then you're actually looking at the product topology.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.