The Unapologetic Mathematician

Mathematics for the interested outsider

Tychonoff’s Theorem

One of the biggest results in point-set topology is Tychonoff’s Theorem: the fact that the product of any family \{X_i\}_{i\in\mathcal{I}} of compact spaces is again compact. Unsurprisingly, the really tough bit comes in when we look at an infinite product. Our approach will use the dual definition of compactness.

Let’s say that a collection \mathcal{F} of closed sets has the finite intersection hypothesis if all finite intersections of members of the collection are nonempty, so compactness says that any collection satisfying the finite intersection hypothesis has nonempty intersection. We can then form the collection \Omega=\{\mathcal{F}\} of all collections of sets satisfying the finite intersection hypothesis. This can be partially ordered by containment — \mathcal{F}'\leq\mathcal{F} if every set in \mathcal{F}' is also in \mathcal{F}.

Given any particular collection \mathcal{F} we can find a maximal collection containing it by finding the longest increasing chain in \Omega starting at \mathcal{F}. Then we simply take the union of all these collections to find the collection at its top. This is almost exactly the same thing as we did back when we showed that every vector space is a free module! And just like then, we need Zorn’s lemma to tell us that we can manage the trick in general, but if we look closely at how we’re going to use it we’ll see that we can get away without Zorn’s lemma for finite products.

Anyhow, this maximal collection \mathcal{F} has two nice properties: it contains all of its own finite intersections, and it contains any set which intersects each set in \mathcal{F}. These are both true because if \mathcal{F} didn’t contain one of these sets we could throw it in, make \mathcal{F} strictly larger, and still satisfy the finite intersection hypothesis.

Now let’s assume that \mathcal{F} is a collection of closed subsets of \prod\limits_{i\in\mathcal{I}}X_i satisfying the finite intersection hypothesis. We can then get a maximal collection \mathcal{G} containing \mathcal{F}. Then given an index i\in\mathcal{I} we can consider the collection \{\overline{\pi_i(G)}\}_{G\in\mathcal{G}} of closed subsets of X_i and see that it, too, satisfies the finite intersection hypothesis. Thus by compactness of X_i the intersection of this collection is nonempty. Letting U_i be a closed set containing one of these intersection points x_i, we see that the preimage \pi_i^{-1}(U_i) meets every G\in\mathcal{G}, and so must itself be in \mathcal{G}.

Okay, so let’s take the point x_i for each index and consider the point p in \prod\limits_{i\in\mathcal{I}}X_i with i-th coordinate x_i. Then pick some set V=\prod\limits_{i\in\mathcal{I}}V_i containing p from the base for the product topology. For all but a finite number of the i, V_i=X_i. For those finite number where it’s smaller, the closure of V_i contains the point x_i\in X_i, and so \pi_i^{-1}(V_i) is in \mathcal{G}. So their finite intersection must be nonempty, and so is V itself!

Now, since V is in \mathcal{G}, it must intersect each of the closed sets in the original collection \mathcal{F}. Since the only constraint on V is that it contain p, this point must be a limit point of each of the sets in \mathcal{F}. And because they’re closed, they must contain all of their limit points. Thus the intersection of all the sets in \mathcal{F} is nonempty, and the product space is compact!

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January 17, 2008 - Posted by | Point-Set Topology, Topology

5 Comments »

  1. [...] Tychonoff’s Theorem tells us that products of closed intervals are also compact. In particular, the closed cube is [...]

    Pingback by The Heine-Borel Theorem « The Unapologetic Mathematician | January 21, 2008 | Reply

  2. The two last paragraphs are not much clear to me, in particular V that seems to be defined as a random open set from the base…

    Comment by edriv | March 28, 2008 | Reply

  3. Oh.. V should be picked to contain the point p

    Comment by John Armstrong | March 29, 2008 | Reply

  4. [...] of choice — was essential when we needed to show that every vector space has a basis, or Tychonoff’s theorem, or that exact sequences of vector spaces split. So it’s sort of a mixed bag. In practice, [...]

    Pingback by Non-Lebesgue Measurable Sets « The Unapologetic Mathematician | April 24, 2010 | Reply

  5. [...] product of one copy of for every element of . Since each copy of is a compact Hausdorff space, Tychonoff’s theorem tells us that is a compact Hausdorff space. If is any finite subring of containing , let be the [...]

    Pingback by Stone’s Representation Theorem I « The Unapologetic Mathematician | August 18, 2010 | Reply


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