Abstract Nonsense

Crushing one theorem at a time

The dreaded product topology

NOTE: All sets and spaces in the following discussion will be assumed to be, without loss of generality, non-empty. Also, if not stated it is assumed that \displaystyle X=\prod_{j\in\mathcal{J}}X_j under the product topology where \left\{X_j\right\}_{j\in\mathcal{J}} is a non-empty collection of topological spaces.

To continue any further we mus make an, admittedly not brief, sojourn into the concept of product topologies.  Now, this isn’t a text-book so I’ll assume that the reader is familiar with the basic concepts of Cartesian products and projections (denote \pi). With this in mind we first make a definition for the arbitrary Cartesian product of sets:

Arbitrary Cartesian Product: Let \left\{X_j\right\}_{j\in\mathcal{J}} be an arbitrary class of non-empty sets, then we define \displaystyle \prod_{j\in\mathcal{J}}X_j=\left\{\bold{x}:\mathcal{J}\mapsto\bigcup_{j\in\mathcal{J}}X_j \text{ }:\text{ } \bold{x}(j)\in X_j,\text{ }\forall j\in\mathcal{J}\right\}.

Now, if \left\{X_j\right\}_{j\in\mathcal{J}} is a non-empty class of topological spaces we define the product topology on \displaystyle X=\prod_{j\in\mathcal{J}}X_j as the topology generated by the subbase \mathfrak{S}=\left\{\pi_j^{-1}(G): G \in\mathfrak{J}_j\right\} (where \mathfrak{J}_j is the topology on X_j,) in the sense that it is the set of all finite intersections of elements of \mathfrak{S} form an open base \mathfrak{B} for X, and thus the topology on X is made up of all arbitrary unions of elements of this open base. We call \mathfrak{S} in the above discussion the defining open subbase for X and \mathfrak{B} the defining open base for X under the product topology.

So what do elements of \mathfrak{S} look like? Well, let G be an open set in X_j then \pi_j^{-1}(G) is (in words) all \bold{x}\in X such that \pi_j(\bold(x))\in G. It is clear that the only limitation then that needs to be made on \bold{x} is that it’s jth coordinate is in G. From this it is clear that \displaystyle \pi_j^{-1}(G)=\prod_{\ell\in\mathcal{J}}E_\ell where \displaystyle \begin{cases} X_j & \mbox{if} \quad \ell\ne j\\ G & \mbox{if} \quad \ell=j\end{cases}. Thus, \mathfrak{S} is made up of all sets of the form \displaystyle \prod_{j\in\mathcal{J}}E_j where E_j=X_j for all but one j

Example: If X_j=[0,1],\text{ }j\in[0.1] under the usual topology then \displaystyle (0,1)\times\prod_{j\in(0,1]}X_j\in\mathfrak{S} whereas \displaystyle (0,1)\times\prod_{j\in(0,1)}X_j\times(0,1)\notin\mathfrak{S}

So, this being said we see that the elements of \mathfrak{B} (being the finite intersection of elements of \mathfrak{S}) are going to be the product of open sets G_j\subseteq X_j such that G_j=X_j for all but finitely many j.

So, let’s start proving some theorems!

Theorem 1: All projections \pi_j:X\mapsto X_j are open mappings.

Proof: Let E\subseteq X be open then \displaystyle E=\bigcup_{k\in\mathcal{K}}B_k where \left\{B_k\right\}_{k\in\mathcal{K}}\subseteq\mathfrak{B}, and so \displaystyle \pi_j(E)=\pi_j\left(\bigcup_{k\in\mathcal{K}}B_k\right)=\bigcup_{k\in\mathcal{K}}\pi_j\left(B_k\right) and since the jth coordinate of every basic open set (whether equal to the full space or not) is an open set in X_j and so \pi_j(E) is the union of open sets inX_j and thus open. \blacksquare

Corollary 1: If N is a neighborhood of \bold{x}\in X then \pi_j(N) is a neighborhood of \pi_j(\bold{x})

Remark: Projections need not be closed mappings. Consider the set E=\left\{(x,y):\text{ }xy=1,\right\} as a subset of \mathbb{R}^2 with the product topology. It is relatively easy E is closed, but \pi_1(E)=\pi_2(E)=\mathbb{R}-\{0\} which is open.

One might ask “Why not just make the open base for a topology on X the set of all arbitrary products of open sets?”. The topology that this is describing is called the box topology. Namely, the box topology has the set of all arbitrary products of open open sets as an open base. It is clear that the box and product topologies coincide for finite cases but do not for infinite cases. So, why did we choose the box topology over the product topology? The answer comes from the following theorem.

Theorem: Let f:M\mapsto X be a mapping of a topological space M into the product space X. Then, f is continuous if and only if \pi_jf:M\mapsto X_j is continuous for all j.



Let G be open in X_j then \left(\pi_j f\right)^{-1}\left(G\right)=f^{-1}\left(\pi_j^{-1}(G)\right) and since \pi_j^{-1}(G) is subbasic open in X and f continuous we see that f^{-1}\left(\pi_j^{-1}(G)\right) is open in M.


Lemma: A mapping f:E\mapsto F (where E and F are arbitrary topological spaces) is continuous if and only if f^{-1}(B) is open for every B\in\mathcal{B}_F where \mathcal{B}_F is an open base for F.


\implies: Let G be open in F then \displaystyle G=\bigcup_{h\in\mathcal{H}}B_h where \left\{B_h\right\}_{h\in\mathcal{H}}\subseteq\mathcal{B}_F. So then, \displaystyle f^{-1}\left( G\right)=\displaystyle f^{-1}\left(\bigcup_{h\in\mathcal{H}}B_h\right)=\bigcup_{h\in\mathcal{H}}f^{-1}\left(B_h\right) which by assumption is the union of open sets in E and thus open. The conclusion follows.

\Leftarrow: This is obvious. \blacksquare

So, suppose that G is basic open in X, then \displaystyle G=\pi_{j_1}^{-1}(G_{j_1})\cap\cdots\cap\pi_{j_m}^{-1}(G_{j_m}) where \left\{j_,\cdots,j_m\right\}\subseteq\mathcal{J} and G_{j_k} is open in X_k,\text{ }1\leqslant k\leqslant m. It follows then that



=f^{-1}\left(\pi_{j_1}^{-1}(G_{j_1})\right)\cap\cdots\cap f^{-1}\left(\pi_{j_m}^{-1}(G_{j_m})\right)

=\left(\pi_{j_1}f\right)^{-1}(G_{j_1})\cap\cdots\cap \left(\pi_{j_m}f\right)^{-1}(G_{j_m})

and since each \pi_jf,\text{ }j\in\mathcal{J} is continuous it follows that f^{-1}(G) is the finite intersection of open sets in M and thus open. The conclusion follows. \blacksquare

Note that the key point here was that G was the finite intersection of inverse images of open sets. If we had the box topology we could have had that G was the infinite intersection of inverse images of open sets we would have no guarantee that f^{-1}(G) would be open since it would be the infinite intersection of open sets…which we all know need not be open.

Now, one may ask “Topologically does it matter which order we product the sets?”. The answer is “no…for finite cases”

Theorem: Let X_1,\cdots,X_n be a finite collection of topological spaces then X_1\times\cdots\times X_n\approx X_{\sigma(1)}\times\cdots\times X_{\sigma(n)} where \sigma is an permutation of \{1,\cdots,n\}

Proof: It is readily verified that the canonical mapping f:X_1\times\cdots\times X_n\mapsto X_{\sigma(1)}\times\cdots\times X_{\sigma(n)} given by (x_1,\cdots,x_n)\mapsto (x_{\sigma(1)},\cdots,x_{\sigma(n)}) is a homeomorphism. \blacksquare


February 20, 2010 - Posted by | General Topology, Topology | ,


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