# The product topology :Its definition and coarseness

Let A be a finite set of topological spaces $$\ X_\alpha$$ set. Now let us consider the product set of this topological spaces P=$$\ \prod_\alpha X_\alpha$$.

Now a topology in P , with all sets of the form $$\ \prod_\alpha U_\alpha$$, where U is open in $$X_\alpha$$ as its basis is called a box topology and a topology in P ,with all sets of the form $$\ \prod_\alpha U_\alpha$$, with $$\ U_\alpha$$ = $$\ X_\alpha$$ for all but for finitely many $$\ \alpha$$ $$\ \epsilon$$ K,where K is some indexing set is called a product topology.

What is the significance of the later mentioned finitely many sets in the definition of the product topology?

The lecture notes I am going through mention that box topology is too coarse to be useful, it generates a lots of open sets, and as such product topology is more preferable. Can I have a rigorous explanation for this?

• The product topology is the smallest topology where the projections functions are all continuous. This means it is the "right" topology for category theory reasons. – Thomas Andrews Feb 6 '16 at 18:10
• @ThomasAndrews : I guess , the functions would be continuous even in box topology . And moreover, we define the product topology to be the coarsest topology such that all the projection mappings are all continuous . I would like to know the reason for this too . – itp dusra Feb 6 '16 at 18:21

A reasonable property to ask of your product topology is that for all $Y$ with functions $f_{\alpha}: Y \rightarrow X_{i}$, the $f_{i}$ are continuous if and only if $\prod f_{i}:Y \rightarrow \prod X_{i}$ is continuous. This has to do with category theory, and the so called universal property of products mentioned by other answers. If you are curious check it out.
It is very easy to see which is the adequate topology: $f_{i}^{-1}(U_{i}) = (\prod_{j} f_{j})^{-1}(\prod_{j\neq i}X_{j} \times U_{i})$ for all open $U_{i}$ in $X_{i}$. So the $f_{i}$ are all continous if and only if all the sets of the form $(\prod_{j} f_{j})^{-1}(\prod_{j\neq i}X_{j} \times U_{i})$ are open in $Y$, and a subbase for the seeked topology on $\prod X_{i}$ is given by the sets having form $\prod_{j\neq i}X_{j} \times U_{i}$. The base generated by taking finite intersections is the product topology; not the box topology, since the intersections are finite.
• I think the map $Y\to \prod X_i$ should be called $\Delta f_i$, not $\prod f_i$ (which would be the map $\prod Y\to \prod X_i$). – tomasz Aug 28 at 12:07