Direct definition of product or box topology?

Question

Both the product topology and the box topology are defined "indirectly".

The box topology of $A,B$ is defined as the topology generated by the base given by the cartesian product of open sets in $A,B$. Can we define the box topology "directly", without talking about a base?

Similarly, the product topology of $A,B$ is defined as the "coarsest topology on $A\times B$ for which the projections are continuous", but can we define it "directly" without reference to "coarsest"?

If not, what makes it so that we can't do this?

Answer 1

Remark. Let $X$ be a set. Let $\mathscr{T}$ (script T) be a collection of topologies on $X$. Then $\bigcap\mathscr{T}$ is a topology on $X$.

Definition. Let $\{(X_i,\mathcal{T}_i)\}$ be a family of topological spaces. Define $$P:=\prod_i X_i:=\{x_{(\cdot)}:(\forall i)[x_i\in X_i]\}.$$ Let $\mathscr{T}$ be the collection of all topologies on $P$ where each element $\mathcal{I}$ of $\mathscr{T}$ satisfies the following property: For each familiy of open sets $\{U_i\}$, where $U_i\in\mathcal{T}_i$, we have $\prod_iU_i\in\mathcal{I}$. The box topology on $P$ is $\bigcap\mathscr{T}$.

For each $i$, define the projection $\pi_i:P\to X_i:x_{(\cdot)}\mapsto x_i$. Let $\mathcal{I}$ be any topology on $P$. Observe $\pi_i$ is continuous if and only if $\{\pi_i^{-1}(U):U\in\mathcal{T}_i\}\subseteq \mathcal{I}$. The product topology on $P$ is $\bigcap\mathscr{T}$, where $\mathscr{T}$ is the collection of all topologies $\mathcal{I}$ on $P$ satisfying the following. $$(\forall i)[\{\pi_i^{-1}(U):U\in\mathcal{T}_i\}\subseteq \mathcal{I}].$$

The topology on $\mathbb{R}$ can be defined directly. Indeed, the open sets of $\mathbb{R}$ is the following set. $$\big\{U\subseteq\mathbb{R}:\big(\exists\{(a_i,b_i)\}_{i\in I}:a_i<b_i\big)\big[U=\bigcup_{i\in I}(a_i,b_i)\big]\big\}.$$

Answer 2

I don't think it is even possible for many common spaces, eg. $\mathbb R$. Open sets on real line are defined by base of open segments.

As other answerers noted, open sets are sums of any number of sets from the base, but actually this is the very notion of base. Also, for me it is nicer not to use any additional indexing set, that is: $$ \text{topology generated by base } \mathcal{B} = \Big\{ \bigcup \beta:\, \beta\subset \mathcal B\Big\} $$ so any open set is a union of a subfamily of $\mathcal B$. Of course, this is not injective: different subfamilies may yield identical union, and usually many do, so this representation is in no way unique.

Note also, that product topology can be defined through a base, which consists of products of base sets: $$ \text{topology in } A\times B = \Big\{\bigcup \beta:\, \beta\subset \big\{ a\times b:\, a\in \text{some base of } A, \ b \in \text{some base of } B\big\}\Big\}. $$ The whole topologies can be used instead of bases, as the whole topology is a base for itself:

$$ \text{topology in } A\times B = \Big\{\bigcup \beta:\, \beta\subset \big\{ a\times b:\, a \text{ is open in } A, \ b \text{ is open in } B\big\}\Big\}. $$

Answer 3

Well in general let's say you have a set $X$ equipped with a topology $\mathcal{T}$, if $\mathcal{T}$ is generated by a basis $\mathcal{B}$, then elements of $\mathcal{T}$ are unions of elements in $\mathcal{B}$

If you want to write that down explicitly you could say that any $U \in \mathcal{T}$ is of the form $$U = \bigcup_{i \in I} A_i$$ where $A_i$ is an element of $\mathcal{B}$ and $I$ is some arbitrary indexing set.

Answer 4

If we have a base we can find all open sets as unions of base elements, and this is most often formulated as a pointwise condition, e.g.:

$O \subseteq X \times Y$ is open iff $$\forall (x,y) \in O: \exists U \text{ open in } X, \exists V \text{ open in } Y: x \in U, y \in V \text{ and } U \times V \subseteq O$$

which gives a verifiable condition on a subset to check openness. This can easily be generalised to arbitrary Cartesian products and both the product and the box topology.