Prove that Basis of Product Topology is Not a Topology

Question

We know that for topological spaces $(X,\mathcal{T}_{X})$ and $(Y,\mathcal{T}_{Y})$ the product topology is the topology on $X\times Y$ generated by the basis $\mathcal{B}=\lbrace U\times V\mid U\in\mathcal{T}_{X}\wedge V\in\mathcal{T}_{Y}\rbrace$. I need to show that $\mathcal{B}$ is not a topology on $X\times Y$. I know the theorem which states that $\mathcal{B}$ is a basis for topology $\mathcal{T}$ on set $X$ iff $\mathcal{B}\subset\mathcal{T}$ and for all $O\in\mathcal{T}$ containing an arbitrary element $x$, there is an element $B\in\mathcal{B}$ such that $x\in B$ and $B\subset O$. So if I want to show that $\mathcal{B}$ is not a basis for the said topology, I need to prove that the negation of the theorem holds. Is the statement:( $\mathcal{B}\not\subset\mathcal{T}$ or there exists some $O\in\mathcal{T}$ containing arbitrary element $x$ such that for all $B\in\mathcal{B}$, we have that $x\notin B$ or $B\not\subset O$) the negation of the said theorem? Any idea how to find such $O$?

Answer 1

I think the question wants you to note that the union axiom is not fulfilled, even for finite unions:

Small example of this: let $X=Y=\{0,1\}$ in the discrete topology both. Then the sets $\{(0,0)\} =\{0\} \times \{0\}$ and $\{(1,1)\}=\{1\}\times \{1\}$ are in $\mathcal B$, but their union $\{(0,0),(1,1)\}$ is not. So $\mathcal B$ is not a topology. It does obey the axioms for being a base for some topology.

Answer 2

The first comment is an example of the fact that you cannot have a general statement, since there are pairs of topological spaces $X,Y$ such that $\mathcal{B}$ is a topology on $X\times Y$.

However, if you need a counterexample just to show that $\mathcal{B}$ is not always a topology, it suffices to take $X=Y=\mathbb R$ and remark that open sets $U,V\subseteq\mathbb R$ are a union of countably many disjoints open intervals, that is $U\times V$ is the union of countably many open rectangles. For example an open disc is an open set of $\mathbb R^2$,namely $\mathbb D^{2}\in\mathcal T$. Nevertheless, is not a union of disjoint rectangles, that is $\mathbb D^{2}\notin\mathcal B$.

Answer 3

Let us prove

$\mathcal B$ is a topology if and only if at least one of $\mathcal T_X , \mathcal T_Y$ is the trivial topology.

1. Let one of $\mathcal T_X , \mathcal T_Y$ be the trivial topology, w.l.o.g. $\mathcal T_X$. Then the non-empty elements of $\mathcal B$ have the form $X \times V$ with $V \in \mathcal T_Y$. It is now easy to verify that $\mathcal B$ is a topology.

2. Let both $\mathcal T_X , \mathcal T_Y$ be non-trivial topologies. Let $U \in \mathcal T_X, U \ne \emptyset, X$ and $V \in \mathcal T_Y, V \ne \emptyset, Y$. Then the two products $U \times Y, X \times V$ are in $\mathcal B$, but their union $U \times Y \cup X \times V$ does not belong to $\mathcal B$ which shows that $\mathcal B$ is not a topology.

Answer 4

Let $X=Y=\{0,1\}$ with the discrete topology. Then every element of $\mathcal B$ contrains $0$, $1$, $2$, or $4$ points, but there is an open subset of $X\times Y$ with exactly $3$ points.