Let $X$ be a topological space which is the union $\cup _\alpha X_\alpha$ of subspaces $X_\alpha$. Recall that we say the topology of $X$ is coherent with $\{ X_\alpha \}$ if a subset $A \subset X$ is closed in $X$ iff $A \cap X_\alpha$ is closed in $X_\alpha$ for each $\alpha$. In this case, does the topology of $X\times Y$ is coherent with $\{ X_\alpha \times Y\}$ where $Y$ is any space? (Here $X\times Y$ of course has the product topology.)
To show this, it suffices to show the following:
If $A \subset X \times Y$ is such that $A\cap (X_\alpha \times Y)$ is closed in $X_\alpha \times Y$ for all $\alpha$, then $A$ is closed in $X\times Y$.
But I have no idea for this.
If this is not true, can I make this true by adding assumptions, for example letting $Y$ be compact?
Best Answer
This can fail. Consider the spaces from Example 2.4.20 from Engelking’s “General topology”. Let $X=\Bbb R\setminus\{1/2,1/3,\dots\}$ with the topology of a subspace of $\Bbb R$. Then the topology of $X$ is coherent with a family of all compact subsets of $X$. Let $Y=q(\Bbb R)$ be the quotient space obtained from $\Bbb R$ by identifying the set of positive integers to a point. Put $A = \{(1/i + \pi/j, q(i + 1/j )): i,j = 2,3,\dots\}\subset X\times Y$. It is easy to check that $(0,q(0))\in\overline{A}\setminus A$, whereas for each compact set $X_\alpha$ of $X$ an intersection $(X_\alpha\times Y)\cap A$ is closed in $X_\alpha\times Y$.