Show that $\bigcup_{t\in I}U_t\times\{t\}$ could be not open although $I$ is open and $U_t$ is open for each $t\in I$.

Question

So let be $I$ an open interval of the real line and thus let $U_t$ for each $t\in I$ an open set of a topological space $X$. So I ask to clarify if the set $$ \bigcup_{t\in I}U_t\times\{t\} $$ is open or not in the product topology. I am sure that the statement is generally false but unfortunately I did not find any counterexample: in particular I'd really like clarify the statement when $X$ is the euclidean plane $\Bbb R^2$. So could someone help me, please?

Answer 1

Taking the example by the OP in the comments: $X=\Bbb R^2$ and $I=(0,42) \subseteq \Bbb R$

$$U_t := \begin{cases} (0,1)^2 & t \in \Bbb Q \cap I\\ (-1,0)^2 & t \in I\setminus \Bbb Q \\ \end{cases}$$

And suppose that $O \times (a,b)$ some non-empty basic open subset of $\bigcup_{t \in I} U_t \times \{t\}$. If $x \in O$ we can find both a rational $q \in (a,b)$ and an irrational $r \in (a,b)$. It follows that $(x,q) \in (0,1)^2 \times \{q\}$ and hence $x \in (0,1)^2$ and also $(x,r) \in (-1,0)^2 \times \{r\}$ and hence $x \in (-1,0)^2$, a contradiction.