# Does open set in $R_d \times R$ containing $R\times 0$ contain any subset $I \times (-\varepsilon, \varepsilon)$, with infinite $I$?

Let $R_d$ be the set of reals with discrete topology, $R$ be the set of reals with natural topology and let $X=R_d \times R$ with product topology. Assume that $R_d \times \{0\} \subset V \subset X$ and $V$ is open. Do there exist an infinite set $I \subset R$ and a number $\varepsilon >0$ such that $I \times (-\varepsilon, \varepsilon) \subset V$ ?

Thanks

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Directly from the definition of product topology we know that for each $x\in\mathbb R$ there is a rational $\varepsilon_x>0$ such that $\{x\}\times(-\varepsilon_x,\varepsilon_x)\subseteq V$.
Now, for each $q\in\mathbb Q$ denote $A_q=\{x\in\mathbb R; \{x\}\times(-q,q)\subseteq V\}$. We have $\mathbb R=\bigcup_{q\in\mathbb Q} A_q$.
If every set $A_q$ would be finite, this would imply that $\mathbb R$ is countable, a contradiction.
(You could do basically the same thing working with the numbers of the form $\frac1n$, $n\in\mathbb N$, instead of rationals.)
You can also note that we would obtain the same contradiction by assuming that all $A_q$'s are countable, hence this shows that there is an $\varepsilon>0$ and an uncountable set $I$ with the property you're asking about.