why ${0}×R$ is not open in $R^2$?

Question

${0}×R $is open in the product topology $R_d×R $(and, hence, in the dictionary topology on R×R ), where R_d is discrete topology on R .

now my question is that why ${0}×R$ is not open in $R^2$

Answer 1

This was a very curious question for me when I was studying Calculus of $\mathbb{R}^n$. First $\{0\}\times\mathbb{R}$ is basically the interval $(-\infty,\infty)$ say $I$ and let $a\in I$. Then $\exists$ no $\varepsilon>0$ such that the open ball (since in $\mathbb{R}^2$ open sets are open balls) $B(a;\varepsilon)\subset I$. Hence $a$ is not an interior point of $I$. So $I$ is not open in $\mathbb{R}^2$.