Show that interval $(a, b)$ is not open in $\mathbb{R}^2$ I know that interval $(a, b)$ is open in $\mathbb{R}$. To show that interval $(a,b)$ is open in $\mathbb{R}$, I have done so: Let it be $x\in (a,b)$. Enough to find an open ball containing the point $x$, and that is included in the interval $(a,b)$. Suffice to get $0<\epsilon\leq \min \{\vert b-x\vert, \vert a-x \vert\}$. In this case $D(x,\epsilon)$ containing the point $x$ and $D(x,\epsilon)\subseteq (a,b)$. I do not know a good act. If it is that I did very well then correct, but I do not know how to prove this fact: To show that interval $(a,b)$ is not open in $\mathbb{R^2}.$
Please if anyone has the opportunity to help, to make verification of keti example, thank you preliminarily

As kahen pointed out, what you want to say is that $$(a,b)\times\{0\}$$ is not open in $\Bbb R^2$. Now, pick any point ${\bf x}=(x,0)$ with $a<x<b$. Then $B({\bf x},\varepsilon)$ contains elements of the form $(x_1,x_2)$ with $x_2\neq 0$ (prove this!), so $B({\bf x},\varepsilon)$ cannot be contained in $(a,b)\times \{0\}$. This means $(a,b)$ is not open, since we've found a point (actually all of them) that are not interior points. In fact, ${\rm int}_{\Bbb R^2}\;(a,b)=\varnothing$.
