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
 A: 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, ${\rm int}_{\Bbb R^2}\;(a,b)=\varnothing$.
A: suppose Interval $(a,b)$ is open  in $\mathbb{R}^2$ then  Its complement $\mathbb{R}^2\setminus (a,b)$ must be closed, so  $\mathbb{R}^2\setminus (a,b)$ contains all its limit points.
but $ x_n=({a+b\over 2},{1\over n})$ is a sequence in $\mathbb{R}^2\setminus (a,b)$ and $x_n\to ({a+b\over 2},0)$ which is a point in open interval $(a,b)$ on $X$ axis so we get a contradiction.
A: Hint $x + \frac{\epsilon}{2}(0,1) \in D_\epsilon (x)$...
As $x + \frac{\epsilon}{2}(0,1)$ has a non-zero $y$-coordinate, it is not in $(a,b)$.
P.S. If $a,b$ are not real numbers, but points in $\mathbb R^2$, you can do the same by replacing $(0,1)$ by an unit vector $u$ such that $u \perp {ab}$.
