# Segment of $\mathbb{R}^2$?

I don't understand this sentence;

The segment $(a,b)$ can be regarded as both a subset of $\mathbb{R}^2$ and an open subset of $\mathbb{R}^1$. If $(a,b)$ is a subset of $\mathbb{R}^2$, it is not open, but it is an open subset of $\mathbb{R}^1$.

What is 'segment $(a,b)$ in $\mathbb{R}^2$'?

-
Imagine the real line $\mathbb{R}$ embedded in $\mathbb{R}^2$, and then imagine the open interval$(a,b)$ (in the real line) embedded in the whole plane $\mathbb{R}^2$. – Old John Jul 23 '12 at 13:44
@John so informally speaking, is it the rectangular region with width $\mathbb{R}$ and height $(a,b)$ in cartesian plane? – Katlus Jul 23 '12 at 13:53
No - I am pretty sure they mean the line from $a$ to $b$ along the $x$-axis. – Old John Jul 23 '12 at 13:55

That sentence is a good example of poor use of mathematical statements. I guess they meant the set $$\{(x,y_0)\in\Bbb R^2\;:\;a< x< b\,\,,\,y_0\in\Bbb R\,\,\text{fixed}\}$$
Taking $\,y_0=0\,$ above gives an interval on the x-axis in the plane...
An open set in the euclidean topology is an open interval. Equivalently we can look at "open balls": a set is open if, for all elements $x$ in the set, there is some neighborhood around $x$ that is still entirely contained in the set. For an interval $(a,b)$ in $\mathbb{R}$ this is true. In $\mathbb{R}^2$ this set of points would be the line from $(a,0)$ to $(b,0)$, non-inclusive. Here it is not true, because open balls in $\mathbb{R}^2$ have height as well as width. A line doesn't have any height whatsoever, so it doesn't contain any open balls and can't possibly be an open set itself.