# 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 $\mathbb{R}^2$'?

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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

## 2 Answers

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...

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Got it. Thank you –  Katlus Jul 23 '12 at 13:58
It should be a < x < b, without equal signs. –  enzotib Jul 23 '12 at 14:05
Right. Thank you –  DonAntonio Jul 23 '12 at 14:19

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.

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