Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathbb{R}^2$'?

share|improve this question
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
add comment

2 Answers

up vote 3 down vote accepted

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...

share|improve this answer
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
add comment

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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.