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.

Let $ x \in \mathbb{R}^m,$ $ x = [x_i], i = 1,2, \dotsm , m.$ Define $\mathbb{R}^m_+ = \lbrace x \in \mathbb{R}^m : x_i \geqslant 0, 1 \leqslant i \leqslant m \rbrace.$ What the boundary set $ \partial \mathbb{R}^m_+$ of the set $ \mathbb{R}^m_+$? Is not it the set $ \lbrace{ 0\rbrace} \in \mathbb{R}^m_+?$

share|improve this question
    
Do it for $m=1,2,3$ then you should see the pattern. –  t.b. Jan 11 '12 at 9:50
2  
It is the set of those points that are zero in at least one coordinate. –  Michael Greinecker Jan 11 '12 at 9:51
1  
Be careful of the "0% accept rate". –  Paul Jan 11 '12 at 9:52
1  
@Paul what should I do to improve the accept rate? –  Zizo Jan 11 '12 at 10:12
2  
@Suso: Clicking your own name, you can look at your own user page. You can find all the questions you have asked in your user page. To accept an answer, you can go to the questions you have asked, and then click the checkmark, as explained in the link t.b provided here: meta.math.stackexchange.com/questions/3286/… –  Paul Jan 11 '12 at 10:42
show 3 more comments

1 Answer

up vote 1 down vote accepted

The definition of boundary I'm assuming is $\partial X = \bar X \smallsetminus X^{\circ}$, where $X^{\circ}$ is the (topological) interior of $X$.

It's not too hard to check that the interior of $\mathbb{R}^m_+$ is the set $$ \{x = (x_1, \dots, x_m) \in \mathbb{R}^m\, :\, x_i > 0\ \text{for all}\ 1 \le i \le m \}$$

This leaves the set of points at least one of whose coordinates is zero as the boundary: any open set around such a point must contain a point with a strictly negative coordinate, which does not lie in $\mathbb{R}^m_+$.

Intuitively, you have an infinite cube one of whose vertices is the origin and the rest of whose vertices are 'at infinity' in the positive directions.

share|improve this answer
1  
It doesn't matter in this example, since $\mathbb{R}^m_+$ is already closed, but the boundary is $\bar{X}\backslash X^\circ$. –  Michael Greinecker Jan 12 '12 at 10:02
    
The definition of boundary is not $X \backslash X^{0}$. Consider e.g. the rationals $\mathbb{Q}\subset \mathbb{R}$. Its boundary is $\mathbb{R}$, while $\mathbb{Q}^{0}=\emptyset$, since $\mathbb{Q}$ contains no intervals. –  Thomas E. Jan 12 '12 at 10:07
    
@ThomasE.: My bad, fixed, thanks. –  Clive Newstead Jan 12 '12 at 10:43
add comment

Your Answer

 
discard

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.