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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_+?$

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Do it for $m=1,2,3$ then you should see the pattern. – t.b. Jan 11 '12 at 9:50
It is the set of those points that are zero in at least one coordinate. – Michael Greinecker Jan 11 '12 at 9:51
What do you mean with the notion $x\in \mathbb{R}^{m}$ and $x\geq 0$? Did you mean that $x_{i}\geq 0$ for all $i\in \{1,...,m\}$? – T. Eskin Jan 11 '12 at 9:57
@Paul: it would probably be much better and much more effective if you explained what that means. E.g. by linking to the faq or to this thread – t.b. Jan 11 '12 at 10:00
@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:… – Paul Jan 11 '12 at 10:42
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.

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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. – T. Eskin Jan 12 '12 at 10:07
@ThomasE.: My bad, fixed, thanks. – Clive Newstead Jan 12 '12 at 10:43

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