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_+?$
Let $\mathbb{R}^m_+ = \lbrace x \in \mathbb{R}^m : x \geqslant 0 \rbrace.$ What is the boundary set of the set $ \mathbb{R}^m_+$?
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general-topology
differential-topology
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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: http://meta.math.stackexchange.com/questions/3286/how-do-i-accept-an-answer – 2012-01-11
1 Answers
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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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0@ThomasE.: My bad, fixed, thanks. – 2012-01-12