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Half-line is given by

$$A=\{x\in R^n \mid x=x_0 + t(a-x_0), \forall t\geq 0\}$$

Is it open or closed set?

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up vote 0 down vote accepted

Clearly the complement of $A$ is open (since the point $x_0$ is not in the complement of $A$). This $A$ is closed.

On the other hand, the set $A$ is not open, so it must be closed (not clopen).

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Why isn't it clopen? Isn't $\infty$ in A? – user23709 May 29 '13 at 13:17
@user23709 It is certainly closed and it's complement is open - so it cannot be open (else its complement would be closed as well)? – gt6989b May 29 '13 at 13:19
But "A set A is closed if its complement is open." is correct, right? – user23709 May 29 '13 at 13:29
@MinimusHeximus - Given that $A$ is at least closed (as was proven before that statement), that $A$ is not open proves it is not clopen, and hence it is just closed. – gt6989b May 29 '13 at 13:29
Thanks to both :) – user23709 May 29 '13 at 13:49

You should consider the different cases:

  1. if $ n = 1 $ then $A=\mathbb{R} $ and $\mathbb{R} $ is your hole space and them open by definition, and closed also by definition since $\varnothing$ is closed.
  2. if $n\geq2$ then clearly $A\neq\mathbb{R} $ and as someone says before, for each point at $A^{C}$ you can find a hole neighborhood which is contained in $A^{C}$ which means $A^{C}$ is open which means $A$ is closed. $A$ is not open since the ball $B_{\varepsilon}\left(x_{0}\right)$ is not contained at $A$ for each $\varepsilon>0$ so $A$ is not open by definition.
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Why is $A= \mathbb{R}$ when $n = 1$? Isn't $A$ equal to $\{x_0\}$ or $(-\infty,x_0]$ or $[x_0,+\infty)$ depending on how $a$ relates to $x_0$? – BU982T May 29 '13 at 14:47
I'm sorry you are right. It's one of the fotms you wrote. which is closed, and not open. – user2359227 May 29 '13 at 14:54

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