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I found a set problem in one of the lecture notes which has asked whether the following set open.

$A $={ $(x,y) | 1<x<3 , y=0$ }

Above set is not open. Since the complement of this set does not include limit point ( let's say $(3,0)$ ) this is not closed. So obviously A is not open.

But that doesn't mean A is a closed set. However I though since A doesn't have point $(1,0)$ or $(3,0)$ ( which are limit points of A (I think) ) included, it cannot be closed. So what I want to know is whether what I thought is correct or not. Whether that set A is closed or neither closed nor open?

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    $\begingroup$ You are correct, apart from your choice of possible limit points: as a subset of $\Bbb R^2$ it is not open, because it does not contain any $\epsilon$-ball at all, and it is not closed, because $\langle 1,0\rangle$ and $\langle 3,0\rangle$ are limit points of it but not in it. $\endgroup$ – Brian M. Scott Nov 1 '16 at 16:59
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Yes, as a subset of $\mathbb{R}$ this is neither open nor closed. The limit points of the set are $(1,0)$ and $(3,0)$ and they are not in the set so the set cannot be closed. However, $(2,0)$ is a limit point of the complement of the set that doesn't fall in the complement of the set, so it's complement cannot be closed either.

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    $\begingroup$ Oh sorry. There was another question which has 2<x<5 there. That's why I mistakenly put those points. I'll correct it. Thank you. $\endgroup$ – Samitha Nanayakkara Nov 1 '16 at 17:06
  • $\begingroup$ @User9125 I have edited my answer to take this into account. I also spotted a minor flaw in what you said which I have corrected. $\endgroup$ – Stella Biderman Nov 1 '16 at 19:54

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