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figure1

I can clearly see that I can create disks around the entire perimeter of this triangle where there's at least one point in the disk that's in the triangle, and outside of the triangle. So why is this not a closed set if it contains all the boundary points around the triangle?

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Such a drawing is normally intended to indicate that points on the dotted line are absent-so this set doesn't actually contain all the boundary point of the triangle that forms its closure.

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But I thought boundary points cannot be on the dotted line because the dotted line means it is open, so you cannot select a point on an open line to choose as a boundary point serving as the center of an open disc. So since there are no boundary points on the dotted line, then it does contain all the boundary points of the triangle, doesn't it? –  nabla blah Feb 24 at 3:55
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Don't forget that a set doesn't have to contain (all of) its boundary. For instance an open set is disjoint from its boundary! To recall the definition: the boundary of the triangle consists of points all discs around which intersect both the interior and the exterior of the triangle. That is, the boundary is the union of the three dark lines in the diagram-the solid ones but also the dotted. –  Kevin Carlson Feb 24 at 4:40
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@user130018: If points not in the set couldn't count as boundary points, then every set would by necessity contain its own boundary, and every set would be closed. –  user2357112 Feb 24 at 5:58
    
Thank you. I think I was confusing the domain of a boundary point; if the triangle set is in $\mathbb{R}^2$, then the domain of the boundary point is in $\mathbb{R}^2$, not the set, right? –  nabla blah Feb 24 at 12:07
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Ah yes, that exactly it-the boundary of a set depends fundamentally on the space the set's embedded in. Your argument for the triangle as a subset of itself would extend to show that every set is closed in itself. –  Kevin Carlson Feb 24 at 14:13

Like @Kevin said, the dotted line is not part of the set. You have to remember that if a set is closed then the interior and the boundary are in the set, moreover: if $A$ is closed then $A=int(A)\cup\partial(A)$, where $\partial (A)$ indicates the boundry.

Now acording to your drawing the dotted line is part of $\partial (A)$, but is not part of $A$ so the equality doesn't happen. Also, this set isn't open either, because if you take a point in the black lines, then for any open set surrounding the point is not entirely submerged in $A$

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