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February 23rd, 2014, 06:08 PM   #1
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Why is this set not closed?

http://i.imgur.com/Z37YCsM.png

So, a set is closed if it contains all of its boundary points, and a boundary point is a point where an open ball around that point has one point inside the ball that's in the set, and one point in the ball that's not in the set.

As seen in the picture, I can create infinitely many open balls around the edges of the triangle, with the balls containing at least one point in the triangle (red point) and one point outside the triangle (blue point). Since the perimeter of the triangle has all the boundary points, this triangle should be closed, but it's not. Why isn't it, if it contains all its boundary points?
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February 23rd, 2014, 08:15 PM   #2
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Re: Why is this set not closed?

Maybe this will help.

http://www.youtube.com/watch?v=FHL4udeLf9Q

My signature (at the bottom of my post) has a link to the entire series of videos about the basics of higher mathematics , if you're interested.

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February 25th, 2014, 04:56 AM   #3
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Re: Why is this set not closed?

Quote:
Originally Posted by E7.5
http://i.imgur.com/Z37YCsM.png

So, a set is closed if it contains all of its boundary points, and a boundary point is a point where an open ball around that point has one point inside the ball that's in the set, and one point in the ball that's not in the set.

As seen in the picture, I can create infinitely many open balls around the edges of the triangle, with the balls containing at least one point in the triangle (red point) and one point outside the triangle (blue point). Since the perimeter of the triangle has all the boundary points, this triangle should be closed, but it's not. Why isn't it, if it contains all its boundary points?
It's impossible to answer this without knowing what notation you are using. Specifically what is the distinction between the solid line and the dashed line? Does the solid line indicate points of the boundary that are in the set and the dashed line boundary points that are not in the set? If so, that should answer your question immediately.
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