Why is the Triangle Set Not Closed?

[E7.5]

Homework Statement

Homework Equations

A set is closed if it contains alll of its boundary points.

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.

The Attempt at a Solution

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?

 

[Dick]

Homework Statement

Homework Equations

A set is closed if it contains alll of its boundary points.

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.

The Attempt at a Solution

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?

If you mean that the set in question is the interior of the triangle with all of the boundary edges included then it is closed. If you mean the interior of the triangle with one boundary edge omitted, (which drawing the dashed line usually means) then it's not closed. What do you mean?

 

[E7.5]

If you mean that the set in question is the interior of the triangle with all of the boundary edges included then it is closed. If you mean the interior of the triangle with one boundary edge omitted, (which drawing the dashed line usually means) then it's not closed. What do you mean?

I mean the interior of the triangle with one edge omitted. Why is it not closed then, according to the fact that a set is closed if it contains all of its boundary points? Doesn't the figure, even with one edge omitted, contain all the boundary points of its edges?

 

[Dick]

I mean the interior of the triangle with one edge omitted. Why is it not closed then, according to the fact that a set is closed if it contains all of its boundary points? Doesn't the figure, even with one edge omitted, contain all the boundary points of its edges?

It doesn't contain the edge you omitted. The edge contains boundary points (as your diagram shows) but the edge isn't in the set.

 

[E7.5]

It doesn't contain the edge you omitted. The edge contains boundary points (as your diagram shows) but the edge isn't in the set.

So is it then not always true that a set is closed if it contains all of its boundary points?

 

[Dick]

So is it then not always true that a set is closed if it contains all of its boundary points?

No, it's always true that a closed set contains all of its boundary points. Your set ISN'T closed because it doesn't contain all of its boundary points. The missing edge contains boundary points, the missing edge isn't in the set. Is it?

 

[E7.5]

No, it's always true that a closed set contains all of its boundary points. Your set ISN'T closed because it doesn't contain all of its boundary points. The missing edge contains boundary points, the missing edge isn't in the set. Is it?

But the missing edge is open, so there cannot be any points on that edge to be the center of a disc acting as a boundary point, can there?

 

[vela]

Wouldn't that imply that all open sets are closed because they don't contain any edge points to act as the center of a disc acting as a boundary point?

 

[Dick]

But the missing edge is open, so there cannot be any points on that edge to be the center of a disc acting as a boundary point, can there?

There are plenty of points on that "missing edge". They just aren't in your set.