# Question about open and closed sets

• March 14th 2011, 03:37 AM
sakodo
Question about open and closed sets
Hi guys,

I have two questions wanting to clarify.

1) My lecture notes say that if a set is open then its complement is closed. However some books also say that the converse is true: that is if a set is closed, then its complement is open. Is this correct?

2) Is a boundary point a limit point? The definition says that a boundary point x which belongs to S is a point if, for all epsilon > 0, the neighborhood of x contains both S and the complement of S. Doesn't this make x a limit point also? Since you can approach x from elements in the set S.

• March 14th 2011, 04:17 AM
Plato
Quote:

Originally Posted by sakodo
1) My lecture notes say that if a set is open then its complement is closed. However some books also say that the converse is true: that is if a set is closed, then its complement is open. Is this correct?

2) Is a boundary point a limit point? The definition says that a boundary point x which belongs to S is a point if, for all epsilon > 0, the neighborhood of x contains both S and the complement of S. Doesn't this make x a limit point also? Since you can approach x from elements in the set S.

Your #1 is correct. If $\mathcal{O}$ is open then $\mathcal{O}^c$ is closed.
If $\mathcal{C}$ is closed then $\mathcal{C}^c$ is open.

For #2 consider $\mathcal{A}=\{0\}\cup [1,2]$.
Then $0$ is a boundary point of $\mathcal{A}$ but not a limit point.
• March 14th 2011, 05:22 AM
HallsofIvy
Quote:

Originally Posted by sakodo
Hi guys,

I have two questions wanting to clarify.

1) My lecture notes say that if a set is open then its complement is closed. However some books also say that the converse is true: that is if a set is closed, then its complement is open. Is this correct?

Indeed, in most texts, a "closed" set is defined as "the complement of an open set" and then it is proved that the complement of a closed set is an open set.

Quote:

2) Is a boundary point a limit point? The definition says that a boundary point x which belongs to S is a point if, for all epsilon > 0, the neighborhood of x contains both S and the complement of S. Doesn't this make x a limit point also? Since you can approach x from elements in the set S.

The definition of "boundary point" is, as you say, that every neighborhood contains points of S and of the complement of S. The definition of "limit point" is that every neighborhood contains points of S other than X itelf. Plato's example gives a point, 0, such that every neighborhood of 0 contains only 0 from the set. It is a boundary point but not a limit point. Such points are called "isolated points" of S.
• March 14th 2011, 05:24 AM
HallsofIvy
Quote:

Originally Posted by Plato
Your #1 is correct. If $\mathcal{O}$ is open then $\mathcal{O}^c$ is closed.
If $\mathcal{C}$ is closed then $\mathcal{C}^c$ is closed.

Plato meant "open" as the last word here.

Quote:

For #2 consider $\mathcal{A}=\{0\}\cup [1,2]$.
Then $0$ is a boundary point of $\mathcal{A}$ but not a limit point.