1. ## 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.

2. 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 $\displaystyle \mathcal{O}$ is open then $\displaystyle \mathcal{O}^c$ is closed.
If $\displaystyle \mathcal{C}$ is closed then $\displaystyle \mathcal{C}^c$ is open.

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

3. 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.

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 $\displaystyle \mathcal{O}$ is open then $\displaystyle \mathcal{O}^c$ is closed.
If $\displaystyle \mathcal{C}$ is closed then $\displaystyle \mathcal{C}^c$ is closed.
For #2 consider $\displaystyle \mathcal{A}=\{0\}\cup [1,2]$.
Then $\displaystyle 0$ is a boundary point of $\displaystyle \mathcal{A}$ but not a limit point.