Open subset of a perfect set.

[kostas230]

Suppose we have a perfect set $E\subset\mathbb{R}^k$. Is there an open set $I\subset E$?

 

[mathman]

I'm rusty at this. However, I understand that a closed interval is a perfect set. Take the closed unit cube in R^k, drop all boundary points leaving an open unit cube.

 

[Office_Shredder]

Sometime's it's true (like mathman's example), and sometimes it's false. For example the Cantor set is a perfect set but contains no open interval inside of it

 

[kostas230]

Yeah, I just found out that. Thank you :)

 

[blue_raver22]

Yes, it is possible for there to be an open subset I of a perfect set E. This is because a perfect set is defined as a set that is equal to its set of limit points, meaning that every point in E is a limit point of E.

Furthermore, an open set is a set where every point has a neighborhood contained within the set. Therefore, if we take any point x\in E, we can find a neighborhood around x that is contained within E, making it an open set.

In summary, since every point in E is a limit point, we can construct a neighborhood around each point that is contained within E, making it an open subset of E.