Show that in $\mathbb R^2$ every closed set is boundary for some open set.
Please give hint .
I think it is also true for all $\mathbb R^n$ . Is this proof similar to the above ?
Show that in $\mathbb R^2$ every closed set is boundary for some open set.
Please give hint .
I think it is also true for all $\mathbb R^n$ . Is this proof similar to the above ?
To summarise the above comments: it is not true that every closed set in $\mathbb{R}^2$ is the boundary of an open set. Such sets have empty interior, and many closed sets in the plane do have non-empty interior.
The elegant argument at this answer shows that any resolvable space $X$ (which means it can be written as the union of two disjoint dense subsets) has the property that every closed set is the boundary of some set. And all Euclidean spaces are resolvable ($\mathbb{Q}^n$ and its complement will do, e.g.).
So the linked statement at wikipedia is false, but something weaker is true, in a more general setting. [Added] The statement has now been reverted to a correct statement, see the link in the comment below!