closed sets which are strictly contained in open set I wanted to ask the following:
Suppose I am in the complex plane with the usual topology.
Suppose I have a closed set A such that A ⊂ B, where B is an open set. I wanted to know if I can always find a closed set A' such that A ⊂ A' ⊂ B?
But with one more demand - That the smallest distance between boundaries of A and A' will be strictly positive. You can visualize my question as: is it possible to "inflate" A a bit so it could still be totally contained in B?
Thank you in advance.
 A: No. Let $B = \{(x,y):\space y < 1/(1+x^2)\}$ and $A$ the $x$-axis.
EDIT: If $A$ is compact, the answer is yes.
Proof: For each point $x\in A$ put a closed ball that is inside $B$ to $x$. Since $A$ is compact it's covered with finitely many of these balls. Take the union of the balls as $A'$. To make sure you "inflate", cover the boundary of $A$ "thickly enough" (i.e make sure you don't have a place where two balls would be just tangent to each other).
This would need some argument to make it rigorous but hope you get the idea.
A: Assuming A compact, then the continuous function $d(A,B)$  takes on a max and a min. Then, if $A \subset B, d(A,B)>0$ , and the result is possible.
I think this should work: Assume A compact, contained in B. Then, for every $a \in A $, there is a ball $B(a,r) \subset B ; r>0$ . Cover the boundary of $B$ by these balls $B(a,r)$ and take a finite subcover. In this finite subcover, take  EDIT $Min r: B(a,r) \subset B $. This $r$ should give you the room to expand ino $B$ while still being contained in it.
