A question about open sets containing a closed disc I'm reading a book on analysis and it says that in $\mathbb{R}^n$ with usual topology, any open set containing a closed ball also contains an open ball which contains the closed ball. But it doesn't seem clear to me. Could anyone tell me why it is so?
 A: Let $U$ be open and $C \subset U$ be compact. Let $B$ be the open unit ball centred at the origin. Let $U_n = C+{1 \over n}B$, which is easily seen to be open and $C \subset U_n$.
Then $U_n \subset U$ for some $n$. To see this, suppose it is not true, then there exists some $u_n \in U_n \setminus U$ for all $n$. By construction, we have
$u_n = c_n + {1 \over n} b_n$, where $c_n \in C, b_n \in B$. Since $c_n \in C$, we have $c_{n_k} \to c \in C$ for some subsequence $n_k$. It follows that $u_{n_k} \to c$. However, since $U^c$ is closed, we have $c \in U^c$ which 
contradicts $c \in U$.
Since the closed ball in $\mathbb{R}^n$ is compact, the result follows.
A: This is not a solution but a suggestion and one that needs verification by someone more qualified.  
Let $ B = \{ y \in \Bbb R^n \ | \ |x - y| \le r \}$ be the supposed closed contained in the given open set $A$. Now consider the points on the boudnary. That is the set $ \overline B = \{ y \in \Bbb R^n \ | \ |x - y| = r \} $. Now for each $x \in \overline B$, $x \in A$ and hence there is $\delta_x$ such that the ball centered at $x$ with radius $\delta_x \gt 0$ is contained in $A $. Now take $\delta = \inf \{ \delta_x \ | \ x \in \overline B\} $. Then the open ball $ \{  y \in \Bbb R^n \ | \ |x - y| \lt r + \delta \} $ contains $B$ and is a subset of $A$. 
