There exists $R>r$ such that the open ball $B(x,R)$ of radius $R$ centered at $x$ such that $B(x,R)\subset \mathcal{U}$

Let $\mathcal{U}$ be an open set of $\mathbb{R}^n$ and $x\in \mathcal{U}$ such that there exists $r>0$ such that the closed ball $\bar{B}(x,r)$ of radius $r$ centered at $x$ such that $\bar{B}(x,r)\subset \mathcal{U}$.

Can we say that "there exists $R>r$ such that the open ball $B(x,R)$ of radius $R$ centered at $x$ such that $B(x,R)\subset \mathcal{U}$".

Solution 1:

Yes, we can. Otherwise, for every $n\in\Bbb N$, there would be some $x_n\in \overline{B\left(x,r+\frac1n\right)}\setminus\mathcal U$. But every $x_n$ belongs to $\overline{B(x,r+1)}$, which is compact, and therefore $(x_n)_{n\in\Bbb N}$ has a subsequence which converges to some $x\in\overline{B(x,r)}$. This is impossible, since every $x^n$ belongs to $\mathcal U^\complement$, which is a closed set, and therefore $x$ should also belong to $\mathcal U^\complement$. But $\overline{B(x,r)}\subset\mathcal U$.