Closed Balls in Open Sets

Question

I am brushing up on some real analysis terminology. Suppose we have an open set $E$ and a closed ball $B(x, r)$. Suppose this $r$ is fixed, for example, $r=0.00001$. My question is, can we find this closed ball $B$ in $E$? Or can we find a closed ball in an open ball?

Answer 1

In $\mathbb{R}$, denoting open balls by $B$ and closed ones by $\overline{B}$, we have

$$\begin{align*} B(x;r) &= \{ z \in \mathbb{R} \mid |x-z| < r \} \\ \overline{B}(x;r) &= \{ z \in \mathbb{R} \mid |x-z| \le r \} \end{align*}$$

Thus, given an open ball of radius $r>0$, it is easy to find a closed ball inside: let it have the same center and radius $\rho = r/2$.

For instance, taking $x = 0, r = 1$ gives $$ B(0;1) = (-1,1) \qquad \overline{B}(0;\rho) = \overline{B} \left( 0 ; \frac 1 2 \right) = \left[ - \frac 1 2 , \frac 1 2 \right] \subseteq (-1,1)$$

A generic open set can be written as a union of open balls, so to handle your question for $E$ a generic open set, simply use the above for any particular open ball in said union.

Now, can we find a particular closed ball, or one of a particular radius for any given open set? Not necessarily. If we want a closed ball of radius $r=10$, it's going to be impossible to find one in $B(0;1) = (-1,1)$, for the obvious reason it just is too big. In general, seeking one of radius $r$, just use $r/2$ for your ball.

Answer 2

If your question is "can we always find a closed ball of radius $r = 10^{-5}$ contained in an open set of the real line?", then the answer is no. Let $\varepsilon = 10^{-10000000000}$ (just to make a point). Then obviously the open set $(\pi - \varepsilon, \pi + \varepsilon)$ does not contain any ball of radius $r = 10^{-5}$ (as you fixed it). It can only contain closed balls of radius $< \frac{\varepsilon}{2}$.

On the other hand, if your question is "can we always find a closed ball of radius $\varepsilon$ sufficiently small contained in an open set?", then the answer is yes almost by definition. If $E$ is an open set and $x_0 \in E$ is any point of $E$, then by definition there exists $\delta> 0$ such that the open ball of radius $\delta$ centered at $x_0$ is fully contained in $E$. Then it's clear that the closed ball of radius $\varepsilon \doteq \frac{\delta}{2}$ centered at $x_0$ is fully contained in $E$.