$\lim\limits_{n\rightarrow\infty} B_R\backslash B_{1/n}$ versus $\lim\limits_{n\rightarrow\infty} B_R\backslash \bar B_{1/n}$

Question

If we define $B_R=\{x\in\Bbb R^d\ :\ \|x\|<R\}$ and $\bar B_R$ its closure

Both sequences of sets $$E_n:= B_R\backslash B_{1/n}\text{ and } E_n':= B_R\backslash \bar B_{1/n}\ ,\ n\in\Bbb N$$ are nested sets so $\liminf $ and $\limsup $ exist and coincide for both collections.

But is $\lim\limits_{n\rightarrow\infty}E_n'=B_R\backslash\{0\}$ or $B_R$?

Answer 1

$B_R\setminus\{0\}$ for both. As you said, $\liminf$ and $\limsup$ coincide, so you just need to calculate either.

$$\liminf_{n\to\infty} B_R\setminus \overline {B_{\frac1n}}=B_R\setminus\limsup_{n\to\infty} \overline{B_{\frac1n}}=B_R\setminus \bigcap_{n\in\Bbb N_+}\bigcup_{k\ge n} \overline{B_{\frac1k}}=B_R\setminus\bigcap_{n\in\Bbb N_+} \overline{B_{\frac1n}}=B_R\setminus\{0\}$$

The exact same thing without "$\bar{}$" over the ball carries on identically.