The definition of closure.

Question

Recently, i have seen the definition of closure as follows, \begin{equation} \text{cl}~C = \bigcap_{\varepsilon>0}(C + \varepsilon B), \end{equation} where $B$ is Euclidean unit ball: $B=\{x \mid |x| \leq 1\}$. I think it is right but i feel a little confused, because \begin{equation} \bigcap_{\varepsilon>0} \varepsilon B = \lim_{\varepsilon \downarrow 0}\varepsilon B = \{\boldsymbol{0}\}. \end{equation} Therefore, could i say \begin{equation} \begin{aligned} \text{cl}~C &= \bigcap_{\varepsilon > 0}(C + \varepsilon B) \\ &= \lim_{\varepsilon \downarrow 0}(C + \varepsilon B) \\ &= C + \lim_{\varepsilon \downarrow 0} \varepsilon B \\ &= C + \{\boldsymbol{0}\}. \end{aligned} \end{equation} I think it is incorrect but i don't know why, thanks in advance.

Answer 1

This is the part where you made a mistake: $$ \lim_{\varepsilon \downarrow 0}(C + \varepsilon B) = C + \lim_{\varepsilon \downarrow 0} \varepsilon B $$ The operation of taking limit here is not distributive with respect to Minkowski sum.

Answer 2

The easiest way to analyze such a proof is by using a small example. Take $C=(-\infty,0)$, so the closure is $(-\infty,0]$. You say $\lim_{\varepsilon \downarrow 0}((-\infty,0) + \varepsilon B) = (-\infty,0) + \lim_{\varepsilon \downarrow 0}(\varepsilon B)$, which does not seem right.

Answer 3

Writing an intersection $\bigcap_{\varepsilon>0}^\infty A_\varepsilon$ of sets satisfying $A_x \subset A_y$ for $x<y$ as a limit is misleading to say the least. In this case it lead you to the false believe that this intersection is compatible with (Minkowski) sums.

That is, in general: $$ \bigcap_{\varepsilon>0} (A_\varepsilon+B_\varepsilon) \color{red}\neq \bigcap_{\varepsilon>0} A_\varepsilon + \bigcap_{\varepsilon>0} B_\varepsilon. $$

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Consider the example where $C = (0,1)\subset \Bbb R$ and $B=[-1,1]$ is the closed unit ball. Note that $$ \bigcap_{\varepsilon > 0} (C+\varepsilon B) = \bigcap_{\varepsilon > 0} (-\varepsilon, 1+\varepsilon) = [0,1] = \operatorname{cl}\,(0,1) $$ while $$ C + \bigcap_{\varepsilon > 0} \varepsilon B = C + \{0\} = C = (0,1) \color{red}\neq \operatorname{cl}\,(0,1). $$