The definition of closure. 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.
 A: 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.
A: 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.
A: 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:
$$
\bigcap_{\varepsilon>0} (A_\varepsilon+B_\varepsilon) \color{red}\neq \bigcap_{\varepsilon>0} A_\varepsilon + \bigcap_{\varepsilon>0} B_\varepsilon.
$$

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
$$
C + \bigcap_{\varepsilon > 0} \varepsilon B = C + \{0\} = C = (0,1) \color{red}\neq \operatorname{cl}\,(0,1).
$$
