Show that a subset of $\mathbb{C}$ is compact. Let $\Omega$ be a bounded set in $\mathbb{R}^n$ and $s>0$. Set
$$\Omega_s := \bigcup_{a\in\Omega} {B(a,s)}$$
where 
$$B(a,s) = \left\{ z \in \mathbb{C}^n : \sup_{1 \leq i \leq n} {|z_i - a_i|} < s \right\}.$$
Is the closure $\bar{\Omega}_s$ of $\Omega_s$ compact?

I'm guessing the answer is yes. To show it, I need to show that $\bar{\Omega}_s$ is totally bounded, right? And that will follow if $\Omega_s$ is totally bounded. Now, how do I proceed?
 A: First, recall that a set in $\mathbb{R}^n$ is compact if an only if it is closed and bounded; this is the Heine-Borel theorem. You mix a bit $\mathbb{R}^n$ and $\mathbb{C}^n$ but note that for these things it does not really change much since $\mathbb{R}^{2n}$ and $\mathbb{C}^n$ are from a topological/metric space point of view 'essentially the same'. 
Now, $\bar{\Omega}_s$ is certainly closed, as the closure of a set. 
It thus remains to show it is bounded. Since $\Omega$ is bounded, there exists some constant $C>0$ such that $|z_i| < C$ for each $(z_1, \dots, z_n)\in \Omega$. 
From this using the triangle inequality you get that $|z_i| < C+s$ for each $(z_1, \dots, z_n)\in \Omega_s$. 
To finish you can either invoke some results relating the boundednes for sets and their closure. Or, you say, fix some $\epsilon> 0$, say take $\epsilon =1$. 
Then for each element $(z_1, \dots, z_n)$ in $\bar{\Omega}_s$ there is some element $(y_1, \dots, y_n)$ in $\Omega_s$ such that $|z_i - y_i|< 1$. 
Using again the triangle inequality you arrice at a bound of $C+s+1$ for the elements in $\bar{\Omega}_s$. So the set is bounded. 
Hence, it is closed and bounded and thus compact.    
