Prove that a nest of sets has an empty intersection Let $f$ be a real convex function and $S$ an arbitrary closed bounded subset of the relative interior of the effective domain of $f$. Let $B$ be a closed Euclidean unit ball. The nest of sets
$$(S + \varepsilon B) \cap (\mathbb{R}^n \backslash \text{int}(\text{dom} f)), \quad \varepsilon > 0$$ has an empty intersection. Why? Is it because $\text{cl}S=\bigcap \{S + \varepsilon B \big | \varepsilon > 0\}$ and $S \subset \text{int}(\text{dom} f))$? 
 A: Notice that
$$\bigcap_{\varepsilon>0} \left((S+\varepsilon B) \cap (\mathbb R^n\setminus\operatorname{int}(\operatorname{dom} f))\right)
= \left(\bigcap_{\varepsilon>0} (S+\varepsilon B)\right) \cap (\mathbb R^n\setminus\operatorname{int}(\operatorname{dom} f)).$$
As you said, the intersection appearing on the RHS is simply the closure, $\overline S=\bigcap_{\varepsilon>0} (S+\varepsilon B)$. And since $S$ is closed, we have $S=\overline S$.
So we want to show in fact that
$$S\cap (\mathbb R^n\setminus\operatorname{int}(\operatorname{dom} f))$$
which is equivalent to
$$S \subseteq \operatorname{int}(\operatorname{dom} f).$$
A: Proof. 
If not, there exists $x_0 \in \cap_{\xi>0} H_{\xi}$,
where $H_{\xi}= (S+\xi B) ~\cap Q$ with $Q =( int~ domf)^{c} $.
Notice that $S$ is in the interior of domf. 
For any $y\in S$, there exist $\alpha_{y}>0$ such that
$$B(y,0.5\alpha_{y})\subset B(y,\alpha_{y})\subset int ~ dom f.$$
It follows that $$S\subset\cup_{\alpha_{y}>0} B(y,0.5\alpha_{y}).$$
Since $S$ is a closed bounded set of $R^{n}$, $S$ is compact.
Thus, $$S\subset \cup_{i=1}^{m} B(y_i,\beta_i),$$
where $\beta_i = 0.5 \alpha_{y_{i}}$ and $m$ is an positive integer.
Thus, there exist $\xi_0= \min \{\beta_i\mid i= 1,2,\cdots,m\}$ such that 
$$S +\xi_0 B\subset \ int ~dom f .$$
Together with the above inclusion, $x_0 \in H_{\xi_0}$ implies that 
$$x_0 \in int ~dom f.$$
However, $x_0 \in H_{\xi_0} \subset Q$ which shows us a contradiction.
