# 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 $S \subset \text{int}(\text{dom} f))$?

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 $$S \subseteq \operatorname{int}(\operatorname{dom} f).$$

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.