Equality between two definitions of the same set Let $\Omega\subseteq\mathbb{R}^{n}$ be an open set. For every  $\epsilon>0$ we define the next set:
$$\Omega_{\epsilon}:=\left\lbrace x\in\Omega:d(x,\Omega^{\mathsf{c}})>\epsilon\right\rbrace$$
Prove that this is an open set, and that $\Omega_{\epsilon}$$ can also be defined as the following:
$$\Omega_{\epsilon}:=\left\lbrace x\in\Omega:\overline{B_{\epsilon}}(x)\subseteq\Omega\right\rbrace $$
P.D. The answer to the question doesn't seem very complicated but I am stuck proving the equality between the two definitions. For the openness of the set, I defined a $B_{\gamma}(x)$ with $\gamma>\epsilon$ and with that in mind, it is easy to see that the set open, but for the second part Im stuck. I just feel that the two definitions are not equal as for every $\epsilon$ I could give a $\gamma$ that is big enough for it to be out of the second definition. If someone can help me I would really appreciate it.
 A: For Simplicity, let's call:
$$ \Omega^1_\epsilon :=\left\lbrace x\in\Omega:d(x,\Omega^{\mathsf{c}})>\epsilon\right\rbrace$$
for your first set, and
$$ \Omega^2_\epsilon:=\left\lbrace x\in\Omega:\overline{B_{\epsilon}}(x)\subseteq\Omega\right\rbrace $$
for your second set.
For openness, I'm not sure I fully understand how your argument works. Here's how I can see that $\Omega^1_\epsilon$ is open: Suppose by way of contradiction that $\partial\Omega^1_\epsilon\cap\Omega^1_\epsilon \ne \emptyset$. Let $x \in \partial\Omega^1_\epsilon$. Then $d(x,\Omega^c) > \epsilon$.
Fix $\delta > 0$. Then since $x$ is a boundary point, we have that $B_\delta(x) \cap (\Omega_\epsilon^1)^c \ne \emptyset$. Choose some $y_\delta \in B_\delta(x) \cap (\Omega_\epsilon^1)^c$. Then by the triangle inequality,
$$\begin{align}
d(x,\Omega^c) &\le d(x,y_\delta) + d(y_\delta,\Omega^c)\\
&\le \delta + \epsilon
\end{align}$$
And since $\delta$ is arbitrary, we have that
$$ d(x,\Omega^c) \le \epsilon $$
a contradiction. Hence, $\partial\Omega_\epsilon^1\cap\Omega^1_\epsilon = \emptyset$ so that $\Omega_\epsilon^1$ is open.
Next, to show that the two sets are equal:
$(\subseteq)$. Let $x \in \Omega_\epsilon^1$. Let $y \in \overline{B_\epsilon}(x)$. Then,
$$\begin{align}
d(x,y) &\le \epsilon
\end{align}$$
so that if $y \notin \Omega$, then we would contradict $d(x,\Omega^c) > \epsilon$. Hence, $y \in \Omega$ so that $\overline{B_\epsilon}(x) \subseteq \Omega \Rightarrow \Omega_\epsilon^1 \subseteq \Omega_\epsilon^2$.
$(\supseteq)$. Conversely, suppose that $x \in \Omega_\epsilon^2$. Let $y \in \Omega^c$. Suppose by way of contradiction that $d(x,y) \le \epsilon$. Then $y \in \overline{B_\epsilon}(x) \subseteq \Omega$, a contradiction. Hence, $d(x,y) > \epsilon \Rightarrow d(x,\Omega^c) > \epsilon$ since $y \in \Omega^c$ was arbitrary. It follows that $\Omega_\epsilon^2 \subseteq \Omega_\epsilon^1$.
