Exercise. Let $d$ be a pseudo-metric in $X$ and for $A,B \subset X$ we have $d(A,B) = \text{inf}\{d(a,b): a \in A, b \in B\}$. Show that:
There is only one pseudo-metric in $X$ s.t. $\overline{A}=X \Leftrightarrow A \neq \emptyset$. For any other pseudo-metric there are infinitely many others that define the same closure.
My attempt. Let $O: X \times X \rightarrow \mathbb{R}$ a pseudo-metric (trivial to prove) on $X$ s.t. \begin{equation*} O(A,B) = 0, \quad \forall A,B \in X \end{equation*} Let us now prove that the initially presented equivalence is valid.
$(\Rightarrow)$ Assume $\overline{A} = X$ and $A = \emptyset$. Then we obviously have that $\overline{A}=A=\emptyset \neq X$ (assuming that $X \neq \emptyset$) which is a contradiction and thus proving the first implication.
$(\Leftarrow)$ Let us now suppose that $A \neq \emptyset$, i.e., $A$ contains at least one element. Let $a \in A$. For every $a' \in \overline{A}$ we either have that $a' \in A$ or $a' \in A'$. The case where $a' \in A$ obviously implies $a' \in X$, since $A \subset X.$ On the other hand, let $a' \in A'$. Then, by definition: \begin{equation*} \forall r>0, V_r(a') \hspace{.15cm} \cap \hspace{.15cm} A\backslash\{a'\} \neq \emptyset \end{equation*} Putting $r\rightarrow 0$, $a'$ will be as close as we want to an element of $A$, and since $A \subset X$ we can say that $a' \in X.$ We have just proved that $\overline{A} \subset X$.
Doubts. I am stuck on proving that $X \subset \overline{A}$ to complete the proof. But how can I show that a bigger set is contained on a subset of his own? I assume this is where the initially defined pseudo-metric comes in, but I don't know how to do it. Besides this, I am also having trouble proving the unicity of the affirmation (I know the standard process to do so would be assuming there's a metric $d'$ that verifies the equivalence and then showing that $d' = O$ in every possible scenario but I can't get to it). Thanks for all the help in advance.
