# Doubt on a proof involving a pseudo-metric and elementary topology.

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.

• Have you deleted the other question on this? My answer there contained all the elements... Nov 9, 2021 at 23:11
• $x \in \overline{A}$ iff for all $r>0: V_r(x) \cap A \neq \emptyset$. No need for $A\setminus \{a\}$ stuff... Nov 9, 2021 at 23:12

With the pseudo-metric $$O$$, the only open subsets are $$\emptyset$$ and $$X$$. Therefore, the only closed subsets are $$\emptyset$$ and $$X$$ too. So, if $$A\ne\emptyset$$, $$\overline A=X$$, since $$X$$ is the only closed subset containing $$A$$.

And if $$\rho$$ is any other pseudo-metric, then, for each $$\lambda>0$$, $$\lambda\times\rho$$ is another (distinct) pseudo-metric with respect to which the closed sets are the same as with $$\rho$$. Therefore, the closure of a set with respect to $$\rho$$ is equal to its closure with respect to $$\lambda\times\rho$$.

• Why are the only open subsets $\emptyset$ and $X$ ? I don't understand it! Thanks for your time and help
– xyz
Nov 9, 2021 at 23:03
• If $x\in X$, the open ball centered at $x$ with any radius is the whole $X$. Since the only open ball is $X$, the only non-empty open set is $X$. Nov 9, 2021 at 23:05
• Makes sense to me! How would one prove that $\overline{A} = X \Rightarrow A \neq \emptyset$ ? Is my proof about this right? Thanks once again :D
– xyz
Nov 9, 2021 at 23:08
• Since $\overline\emptyset=\emptyset$, if $\overline A$ is anything other than $\emptyset$, $A\ne\emptyset$. Concerning your proof, I do not understand why is it that you write that “we obviously have that $\overline A=A$”. Nov 9, 2021 at 23:11
• Basically I assumed that $A = \emptyset$ and tried to go for a contradiction. In this case, we have (as you said) $\overline{\emptyset} = \emptyset \Leftrightarrow \overline{A} = A$ (this is where it comes from) and thus (assuming $X$ is non-empty) we have a direct contradiction. I am just not sure about $X$ being non-empty. Thanks for your answers! Really clarified me
– xyz
Nov 9, 2021 at 23:14