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.

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

1 Answer 1


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 $\lambda\times\rho$.

  • $\begingroup$ Why are the only open subsets $\emptyset$ and $X$ ? I don't understand it! Thanks for your time and help $\endgroup$
    – xyz
    Nov 9, 2021 at 23:03
  • 2
    $\begingroup$ 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 $X$. $\endgroup$ Nov 9, 2021 at 23:05
  • $\begingroup$ 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 $\endgroup$
    – xyz
    Nov 9, 2021 at 23:08
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    $\begingroup$ 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 $\overline A=A$”. $\endgroup$ Nov 9, 2021 at 23:11
  • $\begingroup$ 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 $\endgroup$
    – xyz
    Nov 9, 2021 at 23:14

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