Is there a meaning to convergence, limits and closedness in pseudo-metric spaces?

Question

A. A sequence ($x_n$) in a metric space $M$ is said to converge to the limit $x \in M$ if the distance between $x_n$ and $x$ converges to 0 as $n$ goes to infinity.

What happens when $M$ is a pseudo-metric space? It seems that every convergent sequence can have many limits, as long as the distance between all the limits is 0. Is this true?

B. A subset $A$ of the metric space $M$ is closed iff every sequence in $A$ that converges to a limit in $M$ has its limit in $A$.

Does this definition have a meaning in a pseudo-metric space? For example, does it make sense to define a subset $A$ as "closed" iff for every sequence in $A$, all its limits in $M$ are also in $A$? Is this (or a different) definition used anywhere?

Answer 1

Ya it may happen. One trivial example : If we define a pseudo-metric on a non-empty set $X$ as $d:X\times X\rightarrow \mathbb{R}$ where $d(x,y)=0$ for all $x,y\in X$. Then any sequence in it will converge to more than one points. But I am not sure whether its true for every pseudo-metric space.

Answer 2

Yes, it is true, in other words, your definition of topology coincides with the topology generated by all pseudoballs $B_r(x):=\{y:\,d(y,x)<r\}$.

For the second part of your question, let $A$ be closed in the sense that it contains all (pseudometric) limits of sequences in $A$ and $x\notin A$. Then there exists $r>0$ s.t. the $r$-pseudoball around $x$ is disjoint from $A$ (otherwise, you would construct a sequence of $x_j\in A$, $\mathrm{dist}(x_j, x)=\frac{1}{j}$ converging to $x$). So, $X\setminus A$ is open. Similarly you show that if $X\setminus A$ is open (in the topology induced by open pseudo-balls), then $A$ contains all limits of sequences.