When is the limit of a function unique?

Question

The limit of $f:X \to Y$ at $x$ is an element $y \in Y$ such that for any open $N$ containing $y$, there is a $V$ containing $x$ such that $f(V/\{x\})\subseteq N$.

For the particular case when $f$ is a sequence, i.e. $f: \mathbb{N}+\{\infty\} \to X$, it is well known that if $X$ is Hausdorff, then the limit is unique. This however is not necessary. My questions are then:

1. For the case of sequences, what are necessary and sufficient conditions for $X$ so that the limit of the sequence is unique?
2. For the general case, what are necessary and sufficient conditions for $X$ and $Y$ so that the limit of the function is unique?

I'd like if these conditions are stated in terms of already well known studied conditions, e.g. separation, countability, etc; but anything is appreciated.

Answer 1

The US property where the limit of a sequence is unique is a quite weak separation axiom.

• As you said, Hausdorff is sufficient. A weaker sufficient condition is what was called $k_2$-Hausdorff here: https://math.stackexchange.com/a/4761212/86887

• $T_1$ is necessary. But basically every other separation axiom I'm aware of that is sufficient for $T_1$ is not necessary for $US$:

  • https://topology.pi-base.org/spaces?q=US%2B~Locally%20Hausdorff
  • https://topology.pi-base.org/spaces?q=US%2B~%24k_2%24-Hausdorff
  • https://topology.pi-base.org/spaces?q=US%2B~Semi-Hausdorff

If you assume spaces are first-countable, then $US$ is equivalent to $T_2$.

Finally, the weakly Hausdorff property satisfies that every continuous image of a compact Hausdorff space is closed. Then I believe such spaces have unique limits for each $f:K\to X$ with $K$ compact Hausdorff.