Do totally open sets exist?

Question

• In the third answer to this question, a justification is given for calling closed sets closed, since they are literally closed under the $\mathbb{N}$-ary operation of taking limits of (convergent) sequences.
• And then open is declared as the complement of closed, but this doesn't seem to follow naturally from the operation motivation, in particular not every convergent sequence converges outside of an open set, so one would be tempted to say it has at least one convergent sequence converging outside of it, but then a set containing some but not all boundary points is still open.
• We could however call a set totally open if every convergent, injective sequence converges outside the set, The natural numbers satisfy this, since they do not contain any injective convergent sequence, but do totally open sets exist in spaces where there are convergent and injective sequences ?. And if such spaces exists, can we classify them ?.

Answer 1

Edit: Never mind, we can do something much simpler. We can take the one-point compactification $X \cup \{ \infty \}$ of any infinite set $X$ (with the discrete topology); then a convergent sequence is either eventually constant, so converges to its constant value in $X$, or eventually leaves every finite subset of $X$, so converges to $\infty$.

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Here is an example I think works but have not checked carefully, which is a variation of the Cauchy completion. Let $X$ be an infinite set and let $\widetilde{X}$ be the quotient of the space $X^{\mathbb{N}}$ of infinite sequences $x_1, x_2, \dots \in X$, equipped with the product topology (equivalently the topology of pointwise convergence), by the equivalence relation that $x \sim y$ iff $x$ and $y$ eventually agree.

The idea is that we have tried to "formally adjoin" limits to all sequences in $X$ in a maximal way. $X$ embeds into $\widetilde{X}$ as the constant sequences, and I believe if I've set things up properly that every sequence in $X$ which is eventually constant converges to its eventual value (in $X$), while every sequence in $X$ which is not eventually constant converges to "itself," regarded as an element of $X^{\mathbb{N}}$ and hence an element of $\widetilde{X}$ which is not in $X$.