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A‎‎ ‎sequential ‎space ‎has ‎unique ‎sequential ‎limits ‎iff ‎each ‎countably ‎compact ‎subset ‎is ‎closed.

‎Proof: ‎If‎ ‎$ \{ x_n \} $ ‎is a‎ ‎sequence ‎converging ‎to ‎two ‎distinc ‎‎$‎x‎$ ‎and ‎‎$‎y‎$‎, then ‎$ ‎\{ x‎ ‎\}‎ ‎\cup ‎\{‎‎ ‎x_n :‎ n‎ ‎\in‎ ‎\omega ‎\} ‎‎$ ‎is a‎ ‎non ‎compact ‎set.

conversely: ‎if ‎‎$‎X‎$ ‎is ‎sequential ‎and ‎has ‎unique ‎sequential ‎limits, ‎‎$‎X$ ‎is ‎‎$‎T_1‎$ .‎ ‎Let‎ $ K‎ ‎‎\subseteq ‎X $‎ ‎be ‎countably ‎compact.If ‎‎$ ‎\{ ‎x_n ‎\} ‎‎$ ‎is a‎ ‎sequence ‎in ‎‎$‎K‎$ ‎and‎ $ ‎\{‎x‎_{‎n}‎ : n‎ \in‎ \omega ‎\} ‎‎\longrightarrow ‎x‎_{0}‎$ ‎then‎ ‎$ ‎\{ ‎x‎_{0} ‎\}‎ ‎\cup ‎\{‎‎ ‎x_n :‎ n‎ ‎\in‎ ‎\omega ‎\} ‎‎$ ‎is ‎sequentially ‎closed ‎and ‎hence ‎closed. ‎Thus ‎‎$‎x_{0}‎$ ‎is ‎the ‎only ‎possible ‎accumulation ‎poin ‎of‎ $ ‎\{ ‎x‎_{‎n}‎ : n‎ \in‎ \omega ‎\} ‎‎$‎. if $ ‎\{ ‎x‎_{‎n}‎ : n‎ \in‎ \omega ‎\} ‎‎$ ‎is ‎infinite‎, ‎$‎x_0 ‎\in ‎K‎$‎. ‎If ‎not ‎for ‎some ‎‎$‎n‎$‎, $ ‎x‎_{0} =‎ ‎x‎_{n}‎‎ ‎\in K ‎$.Hence ‎‎$‎K‎$ ‎is ‎closed.‎

(1) Why" ‎if ‎‎$‎X‎$ ‎is ‎sequential ‎and ‎has ‎unique ‎sequential ‎limits, ‎‎$‎X$ ‎is ‎‎$‎T_1‎$?

(2) Why "if $ ‎\{ ‎x‎_{‎n}‎ : n‎ \in‎ \omega ‎\} ‎‎$ ‎is ‎infinite‎"?

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You have an error in the first part of the proof: $\{x\}\cup\{x_n:n\in\omega\}$ is compact. The point is that it is compact but not closed.

  1. Suppose that $X$ is sequential and has unique sequential limits. Let $x\in X$ be arbitrary, and suppose that $\operatorname{cl}\{x\}\ne\{x\}$. Since $X$ is sequential there are a point $y\in(\operatorname{cl}\{x\})\setminus\{x\}$ and a sequence $\langle x_n:n\in\omega\rangle$ in $\{x\}$ converging to $y$. (You’ve seen this property of sequential spaces used several times in the papers that you’ve been studying.) Obviously $x_n=x$ for each $n\in\omega$, so $\langle x_n:n\in\omega\rangle\to x$. But $X$ has unique sequential limits, so $y=x$, contradicting the choice of $y\in(\operatorname{cl}\{x\})\setminus\{x\}$. Thus, $\operatorname{cl}\{x\}=\{x\}$ for each $x\in X$, and it follows immediately that $X$ is $T_1$.

  2. I’m going to rephrase this part of the argument slightly. We have a sequence $\langle x_n:n\in\omega\rangle$ in $K$ converging to some point $x$. Let $S=\{x_n:n\in\omega\}$. There are two possibilities.

    • $S$ might be finite. In that case there is a $y\in S$ such that $A=\{n\in\omega:x_n=y\}$ is infinite, and $\langle x_n:n\in A\rangle\to y$. But obviously $\langle x_n:n\in A\rangle\to x$ as well, and $X$ has unique sequential limits, so $x=y\in S\subseteq K$.
    • $S$ might be infinite. Since $X$ is $T_1$ and $\langle x_n:n\in\omega\rangle\to x$, $S$ is discrete. $S$ is therefore an infinite discrete subset of $K$, and since $K$ is countably compact, this means that $S$ cannot be a closed subset of $K$. Thus, $S$ must have some accumulation point $y\in K$. But we already know that the only accumulation point of $S$ is $x$, so $x=y\in K$, and we see that in all cases $x\in K$.
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