If the only convergent sequences in a space are eventually constant, is the space necessarily discrete?
Solution
No, the converse does not hold in general. Let $X$ be any uncountable set, and fix a point $p\in X$. Let $$\tau=\{U\subseteq X:p\notin U\}\cup\{X\setminus C:C\text{ is a countable subset of }X\}\;;$$
then $\tau$ is a non-discrete Hausdorff topology on $X$ in which the only convergent sequences are the trivial ones. (In fact $\tau$ is a $T_5$ topology: the space is hereditarily normal.)
Remark: If $X$ is metrizable, however, the converse does hold. Suppose that $x\in X$ is not an isolated point. Then for each $n\in\Bbb N$ there is a point $x_n\in B(x,2^{-n})\setminus\{x\}$, and the sequence $\langle x_n:n\in\Bbb N\rangle$ is clearly a non-trivial sequence converging to $x$.
Solution
No, the converse does not hold in general. Let $X$ be any uncountable set, and fix a point $p\in X$. Let $$\tau=\{U\subseteq X:p\notin U\}\cup\{X\setminus C:C\text{ is a countable subset of }X\}\;;$$
then $\tau$ is a non-discrete Hausdorff topology on $X$ in which the only convergent sequences are the trivial ones. (In fact $\tau$ is a $T_5$ topology: the space is hereditarily normal.)
Remark: If $X$ is metrizable, however, the converse does hold. Suppose that $x\in X$ is not an isolated point. Then for each $n\in\Bbb N$ there is a point $x_n\in B(x,2^{-n})\setminus\{x\}$, and the sequence $\langle x_n:n\in\Bbb N\rangle$ is clearly a non-trivial sequence converging to $x$.
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