= Solution
{c}
A <sequence> $x_n$ converges to $x$ if every open <neighbourhood> of $x$ contains all sufficiently late terms. A <Hausdorff space> separates any two distinct points by disjoint open neighbourhoods.
If a sequence had distinct limits $x,y$ in a <Hausdorff space>, choose disjoint open neighbourhoods $U$ of $x$ and $V$ of $y$. Convergence to each limit implies that all sufficiently late terms lie in both $U$ and $V$, contradicting $U\cap V=\varnothing$. Hence \b[a convergent sequence in a Hausdorff space has a unique limit].
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