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Let $X = \mathbb{N} \times \mathbb{N}$. Define a non-empty set $U \subseteq X$ to be open if for cofinitely many $x \in \mathbb{N}$ the set ${ \{ y \in \mathbb{N} \vert (x,y) \in U}$ U\}$ is cofinite. Construct a sequence in$X$that hits every point in$X$exactly once. In other words, take a bijection$\mathbb{N} \rightarrow X$. Then: •$X$is countable; • every point in$X$is an accumulation point of this sequence, but • the sequence has no convergent subsequences. In particular, this is an example in a countable set that accumulation point of a sequence does not have to be a limit of a subsequence. I call this the Herreshoff topology for the (high-school) student of mine who came up with it. (I could not find it anywhere else, although I do not discard that I did not look hard enough.) 1 [made Community Wiki] My favourite counterexample is purely academic: it does not have any applications, but I think it is pretty. Let$X = \mathbb{N} \times \mathbb{N}$. Define a non-empty set$U \subseteq X$to be open if for cofinitely many$x \in \mathbb{N}$the set${ y \in \mathbb{N} \vert (x,y) \in U}$is cofinite. Construct a sequence in$X$that hits every point in$X$exactly once. In other words, take a bijection$\mathbb{N} \rightarrow X$. Then: •$X$is countable; • every point in$X\$ is an accumulation point of this sequence, but