“Any sequence with an accumulation point also has a subsequence converging to that point” — right? I believed this until yesterday when I stumbled over the following curious, and countable, Hausdorff space.

Endow with the following topology:

It is routine to verify that this is a topology on .

**Observation 1:** No sequence in converges to .

*Proof:* Let be any sequence. We distinguish two cases:

*Case 1:* For all the set is finite. Then we set . This is easily seen to be an open nbhood of and clearly doesn’t converge to .

*Case 2:* There is such that is infinite. Then is an open nbhood of , because for all with we have . Moreover, by the conditition described in the statement of Case 2, the sequence keeps “popping out” of and therefore does not converge to .

**Observation 2:** There is a sequence in such that is an accumulation point of .

*Proof:* Since is countable, there is a bijection from to and we let be that bijection.

So the sequence we constructed in Observation 2 has as an accumulation point, but no subsequence of converges to because of Observation 1!