“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!

I’m a little bit confused when I try to verify your discription: if is the nbhood of , then clearly, is an element of the discrete topology in the first discription. Hence, is open and closed. Since is not empty, then should be and is the only open set containing . If there was nothing wrong in my deduction, the observation 1 did’t hold.

I think the deduction “Since is not empty, then…” isn’t valid.