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Does an open subset containing all points of a concergent sequence but a finite number of them necessarily contain the sequence's limit ?

Ideally, I would like an answer for each of the following :

-general topology

-Haussdorf (T2)

-metric

My intuition is that it doesn't, because even in a metric space, the radius of a ball centered at each point of the sequence and contained in the open subset may be forced to vary over the sequence.

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No. Take, in $\mathbb R$ with the usual topology, the sequence $\left(\frac1n\right)_{n\in\mathbb N}$ and the open set $(0,+\infty)$.

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The answer is no for all three. Consider the sequence $\frac{1}{n} \in \mathbb{R}$, and consider the open set $(0, 2)$. This open set contains all but finitely many of the sequences points (in fact, it contains all the sequences points), but it does not contain the sets limit, 0.

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