several questions about convergence in topological space In a topology class, we define convergence as follow, a sequence $(x_n)$ converges to $p$ if for any open set $U$ containing $p$, we can always find an $N$ such that whenever $n \geq N$, $x_n$ is in the open set $U$. And here are my questions:


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*We know in discrete topology, the only convergent sequence has the property "eventually constant", i.e. the "tail" of this sequence is constant. So if a sequence converges in a co-countable space, should it necessarily be "eventually constant"? (my answer is yes, but I am not sure about this)

*The second one is kind of related to the first question. We know that in a Hausdorff space, a sequence can converge to at most one point. Is the converse still true? If a sequence converges to at most one point, is the space Hausdorff? (I think it is wrong, as any two non-empty open sets in co-countable space intersect with non-empty stuff, and sequence in this space clearly converges to at most one point.)
 A: I can answer 1. affirmatively.
Suppose $X$ is a space with the co-countable topology and let $(a_n)_{n\in\omega}$ be a sequence which converges to a point $l$. Then $\{l\}\cup X \setminus \{a_n :n<\omega\}$ is an open set containing $l$ and hence contains an entire tail of the sequence $\{a_n\}$.  But this can only happen if $a_n=l$ for all $n$ sufficiently large.
I can answer problem 2. (gary's modification) affirmatively if we assume $X$ is first countable:
Suppose $X$ is not Hausdorff.  Let $x_1,x_2\in X$ be distinct such that for every pair of open sets $U,V$ with $x_1\in U$ and $x_2\in V$, we have $U\cap V\neq\varnothing$.  Let $\{U_n:n\in\omega\}$ and $\{V_n:n\in\omega\}$ be local bases of $x_1$ and $x_2$, respectively.  We may assume each collection is decreasing.  For each $n\in\omega$ choose $a_n\in U_n \cap V_n$.  Then the sequence $(a_n)_{n\in\omega}$ converges to both $x_1$ and $x_2$.
Problem 2. is false in general: consider $\omega_1$ in the co-countable topology. 
This space is not Hausdorff: If $\alpha<\beta<\omega_1$ then an open set containing either point must contain an entire final segment of $\omega_1$.  So clearly the intersection of two open sets containing $\alpha,\beta$, respectively, cannot be empty. 
However, we do have uniqueness of limit points: we have already shown that the limit point of a convergent sequence is equal to the eventual constant of the sequence.
A: For #1, the answer is yes. Let $X$ be the space:
Let $x_n \rightarrow x$ , and consider the set $V$ of points in the sequence that are different from $x$. Then $X-V$ is an open set that contains $x$. Since $x_n \rightarrow x$ , the sequence must eventually land in $V$. But the only element of the sequence in $V$ is $x$ itself, so the sequence must be eventually-constant.
For 2), the answer is no: take an indiscrete space with two points {$x,y$}, and the open sets {{$x$},{$x,y$}, {}}. Then every sequence can converge to at most one limit, since {$x$} does not contain $y$, but the space is not Hausdorff, since the two points cannot be separated (meaning there is no open set that contains $y$ but does not contain $x$).   Maybe this is cheating and you should restrict to spaces with infinitely-many points.
For a more substantial answer to #2), see this link: http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist&task=show_msg&msg=4165.0001 ; answer is not by me.
