# Converges ,sequences

Let $X$ be first countable space $(1^0)$:$U \subset X$ is open $\iff$ whenever a sequence ${x_n}$ converges to x in U ,then $x_n$ eventually in U}.

I was able to prove the fist direction , suppose U is open and $x_n$ converges to $x$ ,$x \in U$ U is open implies U is a nhood of x by the definition of converges

hence, since $x_n$ converges to $x$ , so there exist $n_ {\circ}$ such that $n \ge n_ {\circ} \Rightarrow x_n \in U$, so $x_n$is eventually in $U$.

But i have a problem in proving the other direction ,if any one can help me ?

note: $X$ is a first countable then at each $x \in X$ we can find a countable nhood base $\{U_n : n \in \mathscr N \}$ such that $U_1 \supset U_2 \supset U_3 \supset ....$

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thanks for your efforts to format/use LaTeX. Here on Math.SE, MathJax is used and needs formatted text enclosed in dollar signs: e.g., $U_i \supset U_k$ would be $U_i \supset U_k$. For displaystyle, use two dollar signs at start and at end of formatting. –  amWhy Oct 31 '12 at 22:32
than you its my first time and i edit it . –  Miss Independent Oct 31 '12 at 22:39
So suppose, $U$ is not open in $X$, then $X \setminus U$ is not closed, hence there is a point $x \in U \cap \operatorname{cl}(X\setminus U)$. As $X$ is first countable, there is a countable neighbourhood base $(U_n)$, wlog with $U_{n+1}\subset U_n$, of $x$. As $x \in \operatorname{cl}(X \setminus U)$, we have $U_n \cap X \setminus U \ne \emptyset$ for each $n$, let $x_n \in U_n \cap (X \setminus U)$. If $V$ is a neighbourhood of $x$, then there is an $N$ with $U_N \subseteq V$, so $x_n \in U_n\subseteq V$ for all $n \ge N$. As $V$ was arbitrary $x_n \to x$. But $x_n \not\in U$ for all $n$.
Without loss of generality. If $(V_n)$ is an arbitrary countable neighbourhood base, we set $U_n = \bigcap_{k \le n} V_k$. Then $(U_n)$ is a decreasing one as wished. –  martini Oct 31 '12 at 22:46