A question on countable tightness of $X$ For any topological space $X$, if $t(X)>\omega$, then there exists a non-closed set $A \subset X$ such that $\overline {B} \subset A$ for any countable $B \subset A$. I cannot think out such set $A$. Could somebody help me? > countable tightness = $X$ has countable tightness if for any $A \subset X$, whenever $x \in cl(A)$, then $x \in cl(B)$ for any countable $B \subset A$.

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ewcommand{\cl}{\operatorname{cl}}$If $t(X)>\omega$, by definition there are a $B\subseteq X$ and an $x\in\cl B$ such that for all countable $C\subseteq B$, $x\
otin\cl C$. Let $A=\bigcup\\{\cl C:C\subseteq B\text{ and }|C|\le\omega\\}$. Then $x\in(\cl A)\setminus A$, so $A$ is not closed. Let $C\subseteq A$ be countable. For each $y\in C$ there is a countable $C_Y\subseteq B$ such that $y\in\cl C_y$. Let $D=\bigcup_{y\in C}C_y$; then $D$ is a countable subset of $B$, so $\cl D\subseteq A$. But $C\subseteq\cl D$, so $\cl C\subseteq\cl D\subseteq A$, and $A$ has the desired properties.