Recently, I met across a question, i.e., Sequential space implies countable tightness?
sequential space = $X$ is sequential if $A \subset X$ and $A$ is not closed implies that there is a sequence $\{a_n:n\in \omega\}\subset A$ such that $a_n \rightarrow y$ for some $y \in A^c.$
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 $B \subset A$.
Thanks for any help.