Sequential space implies countable tightness? Recently, I met across a question, i.e., Sequential space implies countable tightness?


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*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.
 A: You can see this also from the following facts (which might be useful to know, if you work with these classes of spaces a lot):


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*Each sequential space can be obtained as quotient of topological sum of several copies of convergent sequence. (By convergent sequence I mean the topological space on the set $\{0\}\cup\{1/n; n=1,2,\dots\}$ with the topology inherited from real line.)

*Convergent sequence is a countably tight space.

*Quotients and sums of countably tight spaces are countably tight. (So are subspaces of countably tight spaces, but this is not relevant for this question.)

A: Hint: $\def\cl{\operatorname{cl}}$Suppose $X$ weren't countable tight. Then there were an $A \subseteq X$ such that 
\[ \cl A \supsetneq \bigcup_{\substack{B \subset A\\ B \text{ countable}}} \cl B \]
Then $A$ cannot be closed (as then $\cl A = A = \bigcup_{x \in A}\{x\}\subseteq \bigcup_{x\in A}\cl\{x\}$). Therefore by sequentialness there were a $y \in \cl A \setminus A$, and a sequence $(a_n) \in {}^{\mathbb N}A$ such that $a_n \to y$.
