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

  1. 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 $y \in A^c.$

  2. 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.

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2 Answers 2

up vote 3 down vote accepted

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):

  • 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.)
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It is interesting of your result. –  Simple Mar 14 '13 at 2:40

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 $a_n \to y$.

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And then? Is there a contradiction? –  Paul Oct 28 '12 at 8:54
2  
@Paul: I think the main point is that $D = \bigcup \{ \mathop{cl} B : B \subseteq A \text{ is countable} \}$ is not closed. Then there is a sequence in $D$ which converges to some point $x$ outside $D$. Each member of this sequence is in the closure of a countable subset of $A$, and it can then be shown that $x$ itself is in the closure of a countable subset of $A$, contradicting that it is not in $D$. –  Arthur Fischer Oct 28 '12 at 9:58
    
@ArthurFischer Thanks for the hint:) –  Paul Oct 29 '12 at 3:29

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