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Suppose $(X,\tau)$ is a topological space and that $(X^2,\tau_2)$ is the product space. Now define

$\mathscr S\!_\tau=\{W\in\tau_2|\Delta X^2\subseteq W\}$, where $\Delta X^2=\{(x,y)\in X^2|x=y\}$,

and suppose $(x_n)$ is a sequence in $X$ such that

$\forall S\in\!\mathscr S\!_\tau\exists N\in\mathbb N:(n,m>N\implies(x_n,x_m)\in S)$.

Is it then possible to deduce that $(x_n)$ has a clusterpoint in $X$,

$\exists x\in \!X\;\forall \mathcal O\in\tau: (x\in\mathcal O\implies\forall N\in\mathbb N\,\exists n>N: x_n\in\mathcal O)$?

Lehs
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  • You might want to rewrite your question using more words and less symbols. You may also want to define several of the symbols used as well. For example: $\Delta X^2 = { (x,x) ;|; x \in X}$. – Bill Cook Dec 16 '14 at 22:42
  • What is $\Delta X^2$? And are you sure $\mathbb Q$ with the usual topology isn't a counterexample? – hmakholm left over Monica Dec 16 '14 at 22:43
  • @Henning: $\Delta X^2$ is the diagonal. Lehs is asking whether a sequence in $X$ that is Cauchy with respect to the family of all open nbhds of the diagonal always has a cluster point. – Brian M. Scott Dec 16 '14 at 22:45
  • Note that for $X=\mathbb Q$, an ordinary strictly increasing sequence of rationals $x_i$ with irrational limit $L$ isn't "generalized Cauchy" according to this definition. If we choose irrational $y_1$s such that $$ x_1 < y_1 < x_2 < y_2 < x_3 < \cdots$$ then $$ \mathbb Q^2 \cap \Bigl[ (-\infty,y_1)^2 \cup (y_1,y_2)^2 \cup (y_2,y_3)^2 \cup \cdots \cup (L,\infty)^2 \Bigr] $$ is in $\mathscr S$ but doesn't satisfy the condition. – hmakholm left over Monica Dec 16 '14 at 22:56

1 Answers1

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Yes, at least if $X$ is Hausdorff. We can prove the contrapositive: Assume that $(x_n)_n$ does not have any accumulation points, and we will then prove that it is not quasi-Cauchy.

Let $A=\{x_n\mid n\in \mathbb N\}$. Since the sequence has no accumulation points, every subset of $A$ is closed.

In particular, for every $x_n$ the set $B_n=X\setminus (A\setminus \{x_n\})$ is open, and the union of all the $B_n$s is $X$. Now $$ S = \bigcup_n (B_n)^2 $$ is in $\mathscr S$, but $(x_n)_n$ doesn't satisfy the quasi-Cauchy criterion with respect to this $S$. Namely, $(x_i,x_j)\in S$ only if $x_i=x_j$, and if that happens for all $i,j>N$, then $x_n$ is constant from some point, contradicting the assumption that it has no accumulation points.

hmakholm left over Monica
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