TITLE: A new kind of convergence
QUESTION [6 upvotes]: Let $(X,d)$ be a connected metric space. We say a sequence $\{x_n\}_n \subseteq X$ is T-convergent to $x \in X$ if the following is true: 
$$ \mbox{if} \;  a,b \in X \mbox{ and } d(a,x) < d(x,b) \mbox{ then } \exists N\; \forall n\geq N:d(a,x_n) < d(x_n,b) $$
Does it true that T-convergence implies convergence !?

REPLY [4 votes]: The result is false in general. Let $X=\ell_2$. For $n\in\Bbb N$ let $x^n=\langle x_k^n:k\in\Bbb N\rangle$ such that $$x_k^n=\begin{cases}1,&\text{if }k=n\\0,&\text{otherwise}\;.\end{cases}$$
Let $a=\langle a_k:k\in\Bbb N\rangle\in X$. Then
$$\|a-x^n\|=\left(\sum_{k\ge 0}(a_k-x_k^n)^2\right)^{1/2}=\left((a_n-1)^2+\sum_{{k\ge 0}\atop{k\ne n}}a_k^2\right)^{1/2}\longrightarrow\left(1+\|a\|^2\right)^{1/2}$$
as $n\to\infty$. Let $z\in X$ be the zero sequence, and suppose that $a,b\in X$ with $\|a\|<\|b\|$. Then $$\lim_{n\to\infty}\|a-x^n\|=\left(1+\|a\|^2\right)^{1/2}<\left(1+\|b\|^2\right)^{1/2}=\lim_{n\to\infty}\|b-x^n\|\;,$$
so there is an $m\in\Bbb N$ such that $\|a-x^n\|<\|b-x^n\|$ for all $n\ge m$. It follows that $\langle x^n:n\in\Bbb N\rangle$ $T$-converges to $z$, but $\langle x^n:n\in\Bbb N\rangle$ does not converge to $z$; indeed, it has no convergent subsequence.