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The property you mention does not imply the existence of a convergent subnet: take e.g. the sequence $x_n:=\sqrt n$ on $\mathbb{R}$, as a metric space with the truncated standard distance, $d(x,y):=\min\{|x-y|, 1\}$, which is uniformly equivalent to the standard metric, and makes therefore $\mathbb{R}$ a bounded complete metric space. The property holds with $n_{\epsilon, K}:=\frac{K^2}{4\epsilon^2}\\ ,$ although of course $(x_n)$ has no converging sub-net.

As a side remark, note that the property you wrote is equivalent to: $d(x_n,x_{n+1})\to0$ as $n\to\infty$.

The property you mention does not imply the existence of a convergent subnet: take e.g. the sequence $x_n:=\sqrt n$ on $\mathbb{R}$, as a metric space with the truncated standard distance, $d(x,y):=\min\{|x-y|, 1\}$, which is uniformly equivalent to the standard metric, and makes therefore $\mathbb{R}$ a bounded complete metric space. The property holds with $n_{\epsilon, K}:=\frac{K^2}{4\epsilon^2}\, ,$ although of course $(x_n)$ has no converging sub-net.

As a side remark, note that the property you wrote is equivalent to: $d(x_n,x_{n+1})\to0$ as $n\to\infty$.

The property you mention does not imply the existence of a convergent subnet: take e.g. the sequence $x_n:=\sqrt n$ on $\mathbb{R}$, as a metric space with the truncated standard distance, $d(x,y):=\min\{|x-y|, 1\}$, which is uniformly equivalent to the standard metric, and makes therefore $\mathbb{R}$ a bounded complete metric space. The property holds with $n_{\epsilon, K}:=\frac{K^2}{4\epsilon^2}\\ .$

Also, note that the property you wrote is equivalent to: $d(x_n,x_{n+1})\to0$ as $n\to\infty$.

The property you mention does not imply the existence of a convergent subnet: take e.g. the sequence $x_n:=\sqrt n$ on $\mathbb{R}$, as a metric space with the truncated standard distance, $d(x,y):=\min\{|x-y|, 1\}$, which is uniformly equivalent to the standard metric, and makes therefore $\mathbb{R}$ a bounded complete metric space. The property holds with $n_{\epsilon, K}:=\frac{K^2}{4\epsilon^2}\\ ,$ although of course $(x_n)$ has no converging sub-net.

As a side remark, note that the property you wrote is equivalent to: $d(x_n,x_{n+1})\to0$ as $n\to\infty$.

The property you mention does not imply the existence of a convergent subnet: take e.g. the sequence $x_n:=\sqrt n$ on $\mathbb{R}$, as a metric space with the truncated standard distance, $d(x,y):=\min\{|x-y|, 1\}$, which is uniformly equivalent to the standard metric, and makes therefore $\mathbb{R}$ a bounded complete metric space.

Also, note that that property is equivalent to: $d(x_n,x_{n+1})\to0$ as $n\to\infty$.

The property you mention does not imply the existence of a convergent subnet: take e.g. the sequence $x_n:=\sqrt n$ on $\mathbb{R}$, as a metric space with the truncated standard distance, $d(x,y):=\min\{|x-y|, 1\}$, which is uniformly equivalent to the standard metric, and makes therefore $\mathbb{R}$ a bounded complete metric space. The property holds with $n_{\epsilon, K}:=\frac{K^2}{4\epsilon^2}\\ .$

Also, note that the property you wrote is equivalent to: $d(x_n,x_{n+1})\to0$ as $n\to\infty$.