A set $A$ is ward compact if every sequence $(a_n) $ in $A$ has a quasi-Cauchy subsequence. A sequence $(x_n)$ is said to be quasi Cauchy if $x_n-x_{n+1}\rightarrow 0$. We know that $a_n=\sqrt n$ is a quasi sequence, what about ward compact set?
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Just out of curiosity, why is it called "ward" compact? Is it from the English word ward, or the proper name Ward, or is it a word in some other language? – bof Oct 27 '17 at 08:18
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At first it was forward compact then authors used ward compact. – Ali Oct 27 '17 at 11:14
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Related – Giuseppe Negro Oct 27 '17 at 12:10
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This property is equivalent to $A$ being totally bounded. In the context of $\mathbb{R}$ or $\mathbb{R}^n$, it is equivalent to being bounded.
Indeed, being totally bounded implies that every sequence has a Cauchy subsequence (no "quasi-" needed); see Metric space is totally bounded iff every sequence has Cauchy subsequence.
And if $A$ is not totally bounded, then there exists $\epsilon>0$ such that there is no finite $\epsilon$-net. This makes it possible to keep choosing points at distance $\ge \epsilon$ from each other, thus forming a sequence $x_k$ for which $d(x_k, x_j)\ge \epsilon$ whenever $k\ne j$. Such a sequence has no "quasi-Cauchy" subsequence.