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The definition of sequential compactness states that set $K$ is sequentially compact if $\forall$ sequences $a_n \in K$, $\exists a_{n_i}$, such that this subsequence converges to a point $p \in K$.

I'm thinking of the corollary to the open cover definition of compactness, that set $K$ is closed and bounded, and trying to relate this to the definition of sequential compactness. I can see how sequential compactness would imply that $K$ is closed, since every sequence contains a limit point in $K$ can follow the definition of sequential compactness

My Question How does the bounded property follow from the definition of sequential compactness? If we are given that set $K$ is sequentially compact then how could we show that $K$ is bounded? What restrictions or conditions apply to the definition of sequential compactness?

Thanks

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  • $\begingroup$ Try an example: why is $\mathbb{R}$ not sequentially compact? $\endgroup$ – Eric Wofsey Jul 19 '18 at 22:18
  • $\begingroup$ well we could use the counterexample where we pick infinite sequence $a_i$, where $a_i=i$. so there do not exist any subsequences that converge since they all tend to $\infty$ $\endgroup$ – john fowles Jul 19 '18 at 23:33
  • $\begingroup$ @EricWofsey and I read on wikipedia that sequential compactness implies compactness only for certain topological spaces. Could I treat the definitions are equivalent if I'm only dealing with sets in $\mathbb{R}$, and what about $\mathbb{R^n}$, and other common metric spaces? Could you tell me some spaces where the equivalence fails $\endgroup$ – john fowles Jul 19 '18 at 23:40
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Suppose it were not bounded. Then we see $X$ is not contained in any balls of finite radius. Pick one point in $X$ (this is possible as the empty set is bounded), say $x_0$ then choose a $x_{1} \notin B_{1}(x_0)$. This is possible as otherwise we would get a bounded $X$. Now choose a $x_{2} \notin B_{2}(x_1) \cup B_{1}(x_{1})$ as otherwise we could bound $X$. Keep repeating this and you'll get a sequence that cannot obtain a convergent subsequence, which is your desired contradiction.

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    $\begingroup$ Did you mean $B_{1}(x_1) \cup B_{2}(x_{1})$, instead of $B_{2}(x_1) \cup B_{2}(x_{1})$? $\endgroup$ – john fowles Jul 19 '18 at 23:16
  • $\begingroup$ Yes, my bad. I fixed the typo $\endgroup$ – user518441 Jul 19 '18 at 23:24
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Hint: suppose $K$ is not bounded. This means that there is a point $a_0\in K$, and for each $i$, a point $a_i\in K$ such that the distance between $a_0$ and $a_i$ is larger than $i$. Can $K$ be sequentially compact?

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  • $\begingroup$ What definition of bounded are you using? $\endgroup$ – john fowles Jul 19 '18 at 22:39
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    $\begingroup$ @johnfowles there exists a (finite) "diameter" D such that the distance between any two elements of the set is less than D. $\endgroup$ – Anonymous Jul 19 '18 at 22:48
  • $\begingroup$ So to negate sequential compactness of $K$ I use the sequence $a_i$ that's defined as $|a_0-a_i|>i$, where $i=1,2,...,n$, and need to show that $\forall$ subsequences of $a_i$ they converge to a point that's outside of $K$. I'm not sure how to show that last part $\endgroup$ – john fowles Jul 19 '18 at 23:10

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