- #1

Eclair_de_XII

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- TL;DR Summary
- Let ##\{x_n\}_{n\in\mathbb{N}}## be a bounded sequence and let ##E## be the set of subsequential limits of that sequence. Assume that ##E## is non-empty. Prove that ##E## is bounded and contains both its supremum and its infimum.

Let ##L\in E##. By definition, there is a subsequence ##\{x_{n_k}\}_{k\in\mathbb{N}}## that converges to ##L##. There is a natural number ##N## s.t. if ##n_k\geq N##, ##L\in(x_{n_k}-1,x_{n_k}+1)\subset(\inf\{x_n\}-1,\sup\{x_n\}+1)##. Hence, ##E## is a bounded set.

If ##E## is a finite set, then it has no accumulation points, which means that it must contain both its infimum and supremum.

Now assume otherwise.

Denote ##x=\sup E##, and denote the subsequence of ##\{x_n\}## converging to ##x## as ##\{x_{m_k}\}_{k\in\mathbb{N}}##.

Let ##\epsilon>0##. There is ##N\in\mathbb{N}## s.t. for integers ##n> N##, ##x-L_n<\epsilon##.

If the set ##\{x_n\}\cap(x-L_N,x)## is finite, then each point ##L_n## in the set could not be an accumulation point. For each ##n>N##, simply choose ##\delta=L_n-L_N##; any ball of radius ##\delta## centered at ##L_n## would contain only finitely many points ##x_n##. It follows that each point ##L_n## must be converged onto point-wise. However, as there are only finitely many points ##x_n\in(x-L_N,x)##, and infinitely many subsequential limits ##L_n\in(x-L_N,x)##, this is impossible.

Therefore, there infinitely many points of ##\{x_n\}## in the interval ##(x-L_N,x)##. Consequently, there must exist a subsequence of ##\{x_n\}## that converges to ##x##. By definition, ##x\in E##.

If ##E## is a finite set, then it has no accumulation points, which means that it must contain both its infimum and supremum.

Now assume otherwise.

Denote ##x=\sup E##, and denote the subsequence of ##\{x_n\}## converging to ##x## as ##\{x_{m_k}\}_{k\in\mathbb{N}}##.

Let ##\epsilon>0##. There is ##N\in\mathbb{N}## s.t. for integers ##n> N##, ##x-L_n<\epsilon##.

If the set ##\{x_n\}\cap(x-L_N,x)## is finite, then each point ##L_n## in the set could not be an accumulation point. For each ##n>N##, simply choose ##\delta=L_n-L_N##; any ball of radius ##\delta## centered at ##L_n## would contain only finitely many points ##x_n##. It follows that each point ##L_n## must be converged onto point-wise. However, as there are only finitely many points ##x_n\in(x-L_N,x)##, and infinitely many subsequential limits ##L_n\in(x-L_N,x)##, this is impossible.

Therefore, there infinitely many points of ##\{x_n\}## in the interval ##(x-L_N,x)##. Consequently, there must exist a subsequence of ##\{x_n\}## that converges to ##x##. By definition, ##x\in E##.