Given a sequence of real numbers, $\{ x_n \}_{n=1}^{\infty}$, let $\alpha =$ limsup$x_n$ and $\beta = $ liminf$x_n$.
Prove that there exists a subsequence $\{ x_{n_k}\}$ that converges to $\alpha$ as $k \rightarrow \infty$.
Not sure how to start this without since I'm not given that the subsequence is bounded..
Best Answer
Since $\alpha=\limsup x_n$, by the definition of $\limsup$, there is some $x_{n_1}$ with $|x_{n_1}-\alpha|<{1\over 2}$. (That's the crucial step, so be sure you understand why.)
Similarly, there is some $x_{n_2}$ with $|x_{n_2}-\alpha|<{1\over 2^2}$. Continuing, for each $k\in\mathbb{N}$, there is some $x_{n_k}$ with $|x_{n_k}-\alpha|<{1\over 2^k}$.
Then $\{x_{n_k}\}\subset \{x_n\}$ and $x_{n_k}\to \alpha$ as $k\to\infty$.