Let $S=\{\frac{1}{n}:n\in\ N\}$. Prove directly from the definition that $\ S$ is not sequentially compact in $\ R$.
From the definition that my professor gave, for any sequence in $\{x_n\}$ in $\ S$, there exists a convergent subsequence $\{x_{n_k}\}$ that converges to a limit ${x \in\ S}$.
In order to prove the statement above, can't I just prove that limit of $\ S$ is $0$, but $0$ is not in $\ S$, therefore it is not sequentially compact?
No, because $S$ is a set. What is the limit of a set?
Take the obvious sequence: $\left(\frac1n\right)_{n\in\mathbb{N}}$. It does't converge in $S$, right? Therefore, $S$ is not sequentially compact.