We say that $K \subset \mathbb{R}^n$ is [[Sequentially compactness|sequentially compact]], that is, every sequence $x_{j}$ in $K$ has a [[Convergence|convergent]] subsequence $x_{j_{l}}$ whose limit is in $K$. > [!NOTE] Theorem (Sequentially Compact Subset of Euclidean Space is Necessarily Closed and Bounded) > > > $K$ is sequentially compact if, and only if, $K$ is [[Closed Sets|closed]] and bounded. ###### Proof of Sequentially Compact Subset of Euclidean Space is Necessarily Closed and Bounded ($\implies$) Suppose that $K$ is sequentially compact. To prove that $K$ is closed, we consider a sequence $x_j$ in $K$ which converges to $x \in \mathbb{R}^n$ and then we have to show that $x \in K$. By the sequential compactness of $K$, $x_j$ has a subsequence $x_{j \ell}$ whose limit is in $K$. But since [[Subsequence of Convergent Sequence Converges to Same Limit]], $x=\lim _{j \rightarrow \infty} x_j=\lim _{\ell \rightarrow \infty} x_{j c} \in K$. The proof that $K$ is closed is complete. To prove that $K$ is bounded, assume, for a contradiction, that it is unbounded. Then there exists a sequence $x_j$ in $K$ such that for all $j \in \mathbb{N}$, $\left|x_j\right| \geqslant j$. By the sequential compactness of $K, x_j$ has a subsequence $x_{j \ell}$ whose limit is in $K$. In particular, $x_{j \ell}$ is bounded, i.e., $\exists M>0$ such that for all $\ell \in \mathbb{N}$, $\left|x_{j_{l}}\right| \leqslant M$. But by definition of subsequence, $j_{\ell} \geqslant \ell$ and, by the way the sequence $x_j$ was chosen, $\left|x_{j \ell}\right| \geqslant j_{\ell}$. Therefore, $M \geqslant\left|x_{j \ell}\right| \geqslant j_{\ell} \geqslant \ell, \quad\forall \ell \in \mathbb{N} .$ This clearly cannot hold and we conclude that $K$ must be bounded. $(\impliedby)$ We now assume that $K$ is closed and bounded and prove that it is sequentially compact. So consider an arbitrary sequence $x_j$ in $K$. Since $K$ is bounded, $x_j$ must be bounded and, by the [[Bolzano-Weierstrass Theorem (Sequential Compactness of The Reals)|Bolzano-Weierstrass theorem]], it has a convergent subsequence $x_{j k}$ whose limit $x$ must be in $K$, since $K$ is closed. The proof that $K$ is sequentially compact is complete.