# Statement(s) > [!NOTE] Theorem (Bolzano-Weierstrass) > Any [[Bounded Sequence|bounded sequence]] of real numbers contains a [[Convergence|convergent]] [[Real Subsequence|subsequence]] > [!NOTE] Theorem (Bolzano-Weierstrass for Euclidean Space) > A [[Bounded Sequence|bounded]] sequence $(x_{j})$ in $\mathbb{R}^n$ has a [[Convergence|convergent]] subseqeunce $(x_{j_{k}})$. # Proof(s) ###### Proof of Bolzano-Weierstrass If $(a_{n})$ is bounded then it contains a [[Monotonic Sequence of Real Numbers|monotonic]] [[Real Subsequence|subsequence]] since [[Monotonic Subsequence Theorem]]. This subsequence is bounded both above and below (since $(a_{n})$ is bounded by [[Convergent Real Sequence is Bounded]]) - so it is either increasing and bounded above or decreasing and bounded below. Hence by [[Monotone Bounded Real Sequence is Convergent|monotone convergence]], it must converge. ###### Sketch Proof of Bolzano-Weierstrass for Euclidean Space \[MA270\] The proof of the Bolzano-Weierstrass in Chapter 3 in MA141 is done in $\mathbb{R}$. The argument below is a complete proof in two dimensions, which can be esily generalised to any dimension (primarily by choosing better notation). Let $x_j=\left(x_{j, 1}, \ldots, x_{j, n}\right)$ be a bounded sequence in $\mathbb{R}^n$. Then $x_{j, 1}$ is a bounded sequence in $\mathbb{R}$ and therefore, by the Bolzano-Weierstrass Theorem it has a convergent subsequence $x_{j_k, 1}$ which converges to $x_1^* \in \mathbb{R}$. Since we are only interested in finding a subsequence, we can consider the following sequence, indexed by $k,\left(x_{j_k, 1}, \ldots, x_{j_k, n}\right)$. So far we have constructed a subsequence of the original for which the first coordinate is a convergent sequence. Consider now the sequence $x_{j_k, 2}$. The sequence is of course bounded and therefore, by BolzanoWeierstrass, it has a subsequence $x_{j_k, 2}$ which converges to $x_2^* \in \mathbb{R}$. Notice that since $x_{j_k, 1}$ is convergent, so is $x_{j_{k_l}, 1}$. Therefore, if we consider the sequence indexed by $l,\left(x_{j_{k_l}, 1}, \ldots, x_{j_{k_l}, n}\right)$, we now have convergent sequences in the first two components. It is hopefully clear that, aside from running out letters (and having to resort to cleverer notation), we can repeat this procedure $n$ times, iteratively constructing subsequences to ensure that every component is convergent.