Bolzano-Weierstrass in \({\mathbb R}^n\)

Suppose that \(\{x_k\}_{n=1}^\infty\) is a bounded sequence in \({\mathbb R}^n\). Because all the usual norms we use on \({\mathbb R}^n\) are comparable, let us assume that we are using \(\ell^2\), i.e., that there is some \(M\) such that

where the coordinates of \(x_k\) are denoted \(((x_k)_1,\ldots,(x_k)_n)\).

In particular, this means that for any fixed value of \(\ell \in \{1,\ldots,n\}\), we have \(|(x_k)_\ell| \leq M\) for all \(k\), meaning that in every direction \(\ell\), the sequence of coordinates \(\{(x_k)_\ell\}_{k=1}^\infty\) is bounded.

By the one-dimensional Bolzano-Weierstrass Theorem, there must be an increasing sequence of indices

such that \((x_{k_i^{(1)}})_1\) converges as \(i \rightarrow \infty\); in other words, there is a subsequence along which the first coordinates converge.

Now consider the sequence \((x_{k_i^{(1)}})_2\): it's the same indices but now viewed in the second coordinate direction instead of the first. This is itself a bounded sequence (because it's a subsequence of \(\{(x_k)_2\}_{k=1}^\infty\)). This means that there is an increasing sequence of indices

which are themselves a subsequence of \(x_1^{(1)},k_2^{(1)},\ldots\) such that \((x_{k_i^{(2)}})_2\) converges as \(i \rightarrow \infty\). Because it's a subsequenc of \(x_1^{(1)},k_2^{(1)},\ldots\), it is also true that \((x_{k_i^{(2)}})_1\) converges as \(i \rightarrow \infty\). By induction, for each \(\ell = 1,\ldots,n\), there is an increasing sequence of indices

which for \(\ell > 1\) is a subsequence of \(k_1^{(\ell-1)},k_2^{(\ell-1)},\ldots\) and has the property that

converges whenever \(\ell' \leq \ell\). If we let \((j_1,j_2,\ldots) := (k_{1}^{(n)},k_{2}^{(n)},\ldots)\), then it follows that

exists for each \(\ell' = 1,\ldots,n\). Let \(x \in {\mathbb R}^n\) have the property that for each \(\ell'\), the coordinate \((x)_{\ell'}\) equals the limit (1). Then using the usual properties of limits of real numbers, we know that

which means exactly that \(x_{j_i}\) converges to \(x\) in the \(\ell^2\) norm as \(i \rightarrow \infty\).

Suppose that \(B \subset {\mathbb R}^n\) is a product of closed and bounded intervals \(I_1 \times \cdots I_n\). Given any such box \(B\), we may always subdivide it into exactly \(2^n\) smaller boxes of exactly one half the size by taking each \(I_j\) and dividing it into left and right halves, i.e., \(B\) is the union of boxes \((I_1)_{*_1} \times \cdots (I_n)_{*_n}\) where \((I_j)_{*_j}\) denotes either the left half or right half of \(I_j\) (and the number \(2^n\) comes from the fact that there are exactly \(2^n\) ways to choose left or right in each direction).

Now we mirror the one-dimensional proof of Bolzano-Weierstrass. Every bounded sequence in \({\mathbb R}^n\) must belong to some box \([-M,M]^n\) for an appropriate value of \(M\). Inductively, we may let \(B_0 := [-M,M]^n\) and we may construct a sequence of shrinking boxes

such that \(\operatorname{diam} B_k = 2^{-k} \operatorname{diam} B_0\) for each \(k\) and such that each \(B_k\) has \(x_j \in B_k\) for infinitely many values of \(j\) (just as in the 1D case, if none of the subdivided “children” of \(B_k\) contained \(x_j\) for infinitely many \(j\), then taking the union of all children would force that \(B_k\) itself only contains \(x_j\) for finitely many \(j\)). It is therefore possible to take a subsequence \(x_{k_i}\) such that \(x_{k_i}\) belongs to \(B_i\) for each \(i\). Because projections of the \(B_i\) onto any given coordinate direction yields a nested sequence of closed intervals whose diameter goes to zero, we know that those intervals must have a point in common, and in particular, there must be a point \((x)_\ell\) belonging to \((B_i)_\ell\) (the projection onto direction \(\ell\)) for each \(i\). If \(x\) is the vector \(((x)_1,\ldots,(x)_n)\), it follows that \(x\) belongs to each box \(B_i\). To conclude, we know that the distance from \(x_{k_i}\) to \(x\) is at most \(2^{-i} \operatorname{diam} B_0\) for each \(i\) since both \(x_{k_i}\) and \(x\) belong to \(B_i\) (which has diameter \(2^{-k} \operatorname{diam} B_0\)). This implies that \(x_{k_i} \rightarrow x\) as \(i \rightarrow \infty\).