Bolzano-Weierstrass Theorem

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Next we present the famous Theorem of Bolzano-Weierstraß.

Proof: First we consider the case \(\mathbb{K}=\mathbb{R}\). Since \((a_n)_{n\in\mathbb{N}}\) is bounded, there exist some \(A,B\in\mathbb{R}\) such that for all \(n\in\mathbb{N}\) holds \(A\leq a_n\leq B\). We will now successively construct subintervals \([A_n,B_n]\subset[A,B]\) which still include infinitely many sequence elements of \((a_n)_{n\in\mathbb{N}}\).

Inductively define \(A_0=A\), \(B_0=B\) and for \(k\geq1\),

• \(A_k=A_{k-1}\), \(B_k=\frac{A_{k-1}+B_{k-1}}2\), if the interval \([A_{k-1},\frac{A_{k-1}+B_{k-1}}2]\) contains infinitely many sequence elements of \((a_n)_{n\in\mathbb{N}}\), and

• \(A_k=\frac{A_{k-1}+B_{k-1}}2\), \(B_k=B_{k-1}\), else.

By the construction of \(A_k\) and \(B_k\), we have that each interval \([A_k,B_k]\) has infinitely many sequence elements of \((a_n)_{n\in\mathbb{N}}\). We furthermore have \(B_1-A_1=\frac12(B-A)\), \(B_2-A_2=\frac14(B-A)\), \(\ldots\), \(B_k-A_k=\frac1{2^k}(B-A)\). Moreover, the sequence \((A_n)_{n\in\mathbb{N}}\) is monotonically increasing and bounded from above by \(B\), i.e., it is convergent by the theorem on monotonic an bounded sequences.

The relation \(B_k-A_k=\frac1{2^k}(B-A)\) moreover implies that \((B_n)_{n\in\mathbb{N}}\) is also convergent and has the same limit as \((A_n)_{n\in\mathbb{N}}\). Denote \[a=\lim_{n\to\infty}A_{n}=\lim_{n\to\infty}B_{n}.\] Define a subsequence \((a_{n_k})_{k\in\mathbb{N}}\) by \(n_1=1\) and \(n_k\) with \(n_k>n_{k-1}\) and \(a_{n_k}\in[A_k,B_k]\) (which is possible since \([A_k,B_k]\) contains infinitely many elements of \((a_{n})_{n\in\mathbb{N}}\)). Then \(A_k\leq a_{n_k}\leq B_k\). The theorem on bounded monotonic sequences then implies that \[a=\lim_{k\to\infty}a_{n_k}.\]

Finally we consider the case \(\mathbb{K}=\mathbb{C}\). Write \(a_n=b_n+ic_n\) where \(i\) is the imaginary unit, \(b_n:=\mathbb{R}e(a_n)\) denotes the real part and \(c_n:={\Im}(a_n)\) denotes the imaginary part of \(a_n\). Since \(|a_n|=\sqrt{b_n^2+c_n^2}\geq \max\{|b_n|,|c_n|\}\geq 0\), the boundedness of the complex sequence \((a_n)_{n\in\mathbb{N}}\) implies the boundedness of both real sequences \((b_n)_{n\in\mathbb{N}}\) and \((c_n)_{n\in\mathbb{N}}\).

Then, by the previous, we now that \((b_n)_{n\in\mathbb{N}}\) has a convergent subsequence \((b_{n_k})_{k\in\mathbb{N}}\). Since the subsequence \((c_{n_k})_{k\in\mathbb{N}}\) of the bounded sequence \((c_n)_{n\in\mathbb{N}}\) is also bounded, it also has a convergent subsequence \((c_{n_{k_m}})_{m\in\mathbb{N}}\). The subsequence \((b_{n_{k_m}})_{m\in\mathbb{N}}\) of the convergent sequence \((b_{n_k})_{k\in\mathbb{N}}\) also converges. Hence \((a_{n_{k_m}})_{m\in\mathbb{N}} = (b_{n_{k_m}}+ic_{n_{k_m}})_{m\in\mathbb{N}}\) is a convergent subsequence of \((a_n)_{n\in\mathbb{N}}\) with \(\lim_{m\rightarrow \infty} a_{n_{k_m}} = \lim_{m\rightarrow \infty} b_{n_{k_m}} + i\cdot \lim_{m\rightarrow \infty} c_{n_{k_m}} \ .\) \(\Box\)