Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 1 / 4F / Solution

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence.

For a proof, place all terms in a closed bounded interval $I_1$. Bisect it and choose a closed half $I_2$ containing infinitely many terms. Continue inductively, choosing nested closed intervalswith infinitely many sequence terms and with lengths tending to zero. Choose $n_k>n_{k-1}$ such that $x_{n_k}\in I_k$. The nested interval theorem gives a unique point $x$ in every $I_k$. Because both $x_{n_k}$ and $x$ lie in $I_k$,Thus $(x_{n_k})$ is a convergent subsequence.

Now suppose every convergent subsequence of the bounded sequence $(x_n)$ converges to $L$. If $x_n$ did not converge to $L$, there would be an $ϵ>0$ and a subsequence $(x_{n_k})$ satisfyingfor every $k$. This subsequence is bounded, so Bolzano--Weierstrass gives a convergent subsubsequence. By hypothesis its limit is $L$, contradicting the displayed inequality. Hence the unique subsequential limit of a bounded sequence principle gives

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