Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics II / 1

True. Put $\alpha =\sup A$, $\beta =\sup B$ and $u=\max \{\alpha ,\beta \}$. These suprema exist by completeness of the real numbers, since the sets are nonempty and bounded. Every set element of the union $A\cup B$ is at most $u$, so $u$ is an upper bound. Conversely, every upper bound of $A\cup B$ bounds both $A$ and $B$, and is therefore at least both suprema. Thus

True. Write $\alpha =\sup A$ and $\beta =\inf B$. Every $a\in A$, $b\in B$ satisfies $a-b\le \alpha -\beta$, giving an upper bound for the set $A-B$ of pairwise differences. For every $\epsilon >0$, the definitions of supremum and infimum give $a\in A$ and $b\in B$ withThen $a-b>\alpha -\beta -\epsilon$. Hence no smaller upper bound works, andHere $A-B$ means pairwise subtraction, rather than set difference.

False. Take the finite sets $A=B=\{-2,-1\}$. Their suprema are both $-1$, whereas the set of pairwise products is $A\cdot B=\{1,2,4\}$. ConsequentlyNegative factors can reverse an inequality, which prevents the proposed argument with separate suprema.

For the first sequence, $|a_n|=n^{-3}\to 0$. The squeeze theorem gives $a_n\to 0$, and every subsequence of a convergent sequence has the same limit of a sequence. Thus its set of subsequential limits is $\boxed{\{0\}}$.

For the second sequence, the four possible sine values giveThe even-indexed subsequence is constantly one, and $b_{4k+3}\to 0$, so both zero and one occur as subsequential limits. A convergent subsequence is bounded, so it contains only finitely many terms with $n\equiv 1(\mathrm{mod}4)$. Of the remaining three residue classes, at least one occurs infinitely often. The corresponding further subsequence has limit of a sequence zero or one; the subsequence rule forces the original convergent subsequence to have that same limit of a sequence. Therefore the exact answer is $\boxed{\{0,1\}}$. Divergence to $+\infty$ along $n\equiv 1(\mathrm{mod}4)$ is not a finite subsequential limit.

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence. It is enough to select infinitely many terms of the tangent sequence that stay uniformly bounded.

For each positive integer $k$, choose $n_k=\lfloor k\pi +1/2\rfloor$, using the floor function. Then $|n_k-k\pi |\le 1/2$. Moreover $n_{k+1}-n_k\ge \pi -1>0$, so these are strictly increasing indices. By the period and monotonicity of the tangent on $(-\pi /2,\pi /2)$,Thus $(\tan n_k)$ is a bounded subsequence. Applying the Bolzano-Weierstrass theorem to it gives a convergent subsequence, which is also a subsequence of the original sequence. Such a convergent subsequence therefore exists.