Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 4 / 7E (Numbers and Sets)

For a nonempty set $S\subset \mathbb{R}$ bounded above, $\sup S$ is its least upper bound: every $s\in S$ satisfies $s\le \sup S$, and every $u<\sup S$ fails to be an upper bound.

A sequence $x_n$ converges to $x$ if for every $\epsilon >0$ there is $N$ such that $n\ge N$ implies $|x_n-x|<\epsilon$.

Every convergent sequence is bounded. Conversely, let $(x_n)$ be increasing and bounded above, and set $L=\sup \{x_n:n\ge 1\}$. For any $\epsilon >0$, $L-\epsilon$ is not an upper bound, so some $x_N>L-\epsilon$. Monotonicity then givesfor all $n\ge N$. Hence $|x_n-L|<\epsilon$, proving the monotone bounded sequence criterion.

LetFor $n\ge 3$, the binomial theorem givesConsequently $(1000n^2+4)/1.1^n\to 0$: given $\epsilon >0$, the displayed rational upper bound is below $\epsilon$ for all sufficiently large $n$. Thus, directly from the definition of convergence,

Yes. Put $\alpha =\sup S$ and $\beta =\sup T$. Every $s+t\le \alpha +\beta$, so this is an upper bound. Given $\epsilon >0$, choose $s\in S$ with $s>\alpha -\epsilon /2$ and $t\in T$ with $t>\beta -\epsilon /2$. Then $s+t>\alpha +\beta -\epsilon$. Hence

No. Take $S=T=\{-2,-1\}$. Thenbut $ST=\{1,2,4\}$ and therefore $\sup (ST)=4$.

No, even when every element of $S$ is positive. TakeThen $S^T=\{1/2,1/4\}$, so