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The Archimedean property says that the natural numbers are unbounded in the real numbers. In particular, for every positive real number $a$ there is a positive integer $n$ with $1/n<a$. Otherwise a purported upper bound for all natural numbers contradicts the least-upper-bound property by translating that bound by one.

For a nonempty set $S\subseteq \mathbb{R}$, an upper bound in a partially ordered set is a real number $u$ such that $s\le u$ for every $s\in S$. Its supremum, or least upper bound, is an upper bound $L$ with $L\le u$ for every upper bound $u$. The least-upper-bound property asserts that every nonempty set of real numbers bounded above has a real supremum.

Let $(x_n)$ be a bounded, increasing sequence; here increasing allows $x_{n+1}=x_n$. Apply the least-upper-bound property to its set of terms and put $L=\sup \{x_n:n\ge 1\}$. Given $\epsilon >0$, the number $L-\epsilon$ cannot be an upper bound, so some $x_N$ satisfies $x_N>L-\epsilon$. The monotone sequence property then givesHence $|x_n-L|<\epsilon$ for all $n\ge N$. This proves the monotone bounded sequence result, with limit $\boxed{x_n⟶L}$.

For the two series, define their partial sums $A_N={\sum }_{n=1}^N{a}_n$ and $B_N={\sum }_{n=1}^N{b}_n$. Positivity makes both partial-sum sequences increasing, and termwise comparison gives $0<B_N\le A_N$. Since $(A_N)$ is a convergent sequence, it is bounded above, and therefore so is $(B_N)$. Applying the monotone bounded sequence result already proved shows that $(B_N)$ converges. By the definition of a convergent series, ${\sum }_{n=1}^{\infty }{b}_n$ converges. This is the comparison test for series for positive terms.

The intermediate value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous and $y$ lies between $f(a)$ and $f(b)$, then some $c\in [a,b]$ satisfies $f(c)=y$.

Assume $f(a)<y<f(b)$ and letThe least-upper-bound property makes $c$ well defined. If $f(c)<y$, continuity gives points immediately to the right of $c$ in $S$, contradicting that $c$ is an upper bound. If $f(c)>y$, continuity gives a left neighborhood of $c$ disjoint from $S$, contradicting the definition of the supremum. Thus $f(c)=y$. The endpoint and reversed-order cases follow directly or by replacing $f$ with $-f$.

The mean value theorem states that if $g$ is continuous on $[\alpha ,\beta ]$ and differentiable on $(\alpha ,\beta )$, then some $c\in (\alpha ,\beta )$ satisfiesFor the functions in the question, the definition of the derivative makes $h$ continuous at $a$ and $f$ continuous at $b$, withIf the middle secant slope equals $k$, take $[\alpha ,\beta ]=[a,b]$. If it exceeds $k$, the intermediate value theorem applied to $h$ gives $\beta \in (a,b)$ with $h(\beta )=k$, and we take $\alpha =a$. If it is below $k$, apply the theorem to $f$ to obtain $\alpha \in (a,b)$ with $f(\alpha )=k$, and take $\beta =b$. In every caseThe mean value theorem on this subinterval then produces $c\in (\alpha ,\beta )\subset (a,b)$ withThis is the Darboux theorem for derivatives for the stated pair of derivative values.

First recall from the least-upper-bound property that every real interval is connected. For completeness, if $[a,b]=U\cup V$ were a separation with $a\in U$, let $s$ be the supremum of the points $t$ for which $[a,t]$ remains in $U$. Relative openness rules out both $s\in U$ and $s\in V$: in the first case one can move slightly to the right while staying in $U$, and in the second one finds points of $V$ slightly to the left of $s$, contrary to the definition of $s$. Thus no separation exists.

Part (a) now makes $g([0,1])$ a connected subset of the real line. Such a set is an interval: if it contained $u<v$ but omitted $y\in (u,v)$, its intersections with $(-\infty ,y)$ and $(y,\infty )$ would separate it. Since $g([0,1])$ contains $g(0)$ and $g(1)$, it consequently contains every number between them. Therefore for each such $y$ there is an $x\in [0,1]$ with $\boxed{g(x)=y}$. This derives the intermediate value theorem from connectedness.