Let $U \subseteq \mathbb{R}^{n}$ be a set. What does it mean to say that $U$ is open? Show that if $U$ is open and if $f: U \rightarrow\{0,1\}$ is a continuous function then $f$ is also differentiable, and that its derivative is zero.

Suppose that $g: U \rightarrow \mathbb{R}$ is differentiable and that $\left\|\left.(D g)\right|_{x}\right\| \leqslant M$ for all $x$, where $\left.(D g)\right|_{x}$ denotes the derivative of $g$ at $x$ and $\|\cdot\|$ is the operator norm. Suppose that $\mathbf{a}, \mathbf{b} \in \mathbb{R}^{n}$ and that the line segment $[\mathbf{a}, \mathbf{b}]=\{\lambda \mathbf{a}+(1-\lambda) \mathbf{b}: \lambda \in[0,1]\}$ lies wholly in $U$. Prove that $|g(\mathbf{a})-g(\mathbf{b})| \leqslant M\|\mathbf{a}-\mathbf{b}\|$.

Let $\ell_{1}, \ldots, \ell_{k}$ be (infinite) lines in $\mathbb{R}^{3}$, and write $V=\mathbb{R}^{3} \backslash\left(\ell_{1} \cup \cdots \cup \ell_{k}\right)$. If $\mathbf{a}, \mathbf{b} \in V$, show that there is some $\mathbf{c} \in V$ such that the line segments $[\mathbf{a}, \mathbf{c}]$ and $[\mathbf{c}, \mathbf{b}]$ both lie inside V. [You may assume without proof that $\mathbb{R}^{3}$ may not be written as the union of finitely many planes.]

Show that if $V \rightarrow\{0,1\}$ is a continuous function then $f$ is constant on $V$.