1

Given \(A,B\) two disjoint compact sets in \(\mathbb{R}^n\) prove that there exists continuous function from \(\mathbb{R}^n\to\mathbb{R}\) that sends \(A\) to \(1\) and \(B\) to \(0\).

Explicitly construct \[f(x) = \frac{d(x,B)}{d(x,A) + d(x,B)}\] where \(d(x,A) := \inf\left\{d(x,a):a\in A\right\}\). Easy check: \(f\) satisfies \(f\upharpoonright A = 1\) and \(f\upharpoonright B = 0\). It suffices to show that for any compact set \(A\), the map \[ x \mapsto d(x,A) = \inf\left\{d(x,a):a\in A\right\} \] is continuous. Let \(x_n \to x\), want \(\lim d(x_n, A) = d(x, A)\). Note that for any \(n \in \mathbb{N}, a\in A\), \[\begin{align*} d(x_n,a)\leq d(x_n, x) + d(x, a) \\ d(x_n,A)\leq d(x_n, x) + d(x, A) \end{align*}\] and similarly \[ d(x,A)\leq d(x_n, x) + d(x_n, A) \] which gives \[ \left\lvert d(x,A) - d(x_n,A)\right\rvert \leq d(x_n, x). \tag*{$\square$}\]

2

1. Suppose the graph is closed, we take the preimage of the function \(\operatorname{id}\times f\) (which is continuous and injective by the way, so a homeomorphism) to get \(E\times E\) being closed.

2. image of \(f\) is compact and we express \(G(f)\) as \(E\times \operatorname{range}(f)\).

3. If \(G(f)\) compact then both \(E\) and image \(f(E)\) are compact. Suppose for contradiction \(f\) is not continuous, then there exists a sequence \(x_n \to x\) where \(\lim f(x_n) \ne f(x)\). Now we look at the sequence \(\left(x_n, f\left(x_n\right)\right)\) of elements in \(G(f)\), observe that its limit is not in the graph of \(f\).

4. Counterexample shouldn’t be hard

3 Orthogonal matrix path-connected

hard, do later

4 Riemann function is nowhere differentiable

\[\begin{align*} R&: \mathbb{R}\to\mathbb{R}\\ R(x) &= \begin{cases} \frac1q &\text{if }x=\frac{p}q, (p,q)=1, p>0 \\ 0 &\text{if }x\notin \mathbb{Q} \end{cases} \end{align*}\]

5 Construct stuff

1. continuous function \(\mathbb{R}\to\mathbb{R}\) differentiable at almost all points in \(\mathbb{R}\) except \(a_1,a_2,\dots,a_n\)

   just make it bendy at each \(a_i\)

2. function \(\mathbb{R}\to\mathbb{R}\) (not necessarily continuous) only differentiable at \(a_1,\dots,a_n\)