Factoids #2 (#238)

Jonathan Frech, 26 December 2020

VII) Cardinality coercion: $\mathbb{R}\hookleftarrow\{\mathbb{N}\to\mathbb{N}\}\twoheadrightarrow\mathbb{R}$

Claim. There exist both $\iota:\{\mathbb{N}\to\mathbb{N}\}\hookrightarrow\mathbb{R}$ together with $\pi:\{\mathbb{N}\to\mathbb{N}\}\twoheadrightarrow\mathbb{R}$ .

Proof. Iota. Define $\iota:\{\mathbb{N}\to\mathbb{N}\}\hookrightarrow\mathbb{R}$ via

$$\begin{aligned}
\iota(a):=\quad&\sum_{j=1}^\infty 2^{\left(1-\sum_{i=1}^{j}(1+a(i))\right)}\cdot(2^{a(j)}-1)\\
=\quad&\left(0.\underbrace{1111\dots11}_{\times a(1)}0\underbrace{111\dots11111}_{\times a(2)}0\dots\right)_2
\end{aligned}$$

and observe any sequence’s reconstructibility by dyadic expansion.
Pi. Define $\pi:\{\mathbb{N}\to\mathbb{N}\}\twoheadrightarrow\mathbb{R}$ via

$$\begin{aligned}
\pi(a):=\quad&a(1)+\sum_{j=2}^\infty 2^{1-j}\cdot\delta_{a(j)\in 2\mathbb{Z}}\\
=\quad&a(1)+\left(0.\delta_{a(2)\in 2\mathbb{Z}}\delta_{a(3)\in 2\mathbb{Z}}\delta_{a(4)\in 2\mathbb{Z}}\dots\right)_2
\end{aligned}$$

and observe any real’s constructibility by dyadic expansion.

Thus, $\{\mathbb{N}\to\mathbb{N}\}\cong\mathbb{R}$ in $\mathbf{(SET)}$ is shown.

-=-

VIII) Codomain crumpling: $\{\mathbb{N}_0\to\mathbb{N}_0\}\cong\mathrm{Sq}^\star_\Delta(\mathbb{N}_0)$

Let $\mathrm{Sq}^\star_\Delta(\mathbb{N}_0):=\{\mathbb{N}_0\to\mathbb{N}_0\}\cong\{f:\mathbb{N}_0\to\{\star,\Delta\}:\#f^{-1}(\{\Delta\})=\infty\}$ .

Claim. There exists $\alpha:\{\mathbb{N}_0\to\mathbb{N}_0\}\ {\sfrac\hookrightarrow\twoheadrightarrow}\ \mathrm{Sq}^\star_\Delta(\mathbb{N}_0)$ .

Proof. Define $\alpha:\{\mathbb{N}_0\to\mathbb{N}_0\}\to\mathrm{Sq}^\star_\Delta(\mathbb{N}_0)$ via

$$\alpha(a):=\quad\left(\underbrace{\star,\star,\dots,\star}_{\times a(0)},\Delta,\underbrace{\star,\dots,\star,\star,\star}_{\times a(1)},\Delta,\dots\right)$$

and observe any $\mathbb{N}_0$ -sequence’s reconstructibility by counting $\Delta$ -segmented section’s lengths as well as any $\{\star,\Delta\}$ -sequence’s constructibility by defining $\star$ -run lengths.

IX) A non-isomorphic Lipschitzian bijection

Claim. A bijective Lipschitzian map need not be an isomorphism in $\mathbf{(MET)}$ with Lipschitzian arrows, i. e. its inverse may degrade in regularity.

Proof. Let

$$\begin{aligned}
f:\mathbb{R}_{\geq 0}&\to\partial\mathrm{B}_1(0)\subset\mathbb{R}^2,\\
t&\mapsto(\cos(4\cdot\arctan t),\sin(4\cdot\arctan t)).
\end{aligned}$$

By $\arctan(\mathbb{R}_{\geq 0})=[0,\sfrac\pi 2[$ and the arctangent’s strict isotonicity together with $\partial\mathrm{B}_1(0)$ ’s conventional parametrization, $f$

’s bijectivity is apparent. Since $f$

is furthermore smooth, $f\mid_{[0,t_1]}$ is globally Lipschitzian for all $t_1\geq 0$ . As the local Lipschitz constant approaches zero for $t_1\nearrow\infty$ , one can deduce $f$

’s global Lipschitz continuity.
Now, looking at a neighborhood $N_{(0,0)}$ of $(0,0)\in\mathbb{R}^2$ one finds for any given $t_1\geq 0$ a radius $\varrho>0$ with its ball $\mathrm{B}_\varrho((0,0))\subset N_{(0,0)}$ satisfying $f^{-1}(\mathrm{B}_\varrho((0,0)))\supset [t_1,\infty[$ , proving $f^{-1}$ ’s non-continuity and thereby its violation of the Lipschitzian property.