Jonathan. Frech’s WebBlog

Factoids #2 (#238)

Jonathan Frech,

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

Claim. There exist both $\iota:\{\mathbb{N}\to\mathbb{N}\}\hookrightarrow\mathbb{R}$$\iota:\{\mathbb{N}\to\mathbb{N}\}\hookrightarrow\mathbb{R}$ together with $\pi:\{\mathbb{N}\to\mathbb{N}\}\twoheadrightarrow\mathbb{R}$$\pi:\{\mathbb{N}\to\mathbb{N}\}\twoheadrightarrow\mathbb{R}$.
Proof. Iota. Define $\iota:\{\mathbb{N}\to\mathbb{N}\}\hookrightarrow\mathbb{R}$$\iota:\{\mathbb{N}\to\mathbb{N}\}\hookrightarrow\mathbb{R}$ via
$$\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}$$$$\begin{aligned} \iota(a):=\quad&\sum_{j=1}^\infty 2^{\left(1-\sum_{i=1}^{j}(1+a(i))\right)}\cdot(2^{a(j)}-1)\\ =\quad&\left(0.\underbrace{1111\dots11}_{\times a(1)}0\underbrace{111\dots11111}_{\times a(2)}0\dots\right)_2 \end{aligned}$$
and observe any sequence’s reconstructibility by dyadic expansion.
Pi. Define $\pi:\{\mathbb{N}\to\mathbb{N}\}\twoheadrightarrow\mathbb{R}$$\pi:\{\mathbb{N}\to\mathbb{N}\}\twoheadrightarrow\mathbb{R}$ via
$$\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}$$$$\begin{aligned} \pi(a):=\quad&a(1)+\sum_{j=2}^\infty 2^{1-j}\cdot\delta_{a(j)\in 2\mathbb{Z}}\\ =\quad&a(1)+\left(0.\delta_{a(2)\in 2\mathbb{Z}}\delta_{a(3)\in 2\mathbb{Z}}\delta_{a(4)\in 2\mathbb{Z}}\dots\right)_2 \end{aligned}$$
and observe any real’s constructibility by dyadic expansion.

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

-=-

VIII) Codomain crumpling: $\{\mathbb{N}_0\to\mathbb{N}_0\}\cong\mathrm{Sq}^\star_\Delta(\mathbb{N}_0)$$\{\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\}$$\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)$$\alpha:\{\mathbb{N}_0\to\mathbb{N}_0\}\ {\sfrac\hookrightarrow\twoheadrightarrow}\ \mathrm{Sq}^\star_\Delta(\mathbb{N}_0)$.
Proof. Define $\alpha:\{\mathbb{N}_0\to\mathbb{N}_0\}\to\mathrm{Sq}^\star_\Delta(\mathbb{N}_0)$$\alpha:\{\mathbb{N}_0\to\mathbb{N}_0\}\to\mathrm{Sq}^\star_\Delta(\mathbb{N}_0)$ via
$$\alpha(a):=\quad\left(\underbrace{\star,\star,\dots,\star}_{\times a(0)},\Delta,\underbrace{\star,\dots,\star,\star,\star}_{\times a(1)},\Delta,\dots\right)$$$$\alpha(a):=\quad\left(\underbrace{\star,\star,\dots,\star}_{\times a(0)},\Delta,\underbrace{\star,\dots,\star,\star,\star}_{\times a(1)},\Delta,\dots\right)$$
and observe any $\mathbb{N}_0$$\mathbb{N}_0$-sequence’s reconstructibility by counting $\Delta$$\Delta$-segmented section’s lengths as well as any $\{\star,\Delta\}$$\{\star,\Delta\}$-sequence’s constructibility by defining $\star$$\star$-run lengths.

IX) A non-isomorphic Lipschitz bijection

Claim. A bijective Lipschitz map need not be an isomorphism in $\mathbf{(MET)}$$\mathbf{(MET)}$ with Lipschitz arrows, i. e. its inverse may degrade in regularity.
Proof. Let
$$\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}$$$$\begin{aligned} f:\mathbb{R}_{\geq 0}&\to\partial\mathrm{B}_1(0)\subset\mathbb{R}^2,\\ t&\mapsto(\cos(4\cdot\arctan t),\sin(4\cdot\arctan t)). \end{aligned}$$
By $\arctan(\mathbb{R}_{\geq 0})=[0,\sfrac\pi 2[$$\arctan(\mathbb{R}_{\geq 0})=[0,\sfrac\pi 2[$ and the arctangent’s strict isotonicity together with $\partial\mathrm{B}_1(0)$$\partial\mathrm{B}_1(0)$’s conventional parametrization, $f$$f$’s bijectivity is apparent. Since $f$$f$ is furthermore smooth, $f\mid_{[0,t_1]}$$f\mid_{[0,t_1]}$ is globally Lipschitz for all $t_1\geq 0$$t_1\geq 0$. As the local Lipschitz con­stant approaches zero for $t_1\nearrow\infty$$t_1\nearrow\infty$, one can deduce $f$$f$’s global Lipschitz continuity.
Now, looking at a neighborhood $N_{(0,0)}$$N_{(0,0)}$ of $(0,0)\in\mathbb{R}^2$$(0,0)\in\mathbb{R}^2$ one finds for any given $t_1\geq 0$$t_1\geq 0$ a radius $\varrho>0$$\varrho>0$ with its ball $\mathrm{B}_\varrho((0,0))\subset N_{(0,0)}$$\mathrm{B}_\varrho((0,0))\subset N_{(0,0)}$ satisfying $f^{-1}(\mathrm{B}_\varrho((0,0)))\supset [t_1,\infty[$$f^{-1}(\mathrm{B}_\varrho((0,0)))\supset [t_1,\infty[$, proving $f^{-1}$$f^{-1}$’s non-continuity and thereby its violation of the Lipschitz property.