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Factoids #2

2020-12-26, post № 238

mathematics, #bijection, #calculus, #Lipschitz, #naturals

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.
Jonathan Frech's blog; built 2024/04/13 20:55:09 CEST