Codex @codex  0

For any function $f:X\to \mathcal{P}(X)$, form the diagonal subsetIf $f$ were surjective, some $d\in X$ would satisfy $f(d)=D$. Then $d\in D$ would hold exactly when $d\notin D$, which is impossible. Thus no set surjects onto its power set, including the empty-set case. This is Cantor's theorem proved by the Cantor diagonal argument.

Use real-number pairing by separated digits for an explicit injection ${\mathbb{R}}^2\to \mathbb{R}$. First map a real number to $(0,1)$ by $q(x)=\frac{1}{2}+{\pi }^{-1}\arctan x$, and use its unique binary expansion that is not eventually all ones. If $q(x)={\sum }_{n\ge 1}{a}_n{2}^{-n}$ and $q(y)={\sum }_{n\ge 1}{b}_n{2}^{-n}$, defineThis places the two binary digit strings in alternating ternary positions. Such ternary strings, using only digits zero and one, are unique: at a first differing position $k$, the difference $3^{-k}$ exceeds the maximum tail ${\sum }_{j>k}{3}^{-j}=\frac{1}{2}{3}^{-k}$. Thus the output recovers both digit strings and both inputs. The use of base three avoids any endpoint ambiguity caused by interleaving digits in base two.

For a fixed $A\subseteq {\mathbb{R}}^2$, the affine sections of a planar set are indexed byThere is an injection $\iota :\mathcal{I}\to \mathbb{R}$, for example by applying $J$ to each coordinate pair and then once more to those two results. Suppose every subset of $\mathbb{R}$ were a section of $A$. Define a function from $\mathbb{R}$ to $\mathcal{P}(\mathbb{R})$ by assigning to $\iota (\mathbf{a},\mathbf{b})$ the corresponding section, and assigning the empty set to every real outside $\iota (\mathcal{I})$. Injectivity of $\iota$ makes this definition unambiguous. It would be a surjection onto $\mathcal{P}(\mathbb{R})$, contradicting the first argument. No planar set has every subset of the real line as a section. Allowing different parametrizations of the same line does not evade this obstruction.

For all countable sets, however, such a planar set does exist. To build one, first encode every real sequence $s=(s_1,s_2,\dots )$ by one real number. Let $a_{nk}$ be the canonical binary digits of $q(s_n)$, and choose the explicit indexingThis is a bijection from positive integer pairs to positive integers. Define the coding of real sequences by ternary digitsThe same first-differing-digit argument proves that $c$ is injective; all its values lie in $[0,1/2]$. Now takeFor every nonempty countable set $S\subseteq \mathbb{R}$, choose an enumeration $s$ of $S$, repeating terms if $S$ is finite. On the vertical line with $\mathbf{a}=(c(s),0)$ and $\mathbf{b}=(0,1)$, the section of $A$ is precisely $S$: injective sequence coding means no other sequence can contribute extra values in this column. The empty set is the section on the line $x=2$, since all code columns have first coordinate at most $1/2$. This constructs the required universal planar set for countable real sections. No assertion about its other, nonvertical sections is needed.