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Cantor's diagonal argument proves that the infinite binary sequences form an uncountable set: a sequence obtained by changing the $n$th digit of the $n$th listed sequence differs from every sequence in the list.

A parabola parametrized by $(t,t^2)$ is in bijection with $\mathbb{R}$ and is an uncountable set. Every affine line meets it in at most two points. A countable family of affine lines could therefore cover only countably many points of this parabola, and cannot cover the plane. More generally, an uncountable curve with finite intersection with every permitted member obstructs a countable covering by those members.

Use the parabola obstruction to countable line coverings. The curveis in bijection with the real line and hence is an uncountable set. A vertical line meets $\mathcal{P}$ in one point. A nonvertical line $y=ax+b$ meets it only where $t^2-at-b=0$, so it meets $\mathcal{P}$ in at most two points.

If countably many lines covered the plane, their intersections with $\mathcal{P}$ would cover $\mathcal{P}$. That would express $\mathcal{P}$ as a countable union of countable sets, indeed of sets with at most two elements, contradicting its uncountability. Therefore no countable collection of lines covers the plane.

For the cardinality of integer-periodic function spaces, fix a positive period $n$ first. A periodic function $\mathbb{Z}\to \mathbb{Z}$ with that period is completely and uniquely determined by its $n$ values at $0,\dots ,n-1$, so this class is in bijection with ${\mathbb{Z}}^n$. The integers are a countable set, and repeated pairing shows that a finite Cartesian product of countable sets is countable. Taking the union over positive $n$ and using the countable union of countable sets proves that the integer-domain periodic functions form a countable set.

For the rational domain, even period one leaves infinitely many independent choices. Let $q_k=1/(k+2)$ for $k\in {\mathbb{N}}_0$; these are distinct members of $\mathbb{Q}\cap [0,1)$. For every subset $A\subseteq {\mathbb{N}}_0$, defineThe fractional part is unchanged by adding one, so $f_A$ is an integer-valued periodic function on $\mathbb{Q}$. Distinct subsets give distinct functions because $f_A(q_k)$ records whether $k\in A$. Thus the uncountable power set of ${\mathbb{N}}_0$ injects into these functions. The rational-domain periodic functions form an uncountable set. The difference is that $\mathbb{Z}$ has only finitely many residue classes modulo a fixed integer period, whereas $\mathbb{Q}$ has infinitely many.

Use uncountability by interleaving prescribed binary coordinates. For each arbitrary binary sequence $b=(b_k{)}_{k\ge 1}$, define another binary sequence byThere are infinitely many zero coordinates, namely $3k-2$, so $x\in X$. There are also infinitely many coordinates agreeing with the prescribed binary expansion, namely $3k-1$, so $x\in Y$. The coordinates $3k$ recover $b$, making this map an injection of the uncountable set of all binary sequences into $Y$. If $Y$ were a countable set, composing with an injection into $\mathbb{N}$ would make all binary sequences countable, contrary to part (i). Therefore $\boxed{Y\text{ is uncountable}.}$ The construction works for every prescribed binary sequence $(a_n)$; no special distribution property of the digits of $\sqrt{2}$ is needed.

Suppose $f:S\to \mathcal{P}(S)$ were a surjective function. Form the subsetSince $D$ belongs to the power set, surjectivity would give $D=f(d)$ for some $d\in S$. But then $d\in D$ holds if and only if $d\notin f(d)=D$, a contradiction. Thus no such function is surjective, and in particular there is no bijection from $S$ to its power set. This is Cantor theorem; the argument also applies when $S$ is empty.

For infinite binary sequences, use the bijectionwith $\mathcal{P}(\mathbb{N})$, taking $\mathbb{N}=\{1,2,\dots \}$. Conversely, a subset determines its sequence of membership indicators. The power set is infinite, since it contains all singletons, and Cantor theorem rules out a bijection with $\mathbb{N}$. Hence it is an uncountable set, as is the set of infinite binary sequences. Equivalently, a proposed enumeration $(x_n^{(j)}{)}_{n\ge 1}$ is defeated by the sequence $y_j=1-x_j^{(j)}$, which differs from its $j$th listed sequence at position $j$. This is the Cantor diagonal argument.

Partition the enumeration of $B$ into the pairs $\{b_{2j-1},b_{2j}\}$, for $j\ge 1$. An infinite binary sequence $x$ determines a bijection $h_x:B\to B$: leave the $j$th pair fixed when $x_j=0$, and interchange its two elements when $x_j=1$. Each element stays in its own pair, so this function is a bijection, indeed its own inverse.

Different binary sequences give different bijections, since a changed bit changes the action on the corresponding pair. We have therefore embedded an uncountable set into $S_3$. Hence $S_3$ is uncountable, proving uncountability of permutations of a countably infinite set.

Again let $A=\{a_1,\dots ,a_m\}$ with $m>1$, and enumerate $B=\{b_1,b_2,\dots \}$. For each infinite binary sequence $x=(x_j{)}_{j\ge 1}$, define a function $g_x$ byThe first $m$ values make every $g_x$ a surjective function, while the remaining values recover $x$ uniquely. Hence $x\mapsto g_x$ is an injective function from the uncountable set of binary sequences into $S_2$. This proves $S_2$ is an uncountable set, an instance of uncountability of surjections onto a nontrivial finite set.

False. Take $x=\mathcal{P}(\omega )$, the power set of the first infinite ordinal $\omega$. Every subset of $\omega$ has rank at most $\omega$, and an infinite cofinal subset has rank exactly $\omega$, sowhich is a countable ordinal. But Cantor theorem shows that $\mathcal{P}(\omega )$ is an uncountable set.

Suppose all functions $\mathbb{N}\to \mathbb{N}$ could be listed as $f_1,f_2,\dots$. DefineThen $g$ differs from $f_n$ at $n$ for every $n$, so it is absent from the list. This Cantor's diagonal argument proves that ${\mathbb{N}}^{\mathbb{N}}$ is an uncountable set.

There are uncountably many physically distinct one-qubit unitaries. Part (ii) would require an orthogonal program state for every one, whereas a $K$-qubit register has dimension only $2^K$. This contradiction is the no-programming theorem, so no deterministic gate satisfying (PROG) exists.

Partition the positive natural numbers into three infinite coordinate sets. Fix the coordinates on the first two sets to any prescribed binary values, and leave those on the third set free. Reading the third coordinate set embeds all infinite binary sequences in the resulting family, so the family is an uncountable set by Cantor's diagonal argument. For example, fixing $x_{3k-2}=0$ and $x_{3k-1}=a_{3k-1}$ while leaving $x_{3k}$ free guarantees infinitely many zeros and infinitely many agreements with any prescribed binary sequence $(a_n)$. Infinite free coordinates remain even after both requirements are enforced.