Past exam of the mathematics course of the University of Cambridge / 2014 / ia / Paper 4 / 7E (Numbers and Sets)

A set is a countable set if it admits an injective function into the natural numbers; this includes finite sets and countably infinite sets. A countably infinite set can be listed by a bijection from $\mathbb{N}$.

Let $Z$ be the set of infinite binary sequences. Suppose it were countable and list all its members as $s^{(1)},s^{(2)},\dots$, writing $s_n^{(j)}$ for the $n$th entry of row $j$. A finite listing could be repeated to make such a list as well. Define the binary sequence $y$ by $y_n=1-s_n^{(n)}$. It differs from the $n$th row in coordinate $n$, for every $n$. Thus it belongs to $Z$ but is missing from the list, a contradiction. The set $Z$ is uncountable, by Cantor's diagonal argument.

Use $\mathbb{N}=\{1,2,\dots \}$. If a bijection $f:\mathbb{N}\to \mathbb{N}$ maps $S$ onto the even natural numbers, its restriction is a bijection from $S$ onto an infinite set. Moreover, its restriction to the complement of $S$ maps onto the odd natural numbers. Both sets are therefore infinite.

Conversely, suppose that $S$ and its complement are infinite. Applying the increasing enumeration of an infinite subset of natural numbers to each givesFor completeness, successively choose the least unused member. The process never stops because the set is infinite, and it lists every member: an omitted member would have infinitely many selected natural numbers below it, which is impossible. DefineThis is injective on each list, their images are disjoint, and together their images cover $\mathbb{N}$. Thus it is a bijection and maps $S$ onto $2\mathbb{N}$. Both infinitude conditions are necessary and sufficient. If a convention includes zero in $\mathbb{N}$, index each list from zero and use $2j$, $2j+1$ instead.

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.