Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 2 / 11F (Topics in Analysis)

The Baire category theorem says that in a nonempty complete metric space, a countable intersection of open dense subsets is dense. Equivalently, such a space cannot be a countable union of nowhere dense sets: a set is nowhere dense when the interior of its closure is empty. Indeed the complements of those closures are open dense; their intersection is dense and in particular nonempty, so a countable union of the original sets cannot exhaust the space. The same argument shows that the complement of such a union is dense.

First interpret the decimal expansions as fractional expansions $0.d_1{d}_2\cdots$, with every $d_j\in \{4,5\}$. The resulting set $X_0$ is compact: a diagonal subsequence argument selects successively stabilized digits from any sequence in $X_0$, and the tail bound ${\sum }_{j>m}9\cdot 10^{-j}=10^{-m}$ proves convergence to the corresponding expansion.

It has empty interior. At level $m$ its points lie in $2^m$ decimal intervals, each of length $10^{-m}$, whose total length is $(1/5{)}^m$. An interval of positive length could not lie in all these covers. Thus $X_0$ is closed and nowhere dense. Each translate $X_0+q$ is also nowhere dense, and the rational numbers are countable. The Baire category theorem applied to $\mathbb{R}$ yieldsAny point in this complement has the required property. If the printed description also permits integer digits before the decimal point, there are only countably many such integer parts, and negative expansions merely add reflected translates. This makes $X$ a countable union of translates or reflections of $X_0$. The same countable-union argument therefore still proves that some real $t$ cannot be written as $x+q$.