Bounded partial quotients set is nowhere dense

Question

I've stumbled upon a claim that the set: $$ B_N = \{[a_0;a_1,a_2,...] | \exists n_0 >0\forall n\geq n_0 a_n<N\} $$ for some $N$, is nowhere dense (and closed). Unfortunately, I have found that proving that its complement is open dense set isn't helpful for me either.

Any clues?

Answer 1

Judging by your comment to Henno, this answers the underlying question; I’ll assume material in the introductory discussion in Henno’s answer.

Let $B(m,n)=\{[a_0;a_1,a_2,\dots]:\forall k\ge m(a_k\le n)\}$. If $[a_0;a_1,a_2,\ldots]\notin B(m,n)$, then there is a $k\ge m$ such that $a_k>n$, and $\{[b_0;b_1,b_2,\ldots]:b_i=a_i\text{ for all }i\le k\}$ is a basic open nbhd of $[a_0;a_1,a_2,\ldots]$ disjoint from $B(m,n)$, so $B(m,n)$ is closed. It’s also not hard to see that $B(m,n)$ cannot contain any basic open set. If $\{[b_0;b_1,b_2,\ldots]:b_i=a_i\text{ for all }i\le k\}$ is a basic open nbhd of some $[a_0;a_1,a_2,\ldots]\in B(m,n)$, fix some $j\ge\max\{k+1,m\}$ and let

$$c_i=\begin{cases} b_i,&\text{if }i\ne j\\ n+1,&\text{if }i=j\;; \end{cases}$$

then $$[c_0;c_1,c_2,\ldots]\in\{[b_0;b_1,b_2,\ldots]:b_i=a_i\text{ for all }i\le k\}\setminus B(m,n)\;,$$

and

$$\{[b_0;b_1,b_2,\ldots]:b_i=a_i\text{ for all }i\le k\}\nsubseteq B(m,n)\;.$$

Thus, each $B(m,n)$ is closed and nowhere dense, so their union, your set $B$, is a meagre $F_\sigma$-set.

Answer 2

The set of non-terminating (as you have) continued fractions is just the irrational numbers, and as a topological space the function $h$ mapping an irrational $x$ to its sequence $(a_0, a_1,\ldots,)$ of "components" such that $x = [a_0; a_1, a_2,\dots]$ is a homeomorphism between the irrationals (also called $\mathbb{P}$) and the space $\mathbb{Z} \times \mathbb{N}^\mathbb{N}$ in the standard product topology (also denoted $\omega^\omega$ or $\mathcal{N}$. This is nicely explained here, e.g.

So we can just look at your set as living in $\omega^\omega$, really.

And in the product topology recall that basic open sets are of the form: products of open sets in the component spaces, where all but finitely many of the component open sets equal the whole space; essentially we can have finitely many restricted coordinates in such a basic open set.

We cannot show your set (we fix $N$ in the definition of the set, if I understand you correctly) is closed: if $(a_n)$ is not in your set, this means exactly that there are infinitely many coordinates $k$ with $a_k > N$. Any basic open set around such an $(a_n)$ (or any $(a_n)$, for that matter) will also contain points that are eventually bounded by $N$, because we can only impose restrictions on finitely many coordinates, so all other coordinates could be all $1$, say. So it seems that your set is in fact dense (it intersects all basic open sets), not nowhere dense. But maybe I am misunderstanding the definition of your set. Could you show us a link to the source?