# Decimal expansions and topological connectedness

I'm a bit confused by the following footnote from Moschovakis's Notes on Set Theory, p. 135fn24 (in the note, $\mathcal{N}$ denotes the Baire space). The puzzling part is in bold:

One may think of $\mathcal{N}$ as a "discrete", "digital", or "combinatorial" version of the "continuous" or "analog" $\mathbb{R}$. A real number $x$ is completely determined by a decimal expansion $x(0), x(1), x(2), \dots$, where ($n \mapsto x(n)) \in \mathcal{N}$, but two distinct decimal expansions may compute to the same real number. This is a big "but", it is the key fact behind the so-called topological connectedness of the real line which is of interest in analysis, to be sure, but of little set theoretic consequence. We may view Baire space as a "digital version" of $\mathbb{R}$ because it does not make any such identifications, each point $x \in \mathcal{N}$ determines unambiguously its "digits" $x(0), x(1), \dots$.

I don't understand the bold part. What is the relation between the fact that some real numbers have two distinct decimal expansions and topological connectedness? Does anyone have any clue about the thought behind the quotation?

Consider the real number $1$. The fact that it is the limit of the sequence

$$\langle 1.1,1.01,1.001,1.0001,\ldots\rangle\;,$$

for instance, implies that it must be equal to $1.0000\ldots\;$. The fact that it is the limit of the sequence

$$\langle 0.9,0.99,0.999,0.9999,\ldots\rangle$$

shows that it must be equal to $0.9999\ldots\;$. If this were not the case, i.e., if $0.999\ldots$ were not equal to $1.000\ldots\;$, then

$$L=\{x\in\Bbb R:x\le0.999\ldots\}$$

and

$$R=\{x\in\Bbb R:x\ge1.000\}$$

would be a separation of $R$: $L$ and $R$ would be disjoint closed sets whose union was $\Bbb R$, showing that $\Bbb R$ was not connected.

• Perfect! Thanks a lot for this. – Nagase Apr 6 '16 at 1:30
• @Nagase: You’re very welcome. – Brian M. Scott Apr 6 '16 at 1:31