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**manjohn12** Show that the subset $\displaystyle A $ of terminating sequences is countable (e.g. $\displaystyle a_n = 0 $ for $\displaystyle n >N $).

So suppose $\displaystyle E $ is a countable subset of $\displaystyle A $. Then $\displaystyle E = \{s_1, s_2, \cdots, s_n \} $ where $\displaystyle s_k $ ($\displaystyle k=1 \ldots n $) are terminating sequences. Is it possible to just say that is is impossible to construct a sequence $\displaystyle s $ such that $\displaystyle s \notin E $. Thus not every subset $\displaystyle E $ of $\displaystyle A $ is proper. Hence $\displaystyle A $ is countable.

Or is the only way to do it is to say that $\displaystyle A $ is equivalent to $\displaystyle \mathbb{Q}^{\mathbb{N}} $?