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Is $c$, the Banach space of convergent sequences with the sup norm, separable?

Let $X$ be the set of all sequences which are rational numbers, that converge to some rational number $x$. As the rationals are countably infinite, we need to only show that $X$ is dense in $c$.

That is what I am a bit unsure how to show... help would be appreciated.

user3784030
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    Your set $X$ is uncountable. There are a lot more sequences of rational numbers than there are rational numbers. – user180040 Oct 08 '14 at 01:27

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Hint. Consider the set $S$ of sequences of rational numbers that are eventually constant.

user180040
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  • Would you be able to provide a bit of insight as to how I would show that $S$ is countable and dense in $c$? – user3784030 Oct 08 '14 at 02:20
  • @user3784030 : Start by showing the sequences ${ a_{k}}_{k=1}^{\infty}$ he describes which are constant for $k \ge n$ are countable (involves only the choice of $n$ rational numbers.) Then use the fact that the countable union of countable sets is countable. – Disintegrating By Parts Oct 08 '14 at 09:02
  • @T.A.E. so could I do this by stating that as every rational number is of the form $a\over b$, that there exists at most $n$ of these $a$'s and $n$ of these $b$'s (as after $n$ the values are constant), and therefore there exists a finite combination of these $a,b$, so therefore the sequence of rational numbers is countable? – user3784030 Oct 08 '14 at 09:43
  • @user3784030 : The set of rational numbers is countable; it is a countable union of countable sets $\cup_{n=0}^{\infty}{\cdots,-n/3,-n/2,-n/1,n/1,n/2,n/3,\cdots}$. More fundamentally, you can look at this diagram : http://en.wikipedia.org/wiki/Rational_number#Properties . Then you can build on that result by showing that the sequences user180040 describes are constructed as countable unions of countable sets. – Disintegrating By Parts Oct 08 '14 at 10:18