Denote by $X$ the space of all real sequences convergent to $0$ and equip it with the metric $d(x,y) =\sup \left\{|x_n – y_n|: n \in\mathbb N\right\}$. Prove that $(X,d)$ is separable.
The solution is Exercise $5$ here http://userwikis.fu-berlin.de/download/attachments/460783619/proofs_topologyI.pdf?api=v2
It says that the set of all rational sequences converging to $0$ is countable "by Cantor's diagonal argument". I don't believe this is true, for example to each real number in $(0,1)$, $x = \sum_{n \ge 1} 10^{-n} x_n$, associate the sequence $x_n/n^2$. Isn't this an injection to that set?
My idea was to take the set of rational sequences which vanish after a natural number. I think that we can write this set as (where $Q$ is rational numbers):
$$\cup_{k \ge 1}\left[ \cup_{1 \le n \le k} Q^n \times \prod_{j \ge k+1} \{0\}\right]$$
This set is countable and dense. Does this make sense to you?
Best Answer
You're right, the set of all rational sequences convergent to zero is uncountable. Your proof is good, here's another using Cantor's (other) diagonal argument.
Suppose they were enumerable, denote the sequences as $s_1, s_2, \dots$. Arrange these top to bottom, where $s_{i, j}$ is the $j$th element of the $i$th sequence. Now consider $s_{n, n}$. Let $k_n$ be the sequence defined by
$$k_n = \begin{cases} 1/n & s_{n, n} \neq 1/n\\ 1/(n+1) & s_{n, n} = 1/n \end{cases} $$
then $k_n \to 0$ but $k_n$ is not in the enumeration.
Your idea for the solution is correct, though your notation is a bit confusing. Why not write simply $S = \{(a_1, a_2, \dots) \mid 0 = a_j = a_{j+1} = \dots \text{ for some } j\}$? This is an equally valid and more intuitive explanation of the set.
Proof of denseness: fix $\varepsilon > 0$ and an arbitrary sequence $x$. For some large $N$, we have $|x_n| < \varepsilon$. Take a sequence $y \in S$ that is non-zero for the first $N$ terms and that approximates each of the first $N$ terms of $x$ with less than $\varepsilon$ error. Then $d(x, y) < \varepsilon$.
Proof of countable: $\mathbb{Q}^n$ is countable as it has the same cardinality as the disjoint union of $n$ copies of $\mathbb{Q}$. $S$ is countable since it has the same cardinality as the countable union of $\mathbb{Q}^n$ for each $n \geq 0$.