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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?

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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 $n \geq 0$.

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