There are three spaces of real sequences given:
1) $c$ - a space of convergent sequences,
2) $c_0$ - a space of sequences such that $\forall x_n\in c_0,\ \lim_{n \to\infty} x_n = 0$,
3) $c_{00}$ - a space of sequences with finite amout of numbers different than $0$.

With the supremum norm: $$\sup_{k\in \mathbb{N}}|x_k^n - x_k^m|$$

I am to show whethere they are separable spaces or not.
I think that it will be enough to show that $c_0$ is separable, won't it?

I think so because $c_{00} \subset c_0 \subset c$.

However I don't really know how to show the separability of $c_{00}$. I think it must be something related to rational numbers but I don't know where to go from there.

  • 1
    $\begingroup$ Separable with respect to which topology? $\endgroup$ Mar 10, 2018 at 15:53
  • $\begingroup$ @JoséCarlosSantos I will edit my post. Sorry for that! $\endgroup$
    – Hendrra
    Mar 10, 2018 at 15:55

2 Answers 2


All three spaces $c_{00}, c_0$ and $c$ are indeed separable.

To show that $c_{00}$ is separable, let $S$ be the set of all finitely-supported sequences with rational coordinates. Clearly, $S$ is countable.

Let $(x_n)_{n=1}^\infty \in c_{00}$ be a sequence, and assume that $m \in \mathbb{N}$ is such that $x_k = 0$ for all $k > m$.

Fix $\varepsilon > 0$. For $1 \le n \le m$ take a rational number $q_n \in \mathbb{Q}$ such that $|q_n - x_n| < \frac{\varepsilon}{m}$.

Verify that $(q_1, q_2, \ldots, q_m, 0, 0, \ldots) \in S$ is $\varepsilon$-close to $(x_n)_{n=1}^\infty$. Therefore, $S$ is dense in $c_{00}$ so $c_{00}$ is separable.

It is known that $c_{00}$ is dense in $c_0$ (namely, approximate $(x_n)_{n=1}^\infty \in c_{0}$ with sequences of the form $(x_1, x_2, \ldots, x_n, 0, 0, \ldots) \in c_{00}$).

Now, notice that $S$ is also dense in $c_{0}$. Fix $\varepsilon > 0$. For $x \in c_{0}$ take $y \in c_{00}$ such that $|x-y| < \frac\varepsilon2$ and then take $z \in S$ such that $|y - z| < \frac\varepsilon2$.

Now we have

$$|x-z| \le |x-y| + |y-z| < \varepsilon$$

so $S$ is dense in $c_{0}$. We conclude that $c_{0}$ is separable.

Finally, notice that $c = \operatorname{span}\{(1, 1, \ldots )\} \oplus c_0$, in the sense that every convergent sequence $(x_n)_{n=1}^\infty \in c$ can be written in the form

$$c = \left(\lim_{n\to\infty} x_n \right)\cdot (1, 1, \ldots ) + \underbrace{(x_n)_{n=1}^\infty - \left(\lim_{n\to\infty} x_n \right) \cdot (1, 1, \ldots )}_{\text{converges to $0$}}$$

Use this to conclude that $c$ is also separable.

You can also explicitly show that the set of all eventually constant sequences with rational entries is a countable dense subset of $c$.

  • 1
    $\begingroup$ Today I studied your proof deeply and it's really nice and "tidy". Thanks! :) $\endgroup$
    – Hendrra
    Mar 11, 2018 at 21:18

Another proof that $c$ is separable:

We will show that $(j^{-n})_{j\in\Bbb N},n\in\Bbb N_0$ is dense in $c$.

For any $(x_j)_{j\in\Bbb N}\in c$, for any $ϵ>0,$

Let $f∈C[0,1]$ be such that $f(j^{-1})=x_j\;∀j\in\Bbb N$, then $f(0)=\lim_{j→∞} x_j$.

By Weierstrass's theorem, there is a polynomial $λ_0+λ_1x+⋯+λ_Nx^N$ such that $$\left|f(x)-\sum_{n=0}^N λ_n x^n\right|\ltϵ ∀x∈[0,1]$$ For $x=j^{-1}$ we get $$\left|x_j-\sum_{n=0}^N λ_n j^{-n}\right|\ltϵ  ∀j\in\Bbb N.$$


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