Density of a subset using sequences

Question

Let $E$ be the set of bounded sequences of real numbers. Let $E'$ be the subset of $E$ of sequences converging to $0$ and $E''$ the subset of $E'$ of sequences with finite support. Consider the two norms $N_1(x_n)=\sup_n|x_n|$ and $N_2(x_n)=\dfrac{\sup_n|x_n|}{n+1}$. I want to study the density of :

1. $E''$ in $E$ for $N_1$

2. $E''$ in $E$ for $N_2$

3. $E''$ in $E'$ for $N_1$

For the second one it seems true. Indeed, taking a sequence $x=(x_1,x_2,\cdots,)$ in $E$ we can consider the sequence $y_n$ defined by $y_1=(x_1,0,\cdots)$, $y_2=(x_1,x_2,0,\cdots)$ and so on. So it is clear that for each $n$, $\displaystyle N_2(y_n-x)=\dfrac{\sup_i|(y_n)_i-x_i|}{n+1}\le \dfrac{M}{n+1}$ which goes to zero as $n$ goes to infinity (here $M$ comes from the boundedness of the sequences). So clearly $y_n$ goes to $x$ as $n$ goes to infinity for the norm $N_1$ and so $E''$ is dense in $E$ for the norm $N_2$.

But I think this is not true the cases 1) and 3) but I could not come up with an example to disprove the density of these subsets.

Answer 1

Your guess is incorrect ; density works in case 1) (and therefore in case 3) also).

Here is why. Let $u=(u_n)\in E$. Then $u$ is bounded, so there is an integer $M\gt 0$ such that $|u_n|\leq M$ for all $n$. Let $\varepsilon \gt 0$. Take $N$, an integer such that $N\gt \frac{1}{\varepsilon}$.

Now, define the sequence $v_n=\frac{\lfloor Nu_n \rfloor}{N}$. Then by construction, from $-M \leq u_n \leq M$ we deduce thet each $v_n$ is of the form $\frac{j}{N}$ where $j$ is an integer in $[-MN,MN]$. So $(v_n)$ takes a finite number of values, and hence $(v_n)\in E''$.

Now, $N_1(u_n-v_n)\leq \frac{1}{N} \leq \varepsilon$. Since $\varepsilon$ was arbitrary, we have just shown that $E''$ is dense in $E$ (and threfore in $E'$ also) for $N_1$.

Answer 2

Isn't 3 correct as well? Consider, for $\lbrace x_n \rbrace \in E'$, $$M_1 = \sup( \lbrace \lvert x_n \rvert \ | 1 \leq n \rbrace),$$ $$M_2 = \sup(\lbrace \lvert x_n \rvert \ | 2 \leq n\rbrace),$$

and so on. Then, $\lbrace y_n \rbrace = \lbrace M_1, M_2, ..., M_n, 0, 0, 0,... \rbrace$ is a sequence in $Y_n$ which converges to $\lbrace x_n \rbrace$ for $N_1$.