Closure of the set of null sequences starting from some rank

Question

Let $E$ be the vector space of real bounded sequences $(x_n)_n$. We endow $E$ with the $\sup$ norm. Let $A$ be a subset of $E$ made of all the sequences that are null starting from some rank.

I must show that the closure of $A$ is a subset of $E$ made of all the sequences that converge to 0.

Here is a draft of my proof. I don't know if it is enough to prove the assertion. Furthermore, I find my proof a bit messy. It could be more rigorous.

Let $B = \{ (x_n) \in E : \forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n \geq N, \lvert u_n \rvert < \epsilon \}$ and $A = \{(x_n) \in E : \exists N \in \mathbb{N}, \forall n \geq N, u_n = 0 \}$.

$\text{cl}(A)$ contains the limits of all the convergent sequences of A.

Let $l \in \text{cl}(A)$. Hence, $l$ is the limit of a convergent sequence $(x_n)_n$ of A.

More formally,

\begin{equation*} \forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n \geq N, \lvert\lvert x_n - l \rvert\rvert < \epsilon \end{equation*}

So,

\begin{equation*} \forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n \geq N, \sup_{n \in \mathbb{N}} \lvert x_n - l \rvert < \epsilon \end{equation*}

Yet, starting from some rank $N \in \mathbb{N}$, the terms $(x_n)_n$ are null. Let such rank $N \in \mathbb{N}$. Starting from this rank, it follows that:

\begin{equation*} \forall \epsilon > 0, \forall n \geq N, \sup_{n \in \mathbb{N}} \lvert x_n - l \rvert < \epsilon \end{equation*}

\begin{equation*} \forall \epsilon > 0, \forall n \geq N, \lvert l \rvert < \epsilon \end{equation*}

Or more simply,

\begin{equation*} \forall \epsilon > 0, \lvert l \rvert < \epsilon \end{equation*}

Hence $l = 0$.

So, $\bar{A}$ is made of all the sequences $(x_n)_n$ that converges to 0.

Answer 1

You definitely have the right idea: all sequences that converge to $0$ are sufficiently "close" to an element of $A$.

There are a few instances where I'm not totally sure of your "notation", so I'm just going to write out what I think is a correct answer, and you can judge whether you have essentially the same thing.

Given some $(x_n) \in \bar{A}$, we first have to show that $(x_n)$ converges to $0$. Since $(x_n)$ is in the closure, it follows that for any $\epsilon > 0$, we may choose some $(y_n) \in A$ such that $d((x_n), \ (y_n)) < \epsilon$.

Let $M$ to the number such that for $m > M$, we have $y_m = 0$. It then follows, from the fact that $d((x_n), \ (y_n)) < \epsilon$ we have:

$$|x_m - y_m| = |x_m| < \epsilon$$

for all $m > M$. Hence, by definition, $(x_n)$ converges to $0$.

Now, let $(x_n)$ be a sequence that converges to $0$. As you said, this means that for any $\epsilon > 0$, we can choose some $N$ such that for $n > N$, we have:

$$|x_n| < \epsilon$$

Consider the sequence $(y_n)$ such that $y_j = x_j$ for all $j$ up until $N$, and then $0$ after. Clearly, we will have:

$$|x_n - y_n| = 0 \ \text{for} \ n \leq N$$ $$|x_n - y_n| = |x_n| < \epsilon \ \text{for} \ n > N$$

Therefore, $d((x_n), \ (y_n)) < \epsilon$. It follows that the open $\epsilon$-ball around $(x_n)$ intersects $A$. Hence, $(x_n) \in \bar{A}$.

We have inclusion both ways, so $\bar{A}$ is exactly the set of all sequences that converge to $0$. Let me know if this helps/if you have any more questions!