# Closure of the set of null sequences starting from some rank

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.

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!

• That's crystal clear! Thanks! Feb 5, 2021 at 18:38
• Absolutely, glad to help! Feb 5, 2021 at 18:39