$\lim_{j\xrightarrow{} \infty} |x_j|=0 \Rightarrow [x_j]=[0]$

Question

$x_j$ is a sequence, define its equivalent class as

$$ [x_j]=\{(z_j)_{j\geq1} \mid (z_j - x_j)_{j\geq 1}\in Y\}\\ \text{where} \ Y \ \text{is a linear subspace} $$ Do we have the following statement?

$$ \lim_{j\xrightarrow{} \infty} |x_j|=0 \Rightarrow [x_j]=[0] $$

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Sry if there's any unclear conditions, I'm kind of new to functional analysis, my question comes from this video frame

I'll present all background knowledge here $$ X \text{ is a complete normed linear space}\\ Z=\{(x_j)_{j\geq 1} \mid x_j\in X, \ (x_j) \text{ is CAUCHY}\}\\ Y=\{(x_j)_{j\geq 1}\mid x_j\xrightarrow{} 0\}\\ \overline{X}=Z\mid _Y $$ $x_j$ is a sequence, define its equivalent class as

$$ [x_j]=\{(z_j)_{j\geq1} \mid (z_j - x_j)_{j\geq 1}\in Y\}\\ \text{but I cannot conclude which space } x_j \text{ belongs to} $$

Based on the knowledge above, the professor writes down the following inference $$ |[x_j]|=0\\ \Rightarrow \lim_{j\xrightarrow{} \infty} |x_j|=0 \\ \Rightarrow [x_j]=[0] \\ \Rightarrow (x_j) \in Y $$

My question is how does he derive the following inference in the middle? $$ \lim_{j\xrightarrow{} \infty} |x_j|=0 \Rightarrow [x_j]=[0] $$

Answer 1

$X$ is not a complete space. The point of the construction is to construct the completion of $X$. So here $X$ is a normed vector space.

In a complete space, any Cauchy sequence is convergent. When your space is not complete, you have Cauchy sequences which are not convergent. The point of the construction is to make up limit points for such sequences. The idea is to work on the space of sequences, and identity sequences that "convergent to the same point".

So one takes $$Z=\{(x_k)_k:\ x_k\in X\ \text{ for all }k\}.$$ To identify sequences that "converge to the same point" one needs to take a quotient. For that, let $$ Y=\{(x_j)_j\in Z:\ \lim_j\|x_j\|=0\}. $$ It is easy to show that $Y$ is a subspace of $Z$, and hence one defines the completion of $X$ as $$ \overline X=Z/Y. $$ The elements of $\overline X$ are classes of sequences. The class of the sequence $(x_j)_j$ is denoted by $[(x_j)_j]$ and it consists of those sequences $(z_j)$ such that $(x_j)-(z_j)\in Y$, which is to say $\lim_j x_j-z_j=0$. We want to make $\overline X$ a normed space; for this we define $$\tag1 \|[(x_j)]\|=\lim_j\|x_j\|. $$ Of course one needs to check that this is well defined: but this is easy since $$ |\,\|x_j\|-\|z_j\|\,|\leq\|x_j-z_j\| $$ which shows that if $[(x_j)]=[(z_j)]$ then $\lim\|x_j\|=\lim_j\|z_j\|$.

Next one needs to check that $(1)$ really defines a norm. The homogeneity and the triangle inequality are easy. And if $\|[(x_j)]\|=0$, this means by definition that $\lim_j\|x_j\|=0$; this is exactly $(x_j)\in Y$ (check the definition of $Y$). And in the quotient, $Y$ is the class of $0$. So $\|[(x_j)]\|=0$ implies that $[(x_j)]=[0]$.

What remains is to argue that $\overline X$ is a completion of $X$ and that it is complete. What one does is to embed $X\hookrightarrow\overline X$ by $$ x\longmapsto [(x)] $$ (the class of the sequence which is constantly $x$). It is trivial to check that this map is linear, and moreover $$ \|[(x)]\|=\lim_j\|x\|=\|x\| $$ so the embedding is isometric. As for showing that $\overline X$ is complete, it's not difficult but a bit cumbersome; details can be found here.