Convervence in the normed space.

Question

By definition a sequence convergence in normed space if

$||x_n-x||\le \varepsilon \;\forall n \gt N.$

And this limit must be the element of this space. Does it mean that space must be complate by default?

EDIT: Or can every sequence converges in noncomplate normed space?

Answer 1

Short answer: no.

Long answer: maybe you are confusing the convergence of a sequence with the convergence of all sequences. In general, we say that if $(X,||\cdot ||)$ is a normed space, then a sequence $\{x_n \}_{n\in\mathbb{N}}$ converges to a point $x\in X$ if for all $\varepsilon>0$ there exist $N\in\mathbb{N}$ such that, if $n\geq N$, then $||x_n-x||<\varepsilon$. On the other side, we say that a normed (or more general, metric) space $Y$ is complete if every Cauchy sequence in $Y$ converges to some element in $Y$. With this, is it clear the difference between the two things?

For your last question, the answer is no. If $X$ is a normed space and every sequence converges, in particular every Cauchy sequence converges. Thus, $X$ is complete by definition. (Note that here the word converges meaning converges to some point of the space).