Limit of vectors and limit of components

Question

This question is related to the (closed) topic :

Closed subspace of vector space

Let $(u_n)\in F^{\mathbb{N}}$ be a sequence converging to $l$ and $(e^1,\ldots,e^p)$ be a basis of $F$ (vector space).

There exists $(x_n)_n\in (\mathbb{R}^p)^\mathbb{N}$ so that for all $n$, $$ u_n = x_n^1e^1+\ldots x_n^p e^p. $$

Also $(u_n)$ converges if and only if all the real valued sequence of its components converge.

Then, denote by $x_\star^i$ the limit of $x_n^i,\, i=1\ldots p,$ and $u_\star$ the limit of $u_n$.

Then $u_\star = x_\star^1e^1+\ldots +x_\star^pe^p$.

What is wrong here ?

Answer 1

$\def\R{\mathbb R}$ I see by looking at your other question that you intend $F$ to be a linear subspace of a real Euclidean space $E$. $F$ then has a metric $||\cdot||$ induced by the metric on $E$, and you also give $F$ a fixed basis, which induces a vector space isomorphism $\phi :\R^p \to F$, where $p = \dim F$.

You ask in effect whether the following statements are true:

1. $F$ is closed in $E$.
2. $F$ is complete with respect to the metric inherited from $E$.
3. The vector space isomorphism $\phi :\R^p \to F$ is a homeomorphism, where $\R^p$ is given the standard topology.
4. A sequence in $F$ converges if and only if its coordinate sequences with respect to the given basis converge.

Yes, all four statements are true and they are more or less equivalent. There are several ways to go about proving this. For example, you can go about it this way: The norm on $F$ gives you a norm $||\cdot||'$ on $\R^p$ via $||x||' = ||\phi(x)||$. It is a theorem that any two norms on $\R^p$ are equivalent, and convergence with respect to any norm on $\R^p$ is equivalent to co-ordinatewise convergence. This theorem implies statements 2, 3 and 4, and statement 1 is a consequence.

Addendum: For the equivalence of all norms on a finite dimensional space, see https://math.stackexchange.com/q/2702434. This uses compactness of the unit sphere with respect to one of your favorite norms. Another approach to the whole problem would be to show statement (1) first, as follows. Show that $E = F \oplus F^\perp$ and use finite dimensionality to show $F = (F^\perp)^\perp$. Conclude from this that $F$ is closed.

Answer 2

This statement is not justified:

Also $(u_n)$ converges if and only if all the real valued sequence of its components converge.

How would you prove it?

It is related to the question whether $l \in F$, which is precisely what you are trying to prove.

What you should try to prove is that the finite-dimensional space $F$ is complete (i.e. every Cauchy sequence converges). Then it will follow that $F$ is also closed.