Here is a part of an exercise (from a book) I can't figure out how to solve :
Les $V$ be the set of all functions $f: \mathbb{N} \to \mathbb{R}$. We define also the functions $e_i(n)$ by $e_i(n)=1$ if $i=n$ and $0$ otherwise. We set $B=\lbrace e_i : i \in \mathbb{N} \rbrace$ and $W=span(B)$.
The aim is to show that there is no isomorphism between $V$ and $W$.
I really don't see anything, really... Can you help me please? Thank you very much!
On
RETRACTED, this argument doesn't seem to work. My bad.
No cardinality argument is needed. All you need to do is find a function $f: \mathbb{N} \rightarrow \mathbb{R}$ that's not in the linear span of $\{e_i\}$.
What does an element of $W$ look like? It's some function $g = \sum_{i=1}^n c_i e_i$ for some real numbers $\{c_i\}$. The point being that this is a finite sum by definition of linear span.
So every element of $W$ is a sequence having only finitely many nonzero terms.
Now the answer becomes obvious. The function $f(n) = 1$ is a sequence containing infinitely many nonzero terms; so it can not possibly be a linear combination of any of the $\{e_i\}$'s, so these two vector spaces are not the same.
On
HINT: Show that the [algebraic] dual of $W$ is isomorphic to $V$, and use the fact that $W$ is infinitely dimensional.
(Other hints might include showing that every subspace of $W$ is countable or finitely generated; whereas $V$ has subspaces of infinite dimension which are Banach spaces. Then use the fact that a Banach space cannot have a countable dimension, for example by using the Baire Category Theorem.)
For any $I \subset \mathbb{N}$, define $f_I : \mathbb{N}\to \mathbb{R}$ by $f_I(n)=1$ if $n \in I$ and $0$ otherwise. You can show that $\{ f_I \mid \emptyset \neq I \subset \mathbb{N}\}$ is a family of linearly independent functions of cardinality $|\mathfrak{P}(\mathbb{N})|= |\mathbb{R}|$. Therefore, the dimension of $V$ is uncountable whereas $\dim(W)$ is clearly countable.