completeness of a sequence space

Question

Consider the following space: $X(P):=\{(x_n)_{n\geq0}\in\mathbb{R}^{\mathbb{N}}:(\lambda_nx_n)_{n\geq0}\in l_1, \forall (\lambda_n)_{n\geq0}\in P\}$, where $P$ is a random set of real sequences st. $X(P)$ is separated. Using the familiy $f_{\lambda}:X(P)\rightarrow l_1$ mapping $(x_n)_{n\geq0}\rightarrow (\lambda_nx_n)_{n\geq0}$ we can equip $X(P)$ with the initial topology. Now the question: Is $X(P)$ complete?

I know that using the whole family $(f_{\lambda})_{\lambda\in P}$ one can look at the product $\prod_{\lambda\in P}l_1$ which is complete since $l_1$ is. But even if I could show that $X(P)$ is closed, it is no subset of $\prod_{\lambda\in P}l_1$.

Any idea how to proof/disproof completeness here?

score 3 · Accepted Answer · answered Mar 21 '14 at 09:31

The map

$$F \colon X(P) \to \prod_{\lambda \in P} l_1;\qquad F\left((x_n)_{n\in\mathbb{N}}\right) = \left((\lambda_nx_n)_{n\in\mathbb{N}}\right)_{\lambda\in P}$$

is linear. Since by assumption $X(P)$ is separated, $F$ is injective, and the initial topology on $X(P)$ is just the subspace topology of $F(X(P))$ transported back to $X(P)$ via $F$ (the initial topology with respect to the bijection $F\colon X(P) \to F(X(P))$).

So we need to see whether $F(X(P))$ is a closed subspace of the product.

For all $\lambda,\mu \in P$, $n\in\mathbb{N}$ and $a,b \in \mathbb{R}$, the subspace

$$S(\lambda,\mu,n,a,b) = \left\{(\xi_\lambda) \in \prod_{\lambda\in P}l_1 : a\cdot(\xi_\lambda)_n = b\cdot (\xi_\mu)_n\right\}$$

is closed (it is the kernel of the continuous linear functional $\xi \mapsto a(\xi_\lambda)_n - b(\xi_\mu)_n$). Hence

$$M = \bigcap_{\lambda,\mu,n} S(\lambda,\mu,n,\mu_n,\lambda_n)$$

is closed. It remains to see that $M = F(X(P))$.

For each $n$, choose a $\lambda^{(n)} \in P$ with $\lambda^{(n)}_n \neq 0$. Such a $\lambda^{(n)}$ must exist, since otherwise $F(e_n) = 0$. Then, for $\eta\in M$, define $x \in X(P)$ by

$$x_n := \frac{1}{\lambda^{(n)}_n}\cdot (\eta_{\lambda^{(n)}})_n.$$

Check that indeed $x \in X(P)$, and $F(x) = \eta$.

Just to check if i got it right. The idea is to show that $F(X(P))$ is closed in the "$l_1$-product", hence the preimage under $F$ is a closed subspace of $\mathbb{R}^{\mathbb{N}}$, a complete space (since product of $\mathbb{R}$) and therefore itself complete? Or is it somehow crucial that $F$ is bijective? — Mr. Barrrington, Mar 22 '14 at 10:23
Since we endow $X(P)$ with the initial topology, and not the subspace topology induced by $\mathbb{R}^{\mathbb{N}}$, it's not important whether $X(P)$ is closed in the subspace topology (it isn't usually; since it contains the subspace of sequences with only finitely many nonzero terms it is dense, so closed only if the entire space, and that happens only if all $\lambda\in P$ have only finitely many nonzero terms). We need that $F$ is bijective, since that means $X(P)$ gets the subspace topology induced by $\prod l_1$ on $F(X(P))$. — Daniel Fischer, Mar 22 '14 at 11:01

completeness of a sequence space

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