Is the union of l^p a Banach space under some norm? As a set of sequences, take the union of $\ell^p$, $p\geq 1$. As $p$ increases, the $\ell ^p$ space is larger, with strict inclusion.
However, this infinite union is strictly contained in $c_0$, consider
$x_n = 1/\log (n+1)$, so the usual $c_0$ norm will not yield a closed space, as $c_{00}$ is dense within any $\ell ^p$.
Is there some norm which turns the union of l^p into a Banach space, presumably with $c_{00} $dense?
 A: Although such a norms exists by the axiom of choice, it is an interesting question what nice properties the unit vector basis $(e_n)_n$ could possibly have. From my point of view, it cannot have very nice ones.
For example, $(e_n)_n$ cannot be a Schauder basis. It cannot even be the spatial component of a Markushevich basis. Let's see why not.
Let $X = \cup_{1\leq p<\infty}\ell_p$, $\|\cdot\|_X$ be a complete norm for $X$, and let us assume that $(e_n,f_n)_n$ is an M-basis for $(X,\|\cdot\|_X)$.
For every $1\leq p < \infty$ the inclusion map $I_p:\ell_p\to X$ is well defined. By the closed graph theorem it is also bounded. Indeed, if $(x_n)_n$ converges to zero in $\ell_p$ and $(x_n)_n$ converges to some $x\in X$ (with respect to $\|\cdot\|_X$) then $x$ must be zero. Here we used the fact that $(e_n)_n$ is the spatial component of an $M$-basis and completeness. Both will be used again later.
But it is also the case that for no $1\leq p<\infty$, $\ell_p$ the map $I_p$ is an isomorphism, otherwise your space $X$ would be just $\ell_p$. A small argument (using automatic continuity of linear maps on finite dimensional spaces) tells us that for any $n\in\mathbb{N}$ if we restrict $I_p$ to the tail-space $Y_n$ of $\ell_p$ (i.e., the subspace of all vectors in $\ell_p$ with the first $n$ coordinates zero), then it cannot be an isomorphism. The conclusion is that for any $1\leq p<\infty$, $\varepsilon >0$, and $n\in\mathbb{N}$, there exist $k\in\mathbb{N}$ and a scalar sequence $(a_i)_{i=n+1}^{n+k}$ so that if we define $x = \sum_{i=n+1}^{n+k}a_ie_i$ then $\|x\|_p = 1$ while $\|x\|_X\leq \varepsilon$.
Now we are almost finished. Take $p_n = n$ and inductively pick $(a_i)_{i=k_{n-1}+1}^{k_n}$ so that if $x_n = \sum_{i=k_{n-1}+1}^{k_n}a_ie_i$, then $\|x_n\|_{p_n} = 1$, yet $\|x_n\|_X\leq 2^{-n}$.
We use completeness again and put $x = \sum_{n=1}^\infty x_n$. For every $i\in\mathbb{N}$, the $i$'th coordinate of $x$ (i.e., $f_i(x)$) is $a_i$. By assumption, there must exist some $1\leq p<\infty$ so that $(a_i)_i\in\ell_p$. This is absurd because for every $1\leq p<\infty$, if we pick $n$ with $p_n\geq p$ then
$$\|(a_i)_i\|_p \geq \|(a_i)\|_{p_n} \geq \|(a_i)_{i\geq k_{n-1}+1}\|_{p_n} = (\sum_{j=n}^\infty \|x_j\|_{p_n}^{p_n})^{1/p_n} \geq (\sum_{j=n}^\infty \|x_j\|_{p_j}^{p_n})^{1/p_n} = \infty.$$
