# Complete Inequivalent Norms

So I know of the following result:

If $X$ is infinite dimensional, then number of inequivalent norms on $X$ is $2^{\dim X}$.

I was wondering if there is a similar result for complete norms. Basically I want to know how many complete inequivalent norms exist on an infinite dimensional space.

Edit 1: Is it at least possible to get a bound?

Edit 2: I now know an example of a space with no complete norms. So I would like to change my question to: If there exist a complete norm on an infinite dimensional vector space, then can we get an upper bound to number of inequivalent complete norms?

I also know that given a complete norm, we can construct an inequivalent complete norm. How to proceed further?

Any hints are appreciated.

• Of course we must use AC here. In naked ZF, we cannot even exhibit a real vector space with two inequivalent complete norms. – GEdgar Sep 30 '16 at 17:49
• What can we do using AC? – Sahiba Arora Sep 30 '16 at 17:52
• Note that the bound here definitely fails. For $\dim (X) = \mathbb{N}$, there are no complete norms. – AJY Oct 5 '16 at 16:26
• Yes, that's why I edited to ask that if there are complete norms, then what would the bound be? – Sahiba Arora Oct 5 '16 at 16:27

Assume that $X$ is a real vector space of (infinite) dimension $\kappa$, and that there exists at least one complete norm on $X$. Let us try to show that there are $2^\kappa$ inequivalent complete norms on $X$.
Let us first show that there are at most $2^\kappa$ distinct norms on $X$. Let $(e_i)_{i\in I}$ be a basis of $X$, so that $\vert I\vert=\kappa$. A norm $N$ on $X$ is completely determined by its restriction to $X_{\mathbb Q}$, the set of all linear combinations of the vectors $e_i$ with rational coefficients. (This is because $N$ is determined by its restriction to $E_J:={\rm span}\{e_i;\; i\in J\}$ for any finite subset $J$ of $I$, and that $N$ is continuous on $E_J$ with respect to the unique vector space topology of the finite-dimensional space $E_J$.) Now, $X_{\mathbb Q}$ has the same cardinality as $\bigcup_{J\in\mathcal P_f(I)} \mathbb Q^J$, where $\mathcal P_f(I)$ is the set of all finite subsets of $I$; and since $\kappa$ is infinite, this is equal to $\kappa$. (The set $\mathcal P_f(I)$ has cardinality $\kappa$.) So, there are no more norms on $X$ than there are maps from $\kappa$ to $\mathbb R$; and since the cardinality of $\mathbb R^\kappa$ is equal to $\kappa$ (because $\mathbb R$ has cardinality $2^{\aleph_0}$), the result follows.
Now let us show that there are $2^\kappa$ pairwise inequivalent complete norms on $X$. Let $S$ be a set of cardinality $\kappa$. Then $S\times \mathbb N$ has cardinality $\kappa$ as well because $\kappa$ is infinite. Let $(e_{s,n})_{(s,n)\in S\times \mathbb N}$ be a basis of $X$. For any subset $A$ of $S$, consider the linear map $J_A:X\to X$ defined on the basis as follows: $J_A(e_{s,n})=ne_{s,n}$ if $s\in A$, and $J_A(e_{s,n})=e_{s,n}$ if $s\notin A$. Then $J_A$ is a linear automorphism of $X$ because it sends the basis $(e_{s,n})$ onto another basis of $X$. So the formula $\Vert x\Vert_A:=\Vert J_A(x)\Vert$ defines a norm on $X$, and this is a complete norm because $(X,\Vert\,\cdot\,\Vert_A)$ is by definition isometric with $(X,\Vert\,\cdot\,\Vert)$. Moreover, if $A\neq B$, then the norms $\Vert\,\cdot \Vert_A$ and $\Vert\,\cdot \Vert_A$ are not equivalent. Indeed, assume for example that $A$ is not contained in $B$, and choose $s_0\in A\setminus B$. Then $J_A(e_{s_0,n})=ne_{s_0,n}$ and $J_B(e_{s_0,n})=e_{s_0,n}$, so that $\Vert e_{s_0,n}\Vert_A=n\, \Vert e_{s0,n}\Vert_B$ for all $n\in\mathbb N$, which shows that $\Vert\,\cdot \,\Vert_A$ can not be controlled by $\Vert\,\cdot \,\Vert_B$. Therefore, we have found a family of pairwise inequivalent complete norms indexed by the set $\mathcal P(S)$, which has cardinality $2^\kappa$.