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I was reading a book on differential geometry. In the preliminary section norm on a vector space was defined. Later the norm is used to define Banach spaces. Also, it was mentioned that if there exists a norm on a vector space $E$, we can define a topology on $E$. I was attempting to define a metric in such a way that it generates the discrete topology on $E$. So I need a metric $d$ on $E$ such that $$d(x,y) = \begin{cases} 0 ,\ x=y\\ 1,\ x\ne y \end{cases}$$ I tried to define a norm as $$\|x \| = \begin{cases} 0 ,\ x=0\\ 1,\ x\ne0 \end{cases}$$. But before moving further it is not a norm at all!

So I have this question in my mind. Can we define a norm on any vector space (like we can define two topologies on any set)? If not does there exist a vector space on which we cannot define a norm?

Ted Shifrin
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Madhu
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    If $V$ is a finite-dimensional vector space over $k$, where $k=\mathbb R$ or $k=\mathbb C$, then $V\simeq k^n$ for some $n\in\mathbb N$, and you can lift the standard norm on $k^n$ to a norm on $V$. If $V$ is infinite-dimensional, then it still admits a norm, but the only way I am aware of proving this uses the axiom of choice. – Joe Sep 29 '24 at 12:29
  • If the dimension of $V$ is an infinite cardinal $\kappa$, then $V$ is isomorphic to the direct sum of $\kappa$ copies of $k$, and you can give the direct sum a norm by declaring the norm of $(v_i)_{i\in\kappa}$ to be the supremum of the $|v_i|$ (this supremum exists because only finitely many $v_i$ are nonzero). – Joe Sep 29 '24 at 12:36
  • What about a vector space over a finite field? – J. W. Tanner Sep 29 '24 at 15:02

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