Normed and topological vector spaces

[logarithmic]

Can someone explain how topological vector spaces are more general than normed ones. So this means that if I have a normed vector space, X, it would have to induce a topology. I was thinking the base for this topology would consist of sets of the form \{y\in X: ||x-y||<\epsilon, \textrm{for some x$x\in X$ and $\epsilon>0$}\}, which is the analogue of the open ball centred at y with radius epsilon. Is this correct?

 

[g_edgar]

Yes, given a normed space, your definition is the standard way to define a topology, and the result is a topological vector space.

Now consider the set \mathbb{R}^\mathbb{N} of all sequences of real numbers. With the obvious vector operations and the topology of pointwise convergence this becomes a topological vector space. But it can be shown that there is no norm for this space that produces this topology.

 

[HallsofIvy]

On the other hand, consider any vector space, V, with the topology in which the only open sets are V itself and the empty set. This is a topological vector space which cannot be derived from any metric, much less a norm.