0
$\begingroup$

In the notion of a topological vector space, we define such as a vector space $X$ (over a field $\mathbb{K}$) with topology $\mathscr{T}$ such that

$$\iota_+: (X,\mathscr{T}) \times (X, \mathscr{T}) \to (X,\mathscr{T})$$ $$\iota_+: (x,y) \mapsto x+y$$ and $$\iota_\cdot: (\mathbb{K},\mathscr{U}) \times (X, \mathscr{T}) \to (X,\mathscr{T})$$ $$\iota_\cdot: (\alpha,x) \mapsto \alpha x$$ are continuous.
Now, my question may seem rather dim, but I have not seen an explicit reason for Why we require these mappings to be continuous. What does this ensure, and why is it of importance. Naturally this is true in Banach and Hilbert spaces, since a map between norm spaces is equivalent to it being bounded by some $C$ so we may just take $C= \alpha$ and $C= n\max\{x_i\}_{i=1}^{n}$.

$\endgroup$
  • 2
    $\begingroup$ In general, when we attach the word "topological" to some algebrabic structure (Vec Space, Group, Rings, etc.), we demand the underlying algebraic operations are continuous. The more rigorous answer is the that that a topological vector space $(X,\tau)$ belongs to the category of vector spaces and and the category topological spaces. The structure preserving mappings (morphisms) for each category are linear transformations and continuous maps. Moreover, each of the vector space operations can be realized as linear mapping in the appropriate setting. $\endgroup$ – matt biesecker May 11 '15 at 19:18
  • $\begingroup$ Flip through your favorite book on topological vector spaces, and see how many of the theorems would fail without this assumption. (Hint: pretty much all of them.) $\endgroup$ – Nate Eldredge May 11 '15 at 19:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.