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I see in a book the following example as a motivation to define topological vector space.

$$+: \Bbb R^n \times \Bbb R^n \to \Bbb R^n$$ defined by $(x,y) \to x+y$ and $$.: \Bbb R^n \to \Bbb R^n$$ defined by $x \to \alpha x$ where $\alpha$ is a scalar in the field of the vector space $\Bbb (R^n, +, .)$

and showing that both the maps are continuous.

Then the book defines topological vector space. My question is why such space is important ? (it looks like some generalization, but not getting the reason) To discuss convergence in a vector space we need metric.

Another thing is what is the motivation behind the maps being continuous in the definition of topological vector space ?

Anonymous
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    You don't need a metric to discuss convergence, you need a topology (which is more general since a metric induces a topology). Of course, you sometimes need more assumptions and then add restrictions on the topology. – xavierm02 Aug 04 '14 at 13:21
  • @xavierm02: Thanks. what is the intuition behind addition and scalar multiplication being continuous. – Anonymous Aug 04 '14 at 13:32
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    they are the operations defining the structure of vector space on the given set: it is then natural to ask for continuity of these fundamental operations. If you were dealing with topological groups instead, you would ask for continuity of multiplication and inverse for the same reason. I upvoted the question as it seems ok to me. – Avitus Aug 04 '14 at 13:39
  • @xavierm02, actually, all you really need is a convergence space. – goblin GONE Aug 04 '14 at 14:30

1 Answers1

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Let us summarize the comments.

  1. Topologies define convergence (as pointed out by @xavierm02)

Let $(X, \mathcal T )$ be a topological space and $x_n \in X$ a sequence of points in $X$. We say that the sequence $x_n$ converges to $x \in X$ if for every open set $U \subseteq X$ which contains $x$ there exists an $n_0 \in\mathbb N$ such that for all $n \geq n_0$ the points $x_n$ lie in $U$.

In the case of $\mathbb R^n$ with the usual metric topology, the above definition boils down to the one found on each book of analysis, i.e.

$$\forall \epsilon >0~ \exists ~n_0=n_0(\epsilon)~:~ \forall n\geq n_0\Rightarrow d(x_n,x)<\epsilon;$$

in other words, choosing arbitrarily small open balls around $x$ we can always find elements of the sequence "near" to the limit point $x$. The number of these elements is dependent on the radius of the ball, that is $n_0=n_0(\epsilon)$.

  1. Natural definition of topological vector space

As a vector space is defined by the operations $+$ and $\cdot$, it is natural to topologize it requesting the maps to be continuous. Similar remarks lead to topological groups, rings etc...

Having access to "extra" structure, the process of inducing a topologiy on a vector space can be simplified. In fact, if $(V,\langle\cdot,\cdot\rangle)$ denotes a vector space endowed with scalar product $\langle\cdot,\cdot\rangle$, then

  • $\langle\cdot,\cdot\rangle$ induces a norm $\|\cdot\|$ and a metric $d(\cdot,\cdot)$ via $\|x\|^2:=\langle x,x\rangle$ and $d(x,y)=\|x-y\|$.
  • W.r.t. the metric topology induced by $\langle\cdot,\cdot\rangle$ the space $V$ is a topological vector space ($+$ and $\cdot$ are continuous) and both $\langle\cdot,\cdot\rangle$, $\|\cdot\|$ are continuous.

This is the case of $\mathbb R^n$ in the OP, where the scalar product is the usual one.

  1. And now?

If we move from finite dimensional vector spaces to infinite dimensional ones (like spaces of functions) topologies become even more relevant. Some spaces have topologies induced by norms making them "complete" in a precise sense (people call them Banach spaces); others even admit a scalar product (and are called Hilbert spaces, if complete). Using topology one considers convergence, density / closedness of subspaces, distances, topological linear duals etc.

Avitus
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