Editing Topological vector space (section)

===Finite-dimensional spaces===

By [[F. Riesz's theorem]], a Hausdorff topological vector space is finite-dimensional if and only if it is [[locally compact]], which happens if and only if it has a compact [[Neighborhood (topology)|neighborhood]] of the origin.

Let <math>\mathbb{K}</math> denote <math>\R</math> or <math>\Complex</math> and endow <math>\mathbb{K}</math> with its usual Hausdorff normed [[Euclidean topology]].  Let <math>X</math> be a vector space over <math>\mathbb{K}</math> of finite dimension <math>n := \dim X</math> and so that <math>X</math> is vector space isomorphic to <math>\mathbb{K}^n</math> (explicitly, this means that there exists a [[linear isomorphism]] between the vector spaces <math>X</math> and <math>\mathbb{K}^n</math>). This finite-dimensional vector space <math>X</math> always has a unique {{em|[[Hausdorff space|Hausdorff]]}} vector topology, which makes it TVS-isomorphic to <math>\mathbb{K}^n,</math> where <math>\mathbb{K}^n</math> is endowed with the usual Euclidean topology (which is the same as the [[product topology]]). This Hausdorff vector topology is also the (unique) [[Comparison of topologies|finest]] vector topology on <math>X.</math>  <math>X</math> has a unique vector topology if and only if <math>\dim X = 0.</math> If <math>\dim X \neq 0</math> then although <math>X</math> does not have a unique vector topology, it does have a unique {{em|Hausdorff}} vector topology.

* If <math>\dim X = 0</math> then <math>X = \{0\}</math> has exactly one vector topology: the [[trivial topology]], which in this case (and {{em|only}} in this case) is Hausdorff. The trivial topology on a vector space is Hausdorff if and only if the vector space has dimension <math>0.</math>
* If <math>\dim X = 1</math> then <math>X</math> has two vector topologies: the usual [[Euclidean topology]] and the (non-Hausdorff) trivial topology.
** Since the field <math>\mathbb{K}</math> is itself a <math>1</math>-dimensional topological vector space over <math>\mathbb{K}</math> and since it plays an important role in the definition of topological vector spaces, this dichotomy plays an important role in the definition of an [[absorbing set]] and has consequences that reverberate throughout [[functional analysis]].
{{math proof | title=Proof outline| proof =
The proof of this dichotomy (i.e. that a vector topology is either trivial or isomorphic to <math>\mathbb{K}</math>) is straightforward so only an outline with the important observations is given. As usual, <math>\mathbb{K}</math> is assumed have the (normed) Euclidean topology. Let <math>B_r := \{a \in \mathbb{K} : |a| < r\}</math> for all <math>r > 0.</math> Let <math>X</math> be a <math>1</math>-dimensional vector space over <math>\mathbb{K}.</math> If <math>S \subseteq X</math> and <math>B \subseteq \mathbb{K}</math> is a ball centered at <math>0</math> then <math>B \cdot S = X</math> whenever <math>S</math> contains an "unbounded sequence", by which it is meant a sequence of the form <math>\left(a_i x\right)_{i=1}^{\infty}</math> where <math>0 \neq x \in X</math> and <math>\left(a_i\right)_{i=1}^{\infty} \subseteq \mathbb{K}</math> is unbounded in normed space <math>\mathbb{K}</math> (in the usual sense). Any vector topology on <math>X</math> will be translation invariant and invariant under non-zero scalar multiplication, and for every <math>0 \neq x \in X,</math> the map <math>M_x : \mathbb{K} \to X</math> given by <math>M_x(a) := a x</math> is a continuous linear bijection. Because <math>X = \mathbb{K} x</math> for any such <math>x,</math> every subset of <math>X</math> can be written as <math>F x = M_x(F)</math> for some unique subset <math>F \subseteq \mathbb{K}.</math> And if this vector topology on <math>X</math> has a neighborhood <math>W</math> of the origin that is not equal to all of <math>X,</math> then the continuity of scalar multiplication <math>\mathbb{K} \times X \to X</math> at the origin guarantees the existence of an open ball <math>B_r \subseteq \mathbb{K}</math> centered at <math>0</math> and an open neighborhood <math>S</math> of the origin in <math>X</math> such that <math>B_r \cdot S \subseteq W \neq X,</math> which implies that <math>S</math> does {{em|not}} contain any "unbounded sequence". This implies that for every <math>0 \neq x \in X,</math> there exists some positive integer <math>n</math> such that <math>S \subseteq B_n x.</math> From this, it can be deduced that if <math>X</math> does not carry the trivial topology and if <math>0 \neq x \in X,</math> then for any ball <math>B \subseteq \mathbb{K}</math> center at 0 in <math>\mathbb{K},</math> <math>M_x(B) = B x</math> contains an open neighborhood of the origin in <math>X,</math> which then proves that <math>M_x</math> is a linear [[homeomorphism]]. [[Q.E.D.]] <math>\blacksquare</math>
}}
* If <math>\dim X = n \geq 2</math> then <math>X</math> has {{em|infinitely many}} distinct vector topologies:
** Some of these topologies are now described: Every linear functional <math>f</math> on <math>X,</math> which is vector space isomorphic to <math>\mathbb{K}^n,</math> induces a [[seminorm]] <math>|f| : X \to \R</math> defined by <math>|f|(x) = |f(x)|</math> where <math>\ker f = \ker |f|.</math> Every seminorm induces a ([[Metrizable TVS|pseudometrizable]] [[locally convex]]) vector topology on <math>X</math> and seminorms with distinct kernels induce distinct topologies so that in particular, seminorms on <math>X</math> that are induced by linear functionals with distinct kernels will induce distinct vector topologies on <math>X.</math>
** However, while there are infinitely many vector topologies on <math>X</math> when <math>\dim X \geq 2,</math> there are, {{em|up to TVS-isomorphism}}, only <math>1 + \dim X</math> vector topologies on <math>X.</math> For instance, if <math>n := \dim X = 2</math> then the vector topologies on <math>X</math> consist of the trivial topology, the Hausdorff Euclidean topology, and then the infinitely many remaining non-trivial non-Euclidean vector topologies on <math>X</math> are all TVS-isomorphic to one another.