Finite Dimensional Hausdorff Topological Space

In summary, The theorem states that a Hausdorff topological vector space E is finite dimensional if and only if it admits a precompact neighborhood of zero. This can be proven by showing that a compact neighborhood of zero in E implies that E is a finite dimensional subspace, which is closed and leads to a Hausdorff quotient space. This in turn implies that E is one point, making it a finite dimensional space.
  • #1
creative_
3
0
How do I prove that a Hausdorff topological space E is finite dimensional iff it admits a precompact neighborhood of zero?
 
Physics news on Phys.org
  • #2
i think you mean banach space, not hausdorff space, and then this is a well known theorem of riesz: th.5.9.4., p.109, of dieudonne's foundations of modern analysis.
 
  • #3
Hi mathwonk,

I mean Hausdorff space.
 
  • #4
It is indeed true that a locally compact Hausdorff topological vector space E is finite dimensional.

Proof: Let K be a compact neighborhood of 0. We can assume that K is balanced. Since (1/2)K is a neighborhood of 0, there are finitely many points x1,...,xn such that

[tex]K\subseteq (x_1+\frac{1}{2}K)\cup...\cup (x_n+\frac{1}{2}K)[/tex]

Let M be the finite dimensional subspace spanned by the x1,...,xn. Then M is closed. The quotient space E/M is Hausdorff. Since [itex]K\subseteq M+\frac{1}{2}K[/itex], then [itex]\varphi(K)\subseteq \frac{1}{2}\varphi(K)[/itex]. So (by induction) [itex]\varphi(2^nK)\subseteq \varphi(K)[/itex].

K is balanced, so [itex]E=\bigcup_n 2^n K[/itex]. Thus [itex]\varphi(E)=\varphi(K)[/itex]. Thus E/M is compact. Which implies that E/M is one point. Thus E=M.
 
  • #5
what do you mean by zero in a topological space? oh i see micromass, you assumed he meant a vector space. the argument you gave is a variation of the argument for Riesz's theorem that I referred to. That's a nice example of extending an argument to a more general setting. I didn't know that version.
 
Last edited:
  • #6
Thank you micromass. Your proof makes sense.
 

1. What is a finite dimensional Hausdorff topological space?

A finite dimensional Hausdorff topological space is a mathematical concept that describes a set of points with a particular structure. It is a topological space, which means that it has a defined set of open sets and satisfies certain axioms. The space is also Hausdorff, meaning that any two distinct points have disjoint open neighborhoods.

2. How is a finite dimensional Hausdorff topological space different from other topological spaces?

A finite dimensional Hausdorff topological space is different from other topological spaces because it is both finite dimensional and Hausdorff. Finite dimensional means that the space has a finite number of dimensions, while Hausdorff means that it satisfies a certain separation axiom. Together, these properties make a finite dimensional Hausdorff topological space more specific and restrictive than other topological spaces.

3. What are some examples of finite dimensional Hausdorff topological spaces?

Some examples of finite dimensional Hausdorff topological spaces include finite sets with the discrete topology, Euclidean spaces, and compact subsets of Euclidean spaces. Other examples include the sphere, the torus, and other manifolds with a finite number of dimensions.

4. What are the applications of finite dimensional Hausdorff topological spaces?

Finite dimensional Hausdorff topological spaces have many applications in mathematics, physics, and engineering. They are used in topology and geometry to study properties of spaces with a finite number of dimensions. In physics, they are used to model physical systems with a finite number of degrees of freedom. In engineering, they are used to analyze and design structures with a finite number of dimensions.

5. What are some important properties of finite dimensional Hausdorff topological spaces?

Some important properties of finite dimensional Hausdorff topological spaces include compactness, connectedness, and the existence of a base of open sets. Compactness means that every open cover has a finite subcover, while connectedness means that there are no disjoint open sets that cover the space. The existence of a base of open sets allows for a more convenient and concise way to describe the topology of the space.

Similar threads

  • Topology and Analysis
2
Replies
43
Views
764
Replies
15
Views
1K
Replies
2
Views
245
Replies
2
Views
279
  • Topology and Analysis
Replies
2
Views
1K
Replies
8
Views
2K
  • Topology and Analysis
Replies
5
Views
2K
  • Topology and Analysis
Replies
8
Views
1K
Replies
6
Views
1K
  • Topology and Analysis
Replies
9
Views
2K
Back
Top