Help with Understanding Locally Compact Spaces & Subspaces

  • Thread starter Thread starter winmath
  • Start date Start date
  • Tags Tags
    Compact Subspaces
Click For Summary
A compact space is automatically locally compact because every point can have the entire space as a compact neighborhood. However, the subspace of rational numbers, Q, is not locally compact since it lacks compact neighborhoods for its points. To demonstrate this, one can consider any point in Q and show that any neighborhood includes irrational points, which prevents it from being contained in a compact subspace. The discussion emphasizes that locally compact spaces require every point to have a compact neighborhood, whereas compactness applies to the entire space. Understanding these distinctions is crucial for grasping the properties of locally compact and compact spaces.
winmath
Messages
1
Reaction score
0
hi.. how can we say a compact space automatically a locally compact? how subspace Q of rational numbers is not locally compact? am not able to understand these.. can anyone help me?
 
Physics news on Phys.org
Locally compact is in a certain sense a weaker assertion than compactness. Here is a hint to show that Q is not locally compact. Take a point x and define some neighbourhood. Then show that this neighbourhood is not contained in a compact subspace. To do so consider an irrational point in the neighbourhood. Also as a further hint use limit point compactness. This is one way to prove it. Let me know if you are still stuck.
 
abiyo said:
Locally compact is in a certain sense a weaker assertion than compactness.
Why do you say "in a certain sense"? It's just weaker; locally compact means: every point has a compact neighbourhood. If the whole space is compact, for any point you just take the whole space as compact neighborhood.
winmath said:
how subspace Q of rational numbers is not locally compact?
Just try to find a compact neighborhood of q\in Q.
 
Landau; Thanks for the correction. Locally compactness is just weaker. I don't know what I was thinking.
 

Thread 'About the definition of topological manifold using closed sets'

We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.
View full post »

Similar threads

Undergrad On two definitions of locally compact
  • · Replies 5 ·
Replies
5
Views
3K
Undergrad Tough lemma on locally finite refinement
  • · Replies 8 ·
Replies
8
Views
2K
Undergrad Is [0,1] under the subspace topology Hausdorff, Compact, or Connected?
  • · Replies 15 ·
Replies
15
Views
3K
Undergrad Noncompact locally compact Hausdorff continuous mapping
  • · Replies 14 ·
Replies
14
Views
3K
Undergrad Help Solve Munkres Question: Limit Point Compact Subspace in Hausdorff Space
  • · Replies 8 ·
Replies
8
Views
3K
Undergrad Unit sphere is compact in 1-norm
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad Compact support functions and law of a random variable
  • · Replies 11 ·
Replies
11
Views
3K
Undergrad Is the Set of Integer Outputs of sin(x) Sequentially Compact in ℝ?
  • · Replies 3 ·
Replies
3
Views
3K
Undergrad Is the Projection Restriction to a Linear Subspace a Homeomorphism?
  • · Replies 8 ·
Replies
8
Views
3K
Undergrad A problem of completeness of a metric space
  • · Replies 3 ·
Replies
3
Views
2K