Closed continuous surjective map and Hausdorff space

Click For Summary
SUMMARY

The discussion centers on the proof that if p: X → Y is a closed, continuous, and surjective map with compact preimages for every y in Y, then Y is Hausdorff provided X is Hausdorff. The proof utilizes the properties of compactness and the Hausdorff condition to construct disjoint open sets around points in Y. Specifically, the existence of neighborhoods W1 and W2 around points y1 and y2 in Y is established, demonstrating that these neighborhoods are disjoint, thereby confirming that Y inherits the Hausdorff property from X.

PREREQUISITES
  • Understanding of closed maps in topology
  • Familiarity with Hausdorff spaces
  • Knowledge of compactness in topological spaces
  • Basic concepts of continuous functions in topology
NEXT STEPS
  • Study the properties of compact spaces in topology
  • Explore the implications of continuous functions on topological properties
  • Investigate the relationship between closed maps and compactness
  • Learn about the various types of topological spaces and their properties
USEFUL FOR

Mathematicians, particularly those studying topology, students tackling advanced topology problems, and educators looking for examples of properties of continuous functions and compactness in Hausdorff spaces.

radou
Homework Helper
Messages
3,149
Reaction score
8

Homework Statement



Here's a nice one. I hope it's correct.

Let p : X --> Y be a closed, continuous and surjective map such that p^-1({y}) is compact for every y in Y. If X is Hausdorff, so is Y.

The Attempt at a Solution



Let y1 and y2 in Y. p^-1({y1}) are then p^-1({y2}) disjoint and compact subsets of X. Since X is Hausdorff, for p^-1({y1}) and for any x in p^-1({y2}) there exist disjoint open sets U and V containing p^-1({y1}) and x, respectively. Now find such pair of open sets for p^-1({y1}) and for any x in p^-1({y2}). These sets form open covers for p^-1({y1}) and for p^-1({y2}) respectively, so they have finite subcovers. Take the intersection of all sets from the finite subcover for p^-1({y1}), let's call it U1. Take the union of all sets from the finite subcover for p^-1({y2}), call it U2. U1 and U2 are disjoint.

Now, since U1 and U2 are open sets containing p^-1({y1}) and p^-1({y2}) respectively, there exist neighborhoods W1 of y1 and W2 of y2 such that p^-1(W1) is contained in U1 and p^-1(W2) is contained in U2.

I claim that W1 and W2 are disjoint.

Suppose they were not - let y be an element in their intersection. Then p^-1({y}) is contained both in U1 and U2, contradicting the fact that U1 and U2 are disjoint.
 
Physics news on Phys.org
Your proof is perfect!
 
Excellent! This tradition mustn't go on, since I'll start to think I'm good :D

Btw, the "hint" is extremely useful. I don't see another way we could "generate" an open set with certain required properties in the codomain.
 

Similar threads

Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
12
Views
3K
  • · Replies 17 ·
Replies
17
Views
5K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
4
Views
2K