Open sets and closed sets in product topology

Click For Summary
SUMMARY

The discussion focuses on the properties of open and closed sets within product topology, specifically addressing the projection maps and separability of product spaces. It establishes that the projection maps \( p_a : X \to X_a \) are open maps due to the product topology being the weakest topology that makes these maps continuous. The discussion also confirms that a product set \( S = \prod_{a \in A} S_a \) is closed if and only if each \( S_a \) is closed, and it concludes that if \( |A| \leq |\mathbb{N}| \) and each \( X_a \) is separable, then the product space \( X \) is also separable.

PREREQUISITES
  • Understanding of product topology and its definition.
  • Familiarity with open and closed sets in topological spaces.
  • Knowledge of projection maps in topology.
  • Concept of separability in topological spaces.
NEXT STEPS
  • Study the definition and properties of product topology in detail.
  • Learn about the implications of projection maps in various topological contexts.
  • Explore the concept of separability and its significance in functional analysis.
  • Investigate counterexamples related to countability in product spaces.
USEFUL FOR

Mathematicians, particularly those specializing in topology, students studying advanced topology concepts, and educators preparing coursework on product spaces and their properties.

complexnumber
Messages
61
Reaction score
0

Homework Statement



Let [tex](X_a, \tau_a), a \in A[/tex] be topological spaces, and let [tex]\displaystyle X = \prod_{a \in A} X_a[/tex].

Homework Equations



1. Prove that the projection maps [tex]p_a : X \to X_a[/tex] are open maps.

2. Let [tex]S_a \subseteq X_a[/tex] and let [tex]\displaystyle S = \prod_{a \in A} S_a \subseteq \prod_{a \in A} X_a[/tex]. Prove that [tex]S[/tex] is closed iff each [tex]S_a \subseteq X_a[/tex] is closed.

3. Let [tex]T_a \subseteq X_a[/tex], prove that [tex]\displaystyle \overline{\prod_{a \in A} T_a} = \prod_{a \in A} \overline{T_a}[/tex].

4. If [tex]\abs{A} \leq \abs{\mathbb{N}}[/tex] and each [tex]X_a[/tex] is separable, prove that [tex]X[/tex] is separable.

The Attempt at a Solution



I don't know how to prove open/closed set problems in product topology. Can someone please give me some hint as to how I should approach these proofs? Some hints on each question will be even better.

1. This means that any open subset of the product space [tex]X[/tex] remains open when projected down to the [tex]X_\alpha[/tex].

Is it because the production topology [tex]\tau[/tex] for [tex]X[/tex] is the weakest topology with regard to [tex]\{ p_a :X \to X_a | a \in A \}[/tex]?

2.

3.

4. [tex]X_a[/tex] is separable means there is a countable subset [tex]S_a \subseteq X_a[/tex] such that [tex]\overline{S_a} = X_a[/tex].
 
Physics news on Phys.org
Write out what an open set in the product topology looks like: http://en.wikipedia.org/wiki/Product_topology#Definition (the second paragraph).

Based on this, where does the projection map send an arbitrary open set?

For (4), use (3): take the product of the Sa and show that it's closure is X.
 
VeeEight said:
Write out what an open set in the product topology looks like: http://en.wikipedia.org/wiki/Product_topology#Definition (the second paragraph).

Based on this, where does the projection map send an arbitrary open set?

For (4), use (3): take the product of the Sa and show that it's closure is X.

Thank you very much for your reply. Here are my answers so far based on your suggestions.

1. The product topology [tex]\tau[/tex] is generated from base [tex]\mathfrak{B}[/tex] consisting of product sets [tex]\displaystyle \prod_{a \in A} U_a[/tex] where only finitely many factors are not [tex]X_a[/tex] and the remaining factors are open sets in [tex]X_a[/tex]. Therefore the project [tex]p_a[/tex] projects an open set [tex]S \subseteq X[/tex] to either [tex]X_a[/tex] or some open subset [tex]S_a \subset X_a[/tex].

2.

3.

4. [tex]X_a[/tex] is separable means there is a countable subset [tex]S_a \subseteq X_a[/tex] such that [tex]\overline{S_a} = X_a[/tex]. Using previous result, we have
[tex] \begin{align*}<br /> \prod_{a \in A} \overline{S_a} = \prod_{a \in A} X_a = \overline{\prod_{a \in A} S_a} = X<br /> \end{align*}[/tex]
Since [tex]S_a[/tex] is countable and [tex]|A| \leq |\mathbb{N}|[/tex], the cartesian product [tex]\displaystyle \prod_{a \in A} S_a[/tex] is countable. Hence [tex]X[/tex] is separable.

***This could be wrong, if [tex]|A| = |\mathbb{N}|[/tex], then the cartesian product does not have to be countable. So what is the set separable? Should the question say If [tex]|A| < |\mathbb{N}|[/tex] and each [tex]|X_a|[/tex] is separable, prove that [tex]|X|[/tex] is separable?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
6K
Replies
2
Views
2K
Replies
4
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K