Proving a Set in the Order Topology is Closed

In summary, the homework statement is trying to prove that a set is closed, but I don't know how to get started.
  • #1
Poopsilon
294
1
Proving a Set is Closed (Topology)

Homework Statement



Let [itex]Y[/itex] be an ordered set in the order topology with [itex]f,g:X\rightarrow Y[/itex] continuous. Show that the set [itex]A = \{x:f(x)\leq g(x)\}[/itex] is closed in [itex]X[/itex].


Homework Equations





The Attempt at a Solution



I cannot for the life of me figure this out. As far as I can tell one either needs to show the set A gathers its limit points or that it is the pre-image (under f or g) of some closed set in Y. We know the order topology is Hausdorff, so that's something. I just don't even know how to get started on this one. Hopefully someone can help, thanks.
 
Last edited:
Physics news on Phys.org
  • #2
This one is tricky.

Instead of trying to prove directly that A is closed, try to show that the complement, X-A, is open. Show the set of all x such that f(x) > g(x) to be open. You can show this by showing that every point x in X is an interior point. Just find an open set around an arbitrary x.

So let x be an arbitrary element of X such that f(x) > g(x). If there is no such x, then the set is empty, then and it is open trivially. So assume such an x exists and consider f(x) and g(x) in the order topology.

Now draw the number line to represent Y and put f(x) and g(x) in their respective places.

Then find two open sets in Y.

Does that help get you started?
 
  • #3
So going from there we use the fact that the order topology is Hausdorff and find two open disjoint sets one containing f(x) and the other containing g(x), from here I would like to map back under f^-1 and g^-1 to two open sets in X and from here take the intersection of them which would contain x and would be open, but I can't see how we know anything about these two disjoint open sets in Y...
 

1. What is the definition of a closed set in the order topology?

In the order topology, a set is considered closed if its complement is open. This means that all points outside of the set are contained in open intervals that are also contained in the complement of the set.

2. How do you prove that a set is closed in the order topology?

To prove that a set is closed in the order topology, you must show that its complement is open. This can be done by showing that for any point outside of the set, there exists an open interval containing that point that is also contained in the complement of the set.

3. Can a set be both open and closed in the order topology?

Yes, in certain cases a set can be both open and closed in the order topology. This occurs when the set is the entire space, or when the set is empty.

4. Is a closed set in the order topology necessarily closed in other topologies?

No, a closed set in the order topology is not necessarily closed in other topologies. This is because the definition of a closed set can vary depending on the topology being used.

5. What are some common examples of closed sets in the order topology?

Some common examples of closed sets in the order topology include closed intervals, half-closed intervals, and the entire real line.

Similar threads

  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
20
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
4K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
439
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Topology and Analysis
Replies
12
Views
4K
Back
Top