Prove that the bth projection map is continuous and open.

In summary: Basically, the first proof is saying that if an open set U is in the base for the Tychonoff topology on a set X, then the map P_b is continuous. However, if we consider an arbitrary basis element, we can see that P_b(V) is open in X_b. This means that the map is open.
  • #1
ForMyThunder
149
0
I am trying to prove that the bth projection map Pb:[tex]\Pi[/tex]Xa --> Xb is both continuous and open. I have already done the problem but I would like to check it.

1) Continuity:

Consider an open set Ub in Xb, then Pb-1(Ub) is an element of the base for the Tychonoff topology on [tex]\Pi[/tex]Xa. Thus, Pb is continuous.

2) Openness:

Let U be an open set in [tex]\Pi[/tex]Xa and let p be a point in U. Then there exists a neighbourhood V of p contained in U. Thus, Pb(p) [tex]\in[/tex] Pb(V) and Pb(V) [tex]\subset[/tex] Pb(U) and every point Pb(p) of Pb(U) has a neighbourhood Pb(V). Thus, Pb(U) is open and Pb is an open map.

I'm less sure about whether my proof of openness is correct. Could you tell me if my answer is correct or not and where I need to improve in my proofs? This question came from General Topology by Stephen Willard (Section 8). Thank you in advance.
 
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  • #2
I just realized that my proof on openness is entirely wrong. Could anyone help me out on it?
 
  • #3
ForMyThunder said:
2) Openness:

Let U be an open set in [tex]\Pi[/tex]Xa and let p be a point in U. Then there exists a neighbourhood V of p contained in U. Thus, Pb(p) [tex]\in[/tex] Pb(V) and Pb(V) [tex]\subset[/tex] Pb(U) and every point Pb(p) of Pb(U) has a neighbourhood Pb(V). Thus, Pb(U) is open and Pb is an open map.

You can't really know that [itex]P_b(V)[/itex] is open so this doesn't work. To show that a map is open it suffices to show that it maps basis elements to open sets so try considering an arbitrary basis element:
[tex]V = \prod U_a[/tex]
where the collection [itex]\{U_a\}[/tex] consists of open sets ([itex]U_a[/itex] open in [itex]X_a[/itex]). You should be able to show that [itex]P_b(V)[/itex] is open in [itex]X_b[/itex].

The reason that we only need to show it for basis elements is that once we know it's true for them we can take arbitrary unions of them to show that all open sets also map to open sets.
 
  • #4
Yeah, I think I've figured it out now:

An arbitrary basis element is given by the finite intersection [tex]\bigcap[/tex]Pai-1(Vai), where {ai} is a finite subset of the index set and each Vai is an open set in Xai. So then, Pb([tex]\bigcap[/tex]Pai-1(Vai) = [tex]\bigcap[/tex]Pb(Pai-1(Vai)). When b[tex]\neq[/tex]ai for all i, then this equals Xb. When b=ai for some i, this equals Vai. In either case, the image is an open set, thus Pb is an open map.

I wrote this out right after I saw that my first proof was wrong.
 

Related to Prove that the bth projection map is continuous and open.

1. What is the bth projection map?

The bth projection map is a function that takes in a point in a product space and returns the bth coordinate of that point. For example, in a 3-dimensional space, the bth projection map would return the third coordinate of a point.

2. How is continuity defined in a projection map?

A projection map is continuous if small changes in the input result in small changes in the output. In other words, if a point in the original space is close to another point, the corresponding points in the projected space should also be close together.

3. Why is it important to prove that the bth projection map is continuous and open?

Proving that the bth projection map is continuous and open is important because it ensures that the map preserves the topological structure of the original space. This is useful in various mathematical and scientific fields, such as topology and differential geometry.

4. What is the difference between continuity and openness in a projection map?

Continuity refers to the smoothness of the map, while openness refers to the behavior of the map on open sets. A projection map can be continuous without being open, and vice versa.

5. How can one prove that the bth projection map is continuous and open?

There are several methods to prove that a projection map is continuous and open, including using the definition of continuity and openness, using the properties of product spaces, and using theorems such as the Tube Lemma. It is also helpful to have a good understanding of general topology and basic analysis concepts.

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