Prove that the bth projection map is continuous and open.

[ForMyThunder]

I am trying to prove that the bth projection map P_b:\PiX_a --> X_b is both continuous and open. I have already done the problem but I would like to check it.

1) Continuity:

Consider an open set U_b in X_b, then P_b^-1(U_b) is an element of the base for the Tychonoff topology on \PiX_a. Thus, P_b is continuous.

2) Openness:

Let U be an open set in \PiX_a and let p be a point in U. Then there exists a neighbourhood V of p contained in U. Thus, P_b(p) \in P_b(V) and P_b(V) \subset P_b(U) and every point P_b(p) of P_b(U) has a neighbourhood P_b(V). Thus, P_b(U) is open and P_b 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.

 

[ForMyThunder]

I just realized that my proof on openness is entirely wrong. Could anyone help me out on it?

 

[rasmhop]

2) Openness:

Let U be an open set in \PiX_a and let p be a point in U. Then there exists a neighbourhood V of p contained in U. Thus, P_b(p) \in P_b(V) and P_b(V) \subset P_b(U) and every point P_b(p) of P_b(U) has a neighbourhood P_b(V). Thus, P_b(U) is open and P_b is an open map.

You can't really know that P_b(V) 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:
V = \prod U_a
where the collection \{U_a\}[/tex] consists of open sets (U_a open in X_a). You should be able to show that P_b(V) is open in X_b.<br /> <br /> The reason that we only need to show it for basis elements is that once we know it&#039;s true for them we can take arbitrary unions of them to show that all open sets also map to open sets.

 

[ForMyThunder]

Yeah, I think I've figured it out now:

An arbitrary basis element is given by the finite intersection \bigcapP_ai^-1(V_ai), where {a_i} is a finite subset of the index set and each V_ai is an open set in X_ai. So then, P_b(\bigcapP_ai^-1(V_ai) = \bigcapP_b(P_ai^-1(V_ai)). When b\neqa_i for all i, then this equals X_b. When b=a_i for some i, this equals V_ai. In either case, the image is an open set, thus P_b is an open map.

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