Prove that the bth projection map is continuous and open.

I am trying to prove that the bth projection map Pb:$$\Pi$$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 $$\Pi$$Xa. Thus, Pb is continuous.

2) Openness:

Let U be an open set in $$\Pi$$Xa and let p be a point in U. Then there exists a neighbourhood V of p contained in U. Thus, Pb(p) $$\in$$ Pb(V) and Pb(V) $$\subset$$ 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.

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

2) Openness:

Let U be an open set in $$\Pi$$Xa and let p be a point in U. Then there exists a neighbourhood V of p contained in U. Thus, Pb(p) $$\in$$ Pb(V) and Pb(V) $$\subset$$ 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 $P_b(V)$ 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 $\{U_a\}[/tex] consists of open sets ([itex]U_a$ open in $X_a$). You should be able to show that $P_b(V)$ is open in $X_b$.

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.

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

An arbitrary basis element is given by the finite intersection $$\bigcap$$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($$\bigcap$$Pai-1(Vai) = $$\bigcap$$Pb(Pai-1(Vai)). When b$$\neq$$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.