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.