31st Putnam 1970

S is a closed subset of the real plane. Its projection onto the x-axis is bounded. Show that its projection onto the y-axis is closed.

Solution

Let Y be the projection onto the y-axis and X the projection on the x-axis. Let { y_n} be any Cauchy sequence in Y. Then { y_n } must converge to some real y. We have to show that y is in Y. For each n take any x_n such that (x_n, y_n) is in S. Then { x_n } lies in X, which is bounded, so { x_n } is bounded. But any bounded sequence has a convergent subsequence. So we can take a subsequence { u_n } of { x_n } which is convergent and therefore Cauchy. Let { v_n } be the corresponding subsequence of { y _n }. Then { v_n } is also Cauchy and hence the sequence of points P_n = (u_n, v_n) in S is Cauchy. But S is closed so P converges to some (u, v) in S. Hence { v_n } converges to v, which is in Y (it is the projection of (u, v) ). But since { y_n } converges to y, its subsequence { v_n } must also converge to y. Hence y = v and y is in Y.

31st Putnam 1970

© John Scholes
jscholes@kalva.demon.co.uk
25 Aug 2001