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.

© John Scholes

jscholes@kalva.demon.co.uk

25 Aug 2001