# What is wrong with this proof?

1. Feb 2, 2008

### JasonJo

I'm trying to prove that a countably infinite cartesian product of connected spaces is connected.

Let X be a connected space and let Y be the countably infinite cartesian product of copies of X, and suppose Y is equipped with the product topology.

So suppose Y is not connected. Let A and B be two disjoint, nonempty, open subsets of Y s.t. A U B = Y. Since A and B are open sets in the product topology, they are both of the form U1 X U2 X .... where each Ui's are not X for only finitely many i and each Ui is open in X. So let k be the sup of the index's for A and B which Ui's is not a copy of X.

So A intersect B = (U1 intersect Y1) X ... X (Uk intersect YK) x X x X... = empty set.
Hence for some j less than k, Uj intersect Wj is empty. Hence Uj and Wj form a seperation of X since A U B is Y, then Ui U Wi = X, and for Uj and Wj, they are disjoint and nonempty.

So where did I go wrong?

2. Feb 2, 2008

### andytoh

Here's how you prove this (and this proof works even if the product is not countable)

-Let a be a fixed point of X (the product space, indexed by I).
-Let K be any finite subset of I. Let Y be the subspace of X consisting of all points whose kth component is the kth component of a (k any element of K). Show that Y is connected.
-Show that the union Z of all the Y's is connected (they share a common point!).
-Show that X equals the closure of Z, and thus X is connected (remember that a point is in the closure of a set iff every neighbourhood of the point intersects the set).

This method works assuming that you already know how to prove that the product of 2 connected spaces is connected (and thus by induction a finite product of connected spaces is connected), or you can simply quote that result if it was already taught in class.

Last edited: Feb 2, 2008