This is a homework problem I've been struggling with, so I'd prefer a gentle nudge rather than the absolute solution.

Consider a topological space X that is the disjoint union of two real lines.

ie. X = R + R.

Define the equivalence relation ~ :

(a,0) ~ (b,1) iff a = b, but neither equal to zero.

The quotient space looks like a real line with two separate 0's. Obviously this isn't Hausdorff.

How can I prove this?

My progress is claiming that if itisHausdorff, then there exist disjoint open sets U0 and U1 in the quotient space containing [(0,0)] and [(0,1)] respectively. Then since the pi(x)=[x] function is continuous, there exist open preimages of U0 and U1 in X, and they must also be disjoint.

However, it should be easy to prove that the preimages of U1 and U0 can't be disjoint (a contradiction) and therefore the quotient space isn't Hausdorff.

But I can't really get that argument going.

Any advice?