Product of Hausdorff spaces

1. Nov 30, 2007

mathboy

I know that a product of Hausdorff spaces is Hausdorff. Is the converse also true? That is, if A_1 x A_2 x A_3 x... is Hausdorff, then is each A_i Hausdorff?

2. Dec 1, 2007

olliemath

Yes. Given any subset A of a Hausdorff space X we can give it the subspace topology. Then for x,y in A we can find disjoint open subsets U,V of X containing x and y. By definition of the subspace topology $$U \cap A$$ and $$V \cap A$$ are disjoint and open in A and contain x and y respectively as required (to be completely rigourous you need to show that A_i with the subspace topology inherited from A_1xA_2x... is the same as A_i itself, but this is not hard).

Equivalently you could try proving that for an arbitrary Y and some non-Hausdorff space X, XxY is never Hausdorff.. I don't think this would be too difficault.

3. Dec 1, 2007

andytoh

Notation: PX_i = cartesian product of X_i, i in index set I.

Suppose PX_i is Hausdorff. For some j in I, let a_j and b_j be two distinct points in X_j. Let a and b be two points in PX_i, where the jth component of a and b are a_j and b_j respectively, and all other components of a and b are equal. Then a not=b and so, since PX_i is Hausdorff, there exist disjoint open sets containing a and b, and therefore there exist disjoint basis elements B_a and B_b (of PX_i) containing a and b respectively. Now basis elements in PX_i are of the form B_a = PU_i, B_b = PV_i, where the U_i and V_i are open in X_i (and U_i,V_i = X_i for all but finitely many values of i if the product topology is being used, but that’s not important in this proof). Since a and b only differ in their jth component, so that U_i and V_i cannot be disjoint for any i not= j, then B_a and B_b can only be disjoint if U_j and V_j are disjoint. Thus U_j and V_j are disjoint open sets in X_j containing a_j and b_j respectively, so that X_j is Hausdorff. Since j was arbitrary, then each X_i is Hausdorff for all i in I.

Another way to prove it is to follow olliemath's idea and view X_j as a subspace of PX_i by using PY_i, where Y_j = X_j and Y_i = any fixed point in X_i. I did that proof as well, and both proofs are of the same length (actually the second proof was a bit longer because I had to show that the subspace topology PY_i inherits as a subspace of PX_i is the same as the product/box topology of PY_i with each Y_i viewed as a subspace of X_i).

Last edited: Dec 1, 2007
4. Dec 1, 2007

Singularity

This is in fact a very important and well known result. Indeed this result holds for an arbitrary product (i.e. not just for a countable number of Hausdorff spaces).

5. Dec 2, 2007

mathwonk

actually its false, since the empty set is hausdorff.

6. Dec 2, 2007

Singularity

mathwonk, you raise an interesting point. Question:

If say, P is the empty set with the discrete topology and H is another hausdorff space,
will PxH be homeomorphic to H?

Intuitively I expect this to hold. Let's define f : PxH -> H by f(p,h) = h

It is certainly continuous (it is a projection map) and onto. Opennes seems to hold as well. The only problem that could crop up may be to show 1-1. Perhaps a classic vacuous argument could be employed here.

Supposing that the hypothesis holds, could you please explain why the result posed in the first post fails to hold if the empty set is included.

7. Dec 2, 2007

andytoh

PxH is empty if P is empty because if f belonged to PxH, then f(1) would belong to P and f(2) would belong to H. But f(1) cannot belong to P because P is empty. So there is no such f.

The converse raised by the opening post is true if none of the component spaces is empty, that's the condition that I forgot to stipulate in my proof above.

Last edited: Dec 2, 2007
8. Dec 2, 2007

Singularity

Can a function "belong" to a space? I have never heard of this. Also, your argument only has one component, when it seems like it should have two. Am I missing something fundamental here?

Thanks :)

Last edited: Dec 2, 2007
9. Dec 2, 2007

andytoh

A cartesian product of spaces A_1 x A_2 x ... x A_n can be viewed as the collection of all functions f:{1,...,n} -> U (A_i) such that f(k) belongs to A_k for all k=1,...,n. It's a trivial exercise to show that there is a bijective correspondence between all the "n-tuples" and all such f. And this holds for arbitrary cartesian products as well.

Consequently, A_1 x A_2 x ... x A_n is empty if any of the A_i is empty, since any function would be undefined at i.

Last edited: Dec 2, 2007
10. Dec 3, 2007

Singularity

I still don't fully get it, but will get back to the books!
Anyway, thanks for the assistance.

Ciao