Prove that a quotient topological space is Hausdorff if and only if any two distinct equivalence classes are contained in two disjoint open saturated sets.

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- February 27th 2009, 12:13 AMAmanda1990Quotient spce is HausdorffProve that a quotient topological space is Hausdorff if and only if any two distinct equivalence classes are contained in two disjoint open saturated sets.
- February 27th 2009, 08:45 AMtah
Let be the canonical projection, image by of an open saturated set of is open in the quotient topology. So the converse implication is obvious. For the direct implication, two distinct equivalence classes are contained in two disjoint open sets , by definition 's are open sets in , which are indeed disjoint, saturated and

- February 27th 2009, 10:47 AMtah
- February 27th 2009, 06:01 PMaliceinwonderland
This is a more detailed one that I've tried.

Let X be a space and ~ an equivalence relation on X. Let X/~ denote the set of all equivalence classes determined by ~. Let q be a quotient map of ~ such that . Now, the space induced by a quotient map q is a quotient space of X.

Suppose a quotient topological space is Hausdorff.

Let be distinct points in and U, V be disjoint open sets in containing and , respectively. By definition of a saturated set and Hausdorff property of , and are disjoint open saturated sets containing and in X, respectively. The and are simply equivalence classes in X determined by ~.

(If X/~ is Hausdorff, then X is Hausdorff. But the converse is not necessarily true).

Conversely, suppose any two distinct equivalence classes and in X determined by ~ are contained in two disjoint open saturated sets V and W in X.

Now, for any pair of distinct points and , we have two distinct points q(x) and q(y) in contained in two distinct open sets q(V) and q(W), respectively (The q maps saturated open sets of X to open sets of X/~).

Thus, is Hausdorff.