Quotient spce is Hausdorff

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Feb 26th 2009, 11:13 PM
Amanda1990

Quotient spce is Hausdorff

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.
Feb 27th 2009, 07:45 AM
tah

Quote:

Originally Posted by Amanda1990

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.

Let $\pi$ be the canonical projection, image by $\pi$ of an open saturated set of $X$ is open in the quotient topology. So the converse implication is obvious. For the direct implication, two distinct equivalence classes $[x_1],[x_2]$ are contained in two disjoint open sets $O_1,O_2$ , by definition $\pi^{-1}(O_i)$ 's are open sets in $X$ , which are indeed disjoint, saturated and $[x_i] \subseteq O_i$
Feb 27th 2009, 09:47 AM
tah

Quote:

Originally Posted by tah

Let $\pi$ be the canonical projection, image by $\pi$ of an open saturated set of $X$ is open in the quotient topology. So the converse implication is obvious. For the direct implication, two distinct equivalence classes $[x_1],[x_2]$ are contained in two disjoint open sets $O_1,O_2$ , by definition $\pi^{-1}(O_i)$ 's are open sets in $X$ , which are indeed disjoint, saturated and $[x_i] \subseteq O_i$

Sorry the last expression $[x_i] \subseteq O_i$ should be changed to $x_i \in \pi^{-1}(O_i)$ .
Feb 27th 2009, 05:01 PM
aliceinwonderland

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 $[x]=\{y \in X : x \sim y\}$ determined by ~. Let q be a quotient map of ~ such that $q:X \rightarrow X/\sim$ . Now, the space $X/\sim$ induced by a quotient map q is a quotient space of X.

Suppose a quotient topological space $X/\sim$ is Hausdorff.

Let $[x], [y]$ be distinct points in $X/\sim$ and U, V be disjoint open sets in $X/\sim$ containing $[x]$ and $[y]$ , respectively. By definition of a saturated set and Hausdorff property of $X/\sim$ , $q^{-1}(U)$ and $q^{-1}(V)$ are disjoint open saturated sets containing $q^{-1}([x])$ and $q^{-1}([y])$ in X, respectively. The $q^{-1}([x])$ and $q^{-1}([y])$ 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 $q^{-1}([x])$ and $q^{-1}([y])$ in X determined by ~ are contained in two disjoint open saturated sets V and W in X.

Now, for any pair of distinct points $x \in q^{-1}([x])$ and $y \in q^{-1}([y])$ , we have two distinct points q(x) and q(y) in $X/\sim$ 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, $X/\sim$ is Hausdorff.