# Thread: Quotient spce is Hausdorff

1. ## 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.

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

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

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

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

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

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