Separated topological space Let $\{x_1,x_2,\cdots,x_n\}$ $n$ different points $(n\in \mathbb{N}^*)$ from a separated topological space $(E,\tau).$
How to prove by induction that there exists $n$ neighborhoods $\{U_1,U_2,\cdots,U_n\}$ such that $$\forall i\neq j, U_i\cap U_j=\emptyset: \forall i\in \{1,\cdots,n\}, U_i\in \mathcal{V}_{x_i}$$
Please help me
Thank you
 A: Case $n=2$. $U_1, U_2$ exist by definition of separated space.
Case $n+1$. Consider $U_1', \dots, U_n'$ pairwise disjoint neighbourhoods of $x_1, \dots, x_n$ rispectively. For all $i \in \{ 1, \dots ,n\}$ there exist $V_i \in \mathcal{V}(x_i)$ and $W_i \in \mathcal{V}(x_{n+1})$ disjoint, since $E$ is a separated space.
Call $U_{n+1} = W_1 \cap \dots \cap W_n$ and $U_i = U_i' \cap V_i$ for $i \in \{ 1, \dots ,n\}$.
Then $U_1, \dots, U_n, U_{n+1}$ are pairwise disjoint neighbourhoods of $x_1, \dots, x_n, x_{n+1}$ rispectively.
A: By "separated" you probably mean "Hausdorff". So you want to show that in a Hausdorff space $n$ distinct points have pairwise disjoint open neighborhoods.
I will show you the case $n=3$, the general induction is similar.
Take three points $x,y,z$. Since $X$ is Hausdorff, we may separate $x$ form $y$, $x$ from $z$ and $y$ from $z$. Choose corresponding open neighborhoods. We get two open neighboorhoods of each point, so intersect them for each point. After this, we get disjoint open neighborhoods.
