1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I would start the induction with $n=1$. $\endgroup$ – Martin Brandenburg Nov 1 '14 at 11:56
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.