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This is a question on homework for an introductory point-set topology course.

Let $X$ be Hausdorff and let $x_1,\dots,x_n\in X$ be distinct elements. Prove that there exist open sets $U_1, \dots , U_n$ such that $x_i \in U_i$ for all $i$ and $U_i\cap U_j = \emptyset$ for all $i\neq j$.

First Attempt: Since $X$ is Hausdorff, then for any distinct elements $a,b\in X$ there exist disjoint open neighborhoods $U_a$ and $U_b$ containing $a$ and $b$, respectively. Consequently, for each $x_i$ we can choose open neighborhoods $U_i^j$ of $x_i$ where $j\in \{1,2,\dots,n\}\setminus \{i\}$ such that $x_j \notin U_i^j$ for all $j$. Then $\mathfrak N_i := \cap_j U_i^j$ is a finite intersection of open neighborhoods of $x_i$, so is open and contains $x_i$ because $x_i \in U_i^j$ for all $j$. Furthermore, by construction the set of such intersections $(\mathfrak N_i)_{1\leq i \leq n}$ will be pairwise disjoint, as desired.

Realization: I believe I have ensured that the open neighborhoods don't contain any of the other $x_i$, but I have not necessarily shown that they are all pairwise disjoint.

Question: How can my argument be modified to ensure that $\mathfrak N_i \cap \mathfrak N_j = \emptyset$ for all $i\neq j$?

R. H. Vellstra
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    You already have disjointness. Your $U_i^j$ are picked so that $U_i^j$ and $U_j^i$ are disjoint, presumably. So pick $i\neq j$. Since $\mathfrak{N}_i\subseteq U_i^j$ and $\mathfrak{N}_j\subseteq U_j^i$, you have $\mathfrak{N}_i\cap\mathfrak{N}_j\subseteq U_i^j\cap U_j^i = \varnothing$. – Arturo Magidin Oct 05 '22 at 17:49
  • Yes, I suppose the presumably part is what my question is about. I was unsure what precise verbiage to add, but I think I got it now. – R. H. Vellstra Oct 05 '22 at 17:53
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    "For each pair $i\neq j$, let $U_i^j$ and $U_j^i$ be disjoint open neighborhoods of $x_i$ and $x_j$, respectively. Now let $\mathfrak{N}_k = \cap_j U_i^j$..." – Arturo Magidin Oct 05 '22 at 17:55

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Your approach has the right idea, but it is inappropriate to do it as you write:

for each $x_i$ we can choose open neighborhoods $U_i^j$ of $x_i$ where $j\in \{1,2,\dots,n\}\setminus \{i\}$ such that $x_j \notin U_i^j$ for all $j$.

You do not use the full strength of "Hausdorff" to find your $U_i^j$. As a simple example consider a Hausdorff space $X$ with more than two points and $n = 2$. You can choose $U_1^2 = X \setminus \{x_2\}$ and $U_2^1 = X \setminus \{x_1\}$ and get $\mathfrak N_1 = U_1^2$ and $\mathfrak N_2 = U_2^1$. These sets have nonempty intersection.

Therefore we must modify the approach. For each pair $(i,j)$ with $i \ne j$ choose disjoint open neigborhoods $U_{ij}$ of $x_i$ and $V_{ij}$ of $x_j$.

Now define $$U_i = \bigcap_{j \ne i} U_{ij} , $$ $$V_j =\bigcap_{i \ne j} V_{ij}, $$ $$W_i = U_i \cap V_i .$$ $W_i$ is an open neigborhood of $x_i$ and for $i \ne j$ we have $$W_i \cap W_j = U_i \cap V_i \cap U_j \cap V_j \subset U_i \cap V_j \subset U_{ij} \cap V_{ij} = \emptyset .$$

Kritiker der Elche
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