# From $X=\bigcup_{j=1}^n O_{y_j}$ to $\bigcap_{j=1}^n O_{y_j} \supset X$?

Why $\bigcap_{j=1}^n O_{y_j} \supset X$ in the proof ?

My understanding is $X=\bigcup_{j=1}^n O_{y_j}$ since $\{Oy_j:j\in\{1,...,n\}\}$ is a finite subcover of $X$.

Every Hausdorff space is normal.

$\textbf{Main proof:}$ (Made by D.ex-Machina user)

Let $T$ be a compact Hausdorff topological space and let $X$ and $Y$ be disjoint closed sets in $T$. By the lemma, for any $y\in Y$ there exists a neighborhood $U_y\ni y$ and an open set $O_y$ containing $X$ such that $U_y\cap O_y =\varnothing$.

Since $Y$ is a closed subset of a compact set it is itself compact, and therefore the cover $\left\lbrace U_y\right\rbrace _{y\in Y}$ of $Y$ has a finite subcover $U_{y_1},U_{y_2},\cdots , U_{y_n}$. The open sets

\begin{align*} O^1=\bigcap_{j=1}^n O_{y_j} \supset X, \qquad O^2=\bigcup_{j=1}^n U_{y_j} \supset Y, \end{align*} are then disjoint open sets containing $X$ and $Y$, proving that every compact Hausdorff space is normal.

For each $y\in Y$, the open set $O_{y}$ is such that $X\subseteq O_{y}$. In particular, $X\subseteq O_{y_{j}}$ for all $j=1,...,n$, so $X\subseteq\displaystyle\bigcap_{j=1}^{n}O_{y_{j}}$.

Note that $y_{j}$ is just a specific $y$ taken in $Y$, so $O_{y_{j}}$ is a specific $O_{y}$.

• got it, thanks. – Isa Feb 12 '18 at 22:39
• I was reading the proof again, and I wonder is $O_y$ a neighborhood of $y$ ? – Isa Feb 13 '18 at 21:30
• No, by construction it is not, although it has the subscript of $y$, it is an open neighbourhood of $X$. – user284331 Feb 13 '18 at 21:32
• And why the subscript $y$ was used for $O_y$? – Isa Feb 13 '18 at 21:37
• Because for different $y$, the corresponding open neighbourhood of $X$ may be different, so it is better to have subscript $y$ instead of just $O$. – user284331 Feb 13 '18 at 21:40