why $U_{y_1}\cap ....\cap U_{y_n}$? why not $U_{y_1}\cup ....\cup U_{y_n}?$

Question

I have some confusion about the statement in Munkras Book

Theorem $26.3$: Every compact subspace of a hausdorff space is closed

In the theorem of the proof it is written that the open set $V_{y_1} \cup....\cup V_{y_n} $ contain $Y$ and it is disjoint from the open set $U_{y_1}\cap ....\cap U_{y_n}$ formed by taking the intersection of the corresponding neighbhorhoods of $x_0$

My confusion : Here the Hausdorff condition ensures both $U_y$ and $V_y$ are disjoint neighbhorhoods of the point $x_0$ and $y$

So i think $U_{y_1}\cup ....\cup U_{y_n}$ formed by taking the union of the corresponding neighbhorhoods of $x_0$

Im not getting why $U_{y_1}\cap ....\cap U_{y_n}$? why not $U_{y_1}\cup ....\cup U_{y_n}?$

Answer 1

Because $\bigcap_{j=1}^nU_{y_j}$ and $\bigcup_{j=1}^nV_{y_j}$ are disjoint (since $U_{y_j}$ and $V_{y_j}$ are disjoint for each $j\in\{1,2,\ldots,n\}$). However, there is no reason why $\bigcup_{j=1}^nU_{y_j}$ and $\bigcup_{j=1}^nV_{y_j}$ would be disjoint.