# On Compact Sets

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
In the Principles of Mathematical analysis by Rudin we have the following theorem

If $$\displaystyle \mathbb{K}_{\alpha}$$ is a collection of compact subsets of a metric space $$\displaystyle X$$ such that the intersection of every finite sub collection of $$\displaystyle \mathbb{K}_{\alpha}$$ is nonempty , then $$\displaystyle \cap\, \mathbb{K}_{\alpha}$$ is nonempty .

If I understand correctly then this theorem states that if any finite intersection is nonempty then any arbitrarily intersection is also nonempty , right ?. I was trying to understand the proof but it wasn't so clear for me .

#### caffeinemachine

##### Well-known member
MHB Math Scholar
In the Principles of Mathematical analysis by Rudin we have the following theorem

If $$\displaystyle \mathbb{K}_{\alpha}$$ is a collection of compact subsets of a metric space $$\displaystyle X$$ such that the intersection of every finite sub collection of $$\displaystyle \mathbb{K}_{\alpha}$$ is nonempty , then $$\displaystyle \cap\, \mathbb{K}_{\alpha}$$ is nonempty .

If I understand correctly then this theorem states that if any finite intersection is nonempty then any arbitrarily intersection is also nonempty , right ?. I was trying to understand the proof but it wasn't so clear for me .
Hello Zaid.

Can you point out where in the book is this theorem given?

#### caffeinemachine

##### Well-known member
MHB Math Scholar
I worked out a proof which I think is easy to understand.

We prove a more general result which is:

Let $\mathcal K=\{K_\alpha\}_{\alpha\in J}$ be a family of compact subsets of a Hausdorff space $C$ having the finite intersection property, that is, intersection of any finite subfamily of $\mathcal K$ is non-empty, then $\bigcap_{\alpha\in J} K_\alpha\neq \emptyset$.

Fix $\alpha_0\in J$ and define $C_\beta=K_{\alpha_0}\cap K_\beta$ for all $\beta\in J$. Its easy to show that each $C_\beta$ is a closed subset of $K_{\alpha_0}$ by noting that each $K_\alpha$ is a closed subset of $X$ since compact subsets of Hausdorff spaces are closed. Now clearly $\{C_\beta\}_{\beta\in J}$ has finite intersection property as subspaces of $K_{\alpha_0}$. Since $K_{\alpha_0}$ is compact and each $C_\beta$ is closed in $K_{\alpha_0}$ we know that $\bigcap_{\beta\in J}C_\beta\neq\emptyset$. This leads to the required result.

#### Fernando Revilla

##### Well-known member
MHB Math Helper
If $$\displaystyle \mathbb{K}_{\alpha}$$ is a collection of compact subsets of a metric space $$\displaystyle X$$ such that the intersection of every finite sub collection of $$\displaystyle \mathbb{K}_{\alpha}$$ is nonempty , then $$\displaystyle \cap\, \mathbb{K}_{\alpha}$$ is nonempty.
I don't know which are your doubts. Here is the Rudin's proof (I've added some details):

Consider an element $K_1$ of $\{K_{\alpha}\}$. Suppose $\bigcap K_{\alpha}=\emptyset$ and denote $G_{\alpha}=K_{\alpha}^c$. Then, there in no point in $K_1$ belonging to all $K_{\alpha}$, so $\{G_{\alpha}\}$ is an open cover of $K_1$ (on a metric space every compact set is closed). As $K_1$ is compact there is a finite subcover $\{G_{\alpha_1},\ldots,G_{\alpha_n}\}$ of $\{G_{\alpha}\}$ such that $K_1\subset G_{\alpha_1}\cup\ldots\cup G_{\alpha_n}$. But this implies:
$$K_{1}\cap K_{\alpha_1}\cap\ldots\cap K_{\alpha_n}=\emptyset$$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
I don't know which are your doubts. Here is the Rudin's proof (I've added some details):

Consider an element $K_1$ of $\{K_{\alpha}\}$. Suppose $\bigcap K_{\alpha}=\emptyset$ and denote $G_{\alpha}=K_{\alpha}^c$. Then, there in no point in $K_1$ belonging to all $K_{\alpha}$, so $\{G_{\alpha}\}$ is an open cover of $K_1$ (on a metric space every compact set is closed). As $K_1$ is compact there is a finite subcover $\{G_{\alpha_1},\ldots,G_{\alpha_n}\}$ of $\{G_{\alpha}\}$ such that $K_1\subset G_{\alpha_1}\cup\ldots\cup G_{\alpha_n}$. But this implies:
$$K_{1}\cap K_{\alpha_1}\cap\ldots\cap K_{\alpha_n}=\emptyset$$
$G_{\alpha}=K_{\alpha}^c$ , is this the complement of each compact set ? so since compact sets are closed the complement works as an open cover ?
$G_{\alpha}=K_{\alpha}^c$ , is this the complement of each compact set ? so since compact sets are closed the complement works as an open cover ?