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On Compact Sets

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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 :confused:.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
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 :confused:.
Hello Zaid.

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

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
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
Jan 29, 2012
661
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:
$$K_{1}\cap K_{\alpha_1}\cap\ldots\cap K_{\alpha_n}=\emptyset$$
(contradiction with the hypothesis).
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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:
$$K_{1}\cap K_{\alpha_1}\cap\ldots\cap K_{\alpha_n}=\emptyset$$
(contradiction with the hypothesis).
I know that the proof is easy but I might find difficulties in the notations , first what does
$G_{\alpha}=K_{\alpha}^c$ , is this the complement of each compact set ? so since compact sets are closed the complement works as an open cover ?
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
I know that the proof is easy but I might find difficulties in the notations , first what does
$G_{\alpha}=K_{\alpha}^c$ , is this the complement of each compact set ? so since compact sets are closed the complement works as an open cover ?
Right.