MHB Compactness Theorem: Intersection of Compact Sets

alyafey22
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In the Principles of Mathematical analysis by Rudin we have the following theorem

If $$\mathbb{K}_{\alpha}$$ is a collection of compact subsets of a metric space $$X$$ such that the intersection of every finite sub collection of $$\mathbb{K}_{\alpha}$$ is nonempty , then $$\cap\, \mathbb{K}_{\alpha} $$ 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:.
 
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ZaidAlyafey said:
In the Principles of Mathematical analysis by Rudin we have the following theorem

If $$\mathbb{K}_{\alpha}$$ is a collection of compact subsets of a metric space $$X$$ such that the intersection of every finite sub collection of $$\mathbb{K}_{\alpha}$$ is nonempty , then $$\cap\, \mathbb{K}_{\alpha} $$ 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?
 
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.
 
ZaidAlyafey said:
If $$\mathbb{K}_{\alpha}$$ is a collection of compact subsets of a metric space $$X$$ such that the intersection of every finite sub collection of $$\mathbb{K}_{\alpha}$$ is nonempty , then $$\cap\, \mathbb{K}_{\alpha} $$ 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).
 
Fernando Revilla said:
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 ?
 
ZaidAlyafey said:
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.
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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