I am asking whether a specific construction is a counterexample to Theorem 2.36 in Rudin's "Principles…" book (3rd Ed.), which reads,
2.36 Theorem If $\{K_{\alpha}\}$ is a collection of compact subsets of a metric space $X$ such that the intersection of every finite subcollection of $\{K_{\alpha}\}$ is nonempty, then $\bigcap K_{\alpha}$ is nonempty.
I suggest an extension of an earlier attempted counter-example in this post on the Intersection of compact sets. The earlier attempt fails because the finite number of sets used (just 3 sets) means that all the sets together are also a finite subcollection of the whole. Hence there is a finite subcollection of sets with empty intersection and the counterexample fails. However, it seems to me that the attempt using just 3 sets can be extended to an infinite (countable) number of compact sets.
The original counterexample attempt can be thought of as 3 sausage-shaped connected sets in $\mathbf{R}^2$ each with their ends overlapping any other two sets to form a triangle formation. This pattern can be upgraded to a tetrahedral formation in $\mathbf{R}^3$ for $n=4$, where the sets intersect only at the vertices of the tetrahedron. (Technically, this tetrahedral arrangement can also be mapped to $\mathbf{R}^2$ with some work, but I will confine the discussion to naive embeddings for now.) Inductively, I can continue increasing $n$ to infinity (raising the dimension of the embedding space appropriately). Generalised, for $n=m$ every set $K_i,\:i=1,2,…m$ has distinct intersections with $m−1$ other sets. When $n=\infty$, every finite subcollection $\{K_k: k<n\}$ retains a non-empty intersection.
I.e. if there are $n$ such sets, $K_i,\:i=1,2,…n$, each of which intersects with the $n-1$ others, and then we let $ n = \infty$, then it remains the case that any finite subcollection with $k<n$ has non-empty intersection and yet by construction
$$ \bigcap_{i=1}^{\infty}K_i = \emptyset $$
Rudin's statement of the theorem suggests only that the overall space be metric, though the subsets $K_i$ are compact. So perhaps an infinite dimensional metric space is still OK. It seems to me that inducing from lower dimensional cases, that for any $n$ the naive embedding space is $n−1$ dimensional. But possibly $\mathbf{R}^{\infty−1}=\mathbf{R}^{\infty}$ by convention. So the compact sets would be within $\mathbf{R}^{\infty}$, where $\mathbf{R}^{\infty}$ is the infinite product space, $\mathbf{R} \times \mathbf{R} \times \mathbf{R} \times \ldots $.
By Tychonoff's Theorem any product of compact spaces is compact, even for an infinite product. So if the construction of the $\{K_i\}$ was embedded within compact spaces within the dimensions of the overarching metric space $X$, then even if $X$ is infinite dimensional the sets $K_i$ should also remain compact.
So I am asking if this construction is a good counterexample, or whether it is flawed in some way.
Best Answer
Consider the following case:
$$K_1=[0,1]\cup[4,5]; K_2=[2,3]\cup[4,5]; K_3=[0,3]$$
The intersection of every two is non-empty, but the intersection of the three is empty, so not every finite subcollection has a non-empty intersection, even though every two have non-empty intersection.
You can now add $K_n$ for $n>3$ as $K_n=[0,\frac1n]\cup[2,2+\frac1n]$, while $\bigcap_{n>3} K_n$ is indeed non-empty, one can also note that for every $n$, $K_n\cap K_i\neq\varnothing$ for $i<n$. Still, $K_1\cap K_2\cap K_3=\varnothing$.