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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.

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  • $\begingroup$ @Asaf: While I agree with your example, I was thinking more geometrically (more blobs in spaces). The original counterexample attempt can be thought of as 3 sausages with their ends overlapping any other two sets to form a triangle formation. This pattern can be upgraded to a tetrahedral formation for $n=4$, where the sets intersect only at the vertices of the tetrahedron. I can continue increasing $n$ to infinity (raising the dimension of the space appropriately). When $n = \infty$, every finite subcollection for $n < \infty$ retains a non-empty intersection. Perhaps connectivity matters? $\endgroup$
    – Neej
    Commented Jul 7, 2012 at 14:06
  • $\begingroup$ I am guessing that I misread your question, which is reasonable since I am not a mindreader and it is hard to understand whether or not you are looking for a specific example, or has a particular thing in mind. By your comment it seems that you have a very specific idea that you wish to ask about, but for some reason you did not specify that. As I said mind reading was not in my curriculum, if you have a specific idea that you wish to ask about please state it as clearly as you can. I will be glad to try and give it a go. $\endgroup$
    – Asaf Karagila
    Commented Jul 7, 2012 at 14:35
  • $\begingroup$ @Asaf: Thank you for helping. Apologies if I am not clear. I will try harder. I am asking whether a specific construction is a counterexample to Theorem 2.36 in Rudin. The construction is geometric intersection of the sets as in my first comment. Generalised, for $n = m$ every set has distinct intersections with $m-1$ other sets. Except I should write, "When $n=\infty$, every finite subcollection with $k < n$ retains a non-empty intersection." Yet by construction the intersection of all the sets is empty. I ask for help whether this is a valid counterexample or flawed in some way. $\endgroup$
    – Neej
    Commented Jul 7, 2012 at 15:24
  • $\begingroup$ The construction of the sets seems a bit unclear to me. What is the space itself, that too is very unclear. Remember that $\mathbb R^n$ is not compact for itself for all $n$. You could of course argue the construction in the closed unit ball, but not that in an infinitely dimensional space this unit ball need not be compact. So you would have to first give in details what is the compact metric space and then give in details what are the sets you wish to intersect. In both cases, please put all the relevant information into the question rather than the comments. $\endgroup$
    – Asaf Karagila
    Commented Jul 7, 2012 at 15:30
  • $\begingroup$ The expression $\displaystyle\bigcap_{n=1}^\infty K_n$ should not be thought of as a limit as $n\to\infty$. Limits are defined with $\varepsilon$-$\delta$ definitions, etc. Here, the definition is that $\displaystyle x\in\bigcap_{n=1}^\infty K_n$ iff $\forall n\in\{1,2,3,\ldots\}\ x\in K_n$. $\endgroup$ Commented Jul 7, 2012 at 17:18

1 Answer 1

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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, $K_1\cap K_2\cap K_3=\varnothing$.

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  • $\begingroup$ This first answer by Asaf was a response to my original poorly written question. In respect of the key construct I was trying to express (I hope more accessible in the latest edit of the question) this answer isn't a counterexample to my counterexample. It requires an ordering on the taking of intersections. If I reverse the indexing (counting from $n=\infty$ back down to $n=1$) in this answer then we have a construct similar to my counterexample. So the last statement $K_1 \cap K_2 \cap K_3 = \emptyset $ containing a finite number of sets is dependent on choice of order of intersections. $\endgroup$
    – Neej
    Commented Jul 10, 2012 at 9:54

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