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This question is related to Exercise 2.15 in Baby Rudin, in which the reader is instructed to "[s]how that Theorem 2.36 and its Corollary become false (in $R^1$, for example) if the word 'compact' is replaced by 'closed' or by 'bounded.'"

The relevant theorem and corollary are as follows:

Theorem 2.36: If $\{K_{α}\}$ is a collection of compact subsets of a metric space $X$ such that the intersection of every finite subcollection of $\{K_{α}\}$ is nonempty, then $\{K_{α}\}$ is nonempty.

Corollary: If $\{K_{n}\}$ is a sequence of nonempty compact sets such that $K_{n}$ contains $K_{n+1}, (n = 1, 2, 3, ...)$, then $K_n$ is not empty.

My Question: I realize this question and similar ones have been asked elsewhere on MSE, but I've yet to see a straightforward (to me, at least) answer to my question specifically. Usually, for the "closed" part of the exercise, the following collection of closed sets, $\{K_n \}_{n \in \mathbb{N}}$ is used:

$$K_n = [n, \infty), n \in \mathbb{N}.$$

For, the intersection of every finite subcollection of ${K_n}$ is clearly nonempty. Indeed,

$$ \bigcap_{i=1}^{m} K_{n_{i}} = [N, \infty) $$

where $N = \max\{n_1, n_2, ..., n_m\}$.

Then, the assertion is made that

$$ \bigcap_{n=1}^{\infty} K_n = \emptyset. $$

However, I don't see how the same argument can't simply be used, i.e., that, for example, $N \in \bigcap_{n=1}^{\infty} K_n$ (or maybe, say, $n+1$). Intuitively, this certainly makes sense, I'd say. For, given the archimedean property, it seems that as $n$ approaches infinity, we can always find an $l$ in each increasingly smaller set such that $n < l < \infty$; and so $l \in \bigcap_{n=1}^{\infty} K_n$.

What is it exactly that I'm misunderstanding or overlooking here?

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    $\begingroup$ are you able to find a single element that belongs to all $K_n$? Give it a try and you'll get it. $\endgroup$
    – trying
    Commented Aug 17, 2017 at 16:38
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    $\begingroup$ I think your problem is that you are confusing $\exists l| \forall n $ and $\forall n,\exists l$. $\endgroup$
    – Arnaud D.
    Commented Aug 17, 2017 at 16:43

5 Answers 5

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Let $x$ be a real number and assume that $x\in \bigcap_{n \in \mathbb{N}} K_{n}$, that is $x\in K_{n}$ for any $n\geq 1$. Now let $N$ be an integer such that $N>x$ and $N\geq 1$ (it exists by the Archimedean Principle), then $x\in K_{N}=[N,+\infty)$ which means the $x\geq N$. Contradiction! Therefore $\bigcap_{n \in \mathbb{N}} K_{n}=\emptyset$.

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I suspect your confusion stems from the definition of intersection. Consider the following argument very carefully.

Let $x$ be a real number. Then, by the Archimedean property, there exists an integer $n$ with $n>x.$ Hence, $x\not\in[n, \infty). $ Since $x$ is not in one of the $K_{i}$ (namely, $x\not\in K_{n}$), $x$ is certainly not in all of the $K_{i},$ and therefore is not in the given intersection. But this holds for all real $x,$ so the intersection is empty.

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If $\bigcap_{n\in\mathbb N}K_n\neq \emptyset$ (let say $\ell\in \bigcap_{n\in\mathbb N}K_n$), then $n\leq \ell$ for all $n\in\mathbb N$, and in particular, $\mathbb N$ would be upper-bounded.

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Yes, $\bigcap_{n=1}^N K_n$ is nonempty for each $N$. Indeed, $N\in \bigcap_{n=1}^N K_n$. However $\bigcap_{n=1}^\infty K_n$ is empty because if we say that $x\in\bigcap_{n=1}^\infty K_n$, then we would have $x\in K_n$ for every $n$ and so $x\geq n$ for every $n\in\mathbb{N}$. This is a contradiction to the Archimedian property.

The same argument proving Theorem 2.36 will not work because $[0,\infty)$ is not compact. To see this, note that the sets $([0,n))_n$ form an open cover of $[0,\infty)$ with no finite subcover. It will be instructive to look at the proof of Theorem 2.36 and locate where the compactness hypothesis is used. Then identify why your argument will breakdown without compactness.

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You are not using the definition of intersection; the argument you presented is a plausible reasoning, which can be rejected by logic often times!

Note that $\bigcap_{n}K_{n} = \{ x \mid x \in K_{n}\ \forall n \}$ by definition. If there is some $x$ such that $x$ lies in the intersection, then $x \in K_{n}$ for all $n$. Note that by Archimedian principle (this is a correct application!) there is some integer $N$ such that $N > x$, so $x \notin K_{N}$, a contradiction.

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