Prove the set $K = \{0\} \cup \{\frac{1}{n} \in \mathbb{R} : n \in \mathbb{N} \}$ is compact without Heine-Borel

Question

Prove the set $K = \{0\} \cup \{\frac{1}{n} \in \mathbb{R} : n \in \mathbb{N} \}$ is compact without Heine-Borel

I have completed the question and used the same procedure that was done in this version: Prove the set $K = \{\frac{1}{n} \mid n\in \mathbb{N}\}\cup \{0\}$ is compact. .

My question though is why this proof is valid? If we are showing the set is compact it means that there must exist a finite sub-cover for all open covers. Is the all condition captured by generalizing $U$ to represent any open cover thus meaning all open covers?

Answer 1

The trick is one of the open sets in any cover must contain the $0$. No matter what the open cover is or which open set contains $0$, there is at least one set in the cover, it contains $0$ and it is open. And because that set is open and contains $0$ it must have an open ball around $0$ entirely contained in the set.

And, here is the trick, no matter how small that open ball is, say it is $\epsilon > 0$ in radius, it contains an infinite number of $\frac 1n < \epsilon$ in that ball and in that one set. In fact, there are only a finite number of $\frac 1k$ that set does not contain.

So we have one open set containing $0$ and an infinite number of $\frac 1n$ and we only need some of the other open sets to contain a finite number of the $\frac 1k$ that aren't already contained in our one open set so far. For each $\frac 1k$ not contained, that $\frac 1k$ must be contained in one of the other open sets. We take just that one. We do it for each of the $\frac 1k$ and take a finite number of open sets.

And that's it, we're done. That's a finite subcover. And it doesn't matter what the actual open cover originally was. We were able to pick a finite subcover from it.

Recap: Pick a set with $0$. There always will be one. That set will no matter what, one way or another contain all but a finite number of $\frac 1n$. For the finite $\frac 1k$ not in the set, pick one open set one at a time till done.

This can always be done.

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"If we are showing the set is compact it means that there must exist a finite sub-cover for all open covers."

On third reading: No.. there is not a subcover that works for all possible covers. BUT every open cover will have a finite subcover of it. For example:

$U = \{ (-1,2)\}: K \subset \cup_{O\in U} = (-1,2)$ has a finite subcover. Itself.

But $V = \{(-.001, .001)\}\cup (\cup_{n\in\mathbb N; n>1}\{(\frac 1{n+1},\frac 1{n-1})\}\cup \{(.99, 1.1)\}$ will have the finite subcover:

$\{(-.001, .001), (\frac 1{1001}, \frac 1{999}), (\frac 1{1000}, \frac 1{998}),....., (\frac 13,1), (\frac .99, 1.1)\}$

And $W = \{(-\infty, {10^{-100}})\}\cup_{k=0...\infty}(10^{-k-1}, 10^{-k+1})$ will have the finite subcover:

$\{(-\infty, \frac {10^{-100}}), (10^{-101},10^{-99}), (10^{-100}, 10^{98}), ....,(\frac 1{100},1), (\frac 1{10}, 10)\}$.

There isn't one finite subcover for all open covers. But for every open cover there is a finite subcover for it.

Answer 2

Note the beginning of the proof:

Let $\mathcal{G} = \{G_\alpha \mid \alpha \in A\}$ be any open cover for $K$

(emphasis mine). Since $\mathcal{G}$ is a completely arbitrary open cover, anything we can prove about $\mathcal{G}$ must in fact be true of every open cover.

This is something we do all the time. E.g. to prove the infinitude of primes:

Let $n$ be any natural number. Let $\{p_1, p_2,..., p_k\}$ be the set of all prime numebrs $<n$. It's easy to check that $p_1\cdot p_2\cdot ...\cdot p_k+1$ is not divisible by any $p_i$ ($i<k$), and hence must be divisible by some prime $>n$. This means for every natural number, there is a larger prime number.

I suspect in the current situation it seems more complicated since the subject matter is more abstract, but it's exactly the same underlying logic.

Answer 3

I see a difference between:

there exists a finite subcover for all open covers

and

there exists a finite subcover for any open cover

The former sounds like all the open covers are collected into one, and that cover has a finite subcover. That is not the definition of compact. In fact, it's not likely to be a useful condition for anything in most contexts.

The latter says that given any open cover there is a finite subcover of that cover. That's the correct interpretation of the definition.