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

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?

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 $$. It's easy to check that $$p_1\cdot p_2\cdot ...\cdot p_k+1$$ is not divisible by any $$p_i$$ ($$i), 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.

• Yes it does "seem" more complicated, but it does force one to have to train themselves to really internalize every single word, since small words like any we tend to naturally just breeze over. Thanks for the assistance. Mar 2 '19 at 20:13

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.

• Thank you for your detailed explanation of the ideas being used in this proof. I'm at that stage in my mathematics journey where the level of abstraction is a lot more challenging. I'm finding you need to REALLY understand the objects that you're working with in order to create these solutions. Mar 3 '19 at 20:21

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.

• I wrote it as the first, but actually in my mind am interpreting it as the 2nd....seems like the lesson of the day for me is language. Mar 2 '19 at 20:17
• Just to be clear. There isn't one finite subcover that is a subcover of all open covers. That would be absurd. It's that every open cover will have a finite subcover taken from it, that will work. Different open covers will have different subcover(s) that will work. But EVERY open cover will have some finite subcover that will work. Mar 2 '19 at 20:24
• @fleablood Good point. I elaborated. Mar 3 '19 at 1:46