# Proof that Heine-Borel entails Completeness

I’m reading Royden’s Real Analysis and trying to solve problem 34 from the first chapter, which is to prove the equivalence of the Heine-Borel theorem and the completeness property (that every non-empty set bounded above has a least upper-bound).

I’ve looked at these questions on this topic:

Prove Heine-Borel Theorem and Completeness Axiom are equivalent

Prob. 1, Sec. 27, in Munkres' TOPOLOGY, 2nd ed: How to show that the compactness of every closed interval implies the least upper bound property?

The first link has an answer that I’m concerned is not correct. The argument I summarize as this:

Let $$\emptyset \ne X\subset \mathbb{R}$$ bounded above by $$M$$. If $$X$$ has a maximum then this is the LUB, so assume $$X$$ has no maximum. Then we open-cover $$\overline{X}$$ with the intervals $$(-\infty, x)$$ for every $$x\in \overline{X}$$.

Here is where I think the arguments starts to go wrong: I don’t think this is an open-cover for the set $$\overline X$$. If $$X=(0,1)$$ then $$\overline{X}=[0,1]$$ and none of the open intervals contains 1.

Now the second answer at that link is, I think, very similar to the strategy pursued by the author in the second link. Namely it uses the nested set theorem, which said briefly is the idea that if you have a bounded “quickly” shrinking chain of closed intervals, the infinite intersection contains a unique element. However, that theorem was proved on the assumption of the completeness of the real numbers. So for this argument I worry about circularity.

However, now I’m left not seeing a way forward. My first instincts about how to do this were very similar to the first idea of making an open cover. I tried making $$M$$ the upper limit of the closed interval, but then a finite open cover doesn’t clearly entail anything that I see. The maximum open cover, so to speak, no longer is useful. I thought about intersecting with the set of maxima, but then you don’t get a closed set ... or at least not one that’s useful.

I should note that the proof that Completeness entails Heine-Borel is obviously just the usual proof that Heine-Borel is true. So I have no question about that part.

The argument does work. The only caveat is that you also need that $$X$$ is bounded below (otherwise, the closure won't be compact); but that's not a problem because you can just cut $$X$$ with some big enough interval.
In the linked answer, the argument is not as you say: you state it for $$X$$, but it should be $$\overline X$$. Note that $$\sup X=\sup\overline X,$$which is not hard to prove .
Now, if $$\overline X$$ has a maximum, then it has supremum. So now you assume that $$\overline X$$ has no maximum. In this latter case, indeed the sets $$(-\infty,x)$$ provide an open cover of $$\overline X$$; because if they don't, it means that there exists $$m\in\overline X$$ such that $$m\geq x$$ for all $$x$$, which is what a maximum is. Having obtained an open cover that does not admit a finite subcover, you contradict Heine Borel because $$\overline X$$, being closed and bounded, should be compact.
In your example $$[0,1]$$ lies in the first situation (it has a maximum), so the second situation does not apply. You will not find examples of the second situation; that's precisely what the argument shows: that there are no closed and bounded sets with no maximum.
• I'm confused, I did state "for every $x\in\overline{X}$" and not for $X$. Are you referring to something else? – Addem Apr 24 '20 at 4:32
• Ok never mind, I see it. It's that either $\overline{X}$ does or doesn't have a max, not $X$. I guess I didn't consider the possibility of talking about $\overline{X}$ not having a max because intuitively we know that it does if it's bounded above. But ok, I'll think more about the argument. – Addem Apr 24 '20 at 4:35