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.

  • $\begingroup$ I'm confused, I did state "for every $x\in\overline{X}$" and not for $X$. Are you referring to something else? $\endgroup$ – Addem Apr 24 '20 at 4:32
  • $\begingroup$ 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. $\endgroup$ – Addem Apr 24 '20 at 4:35
  • 1
    $\begingroup$ Exactly. It looks to me that your confusion stemmed from trying to find an example of the impossible case. $\endgroup$ – Martin Argerami Apr 24 '20 at 4:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.