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Heine-Borel Theorem: Every open cover $\mathcal{O}$ of finite interval $[a,b]\subseteq \mathbb{R}$ has finite subcover.

Sketch of Proof: Consider the set $$X=\{x\in[a,b]\colon [a,x] \mbox{ can be covered by finitely many open sets in }\mathcal{O}\}.$$ Then $X$ is a non-empty subset of $\mathbb{R}$ which is bounded above, hence it has supremum. We show that $\sup(X)=b$.

Question: Is there other proof of this theorem which avoids the least upper bound property of $\mathbb{R}$? (In other words, is the Least upper bound property of $\mathbb{R}$ is essential to prove this?)

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    Isn't Heine-Borel Theorem equivalent to the completeness of the reals? I thought so but I cannot find a reference right now. It's not in Real Analysis in Reverse, perhaps because it's not equivalent after all. – lhf Nov 12 '15 at 11:16
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    The result isn't true for the rationals, so it must be necessary. – bilaterus Nov 12 '15 at 11:25
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    @bilaterus: The result "every positive number is a square" can be proven using the least upper bound property, and it is false in $\mathbb{Q}$, but true in the field of real algebraic numbers so then the least upper bound property is not "necessary". – nombre Nov 12 '15 at 11:56
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    Perhaps what you are really asking is this: "Here is a proof which deduces the HB Theorem from the Least Upper Bound property. Is there another proof of HB that uses some other method?" The answer to that question (if indeed you want it, is yes). You can use the fact that monotonic bounded sequences converge, you can use the Bolzano-Weierstrass theorem, you can use Dedekind cuts, etc. – B. S. Thomson Nov 12 '15 at 21:26

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The least upper bound property is "necessary" in the sense that it is implied by this result.

If the segments in a linearly ordered set $F$ are compact with respect to the order topology, then every non-empty bounded subset of $F$ has a least upper bound.

Indeed for $x \in A$ every upper bound of $A$ is in $[x;+\infty[$. Now if $M$ is an upper bound of $A$, then $\overline{A \cap [x;M]}$ is a closed subset of the compact $[x;M]$ so it is compact and since it is non-empty, it has a maximum $\alpha$.*

$\alpha$ is an upper bound of $A$, if $b \in F$ is an upper bound of $A$, then $\overline{A \cap [x;M]} = \overline{A \cap [x;b]}$ so $\alpha \leq b$. Therefore $\alpha$ is the least upper bound of $A$.

*If $(K,<)$ is compact, assuming it has no maximum you can write $K = \bigcup \limits_{x \in K} O_x$ where $O_x = \{y \in K \ | \ y < x\}$. By compacity, there is a finite subcover whose greastest index $x_0$ is such that $K = O_{x_0}$, so $x_0 > x_0$.

nombre
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