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I'm reading the proof of the Heine-Borel theorem on Robert Ash's book. Here, he proved that there is a countable covering and is going to prove that there is a finite subcovering:

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In the following paragraph lies my confusion:

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In the first paragraph, he writes that "by the nested set property, there is an $x\in \cap_{n=1}^{\infty }B_n$". In the second paragraph, he says it follows that $B_m=\emptyset$ for some $m$: Doesn't this contradict the previous assumption? I'm really confused.

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  • $\begingroup$ You are right, the drafting of this argument is really of very poor quality. To which Heine-Borel's theorem are you referring? For me it's the theorem that says that a continuous function on a compact metric set is uniformly continuous. $\endgroup$
    – Thomas
    Commented Feb 15, 2020 at 7:49
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    $\begingroup$ @Thomas Did you read the proof in full before making this comment ? $\endgroup$ Commented Feb 15, 2020 at 7:58
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    $\begingroup$ Here is another way to phrase it: "if there is some $x\in \bigcap_{n=1}^\infty B_n$, we run into a contradiction. It follows that this assumption was wrong, and so $\bigcap_{n=1}^\infty B_n$ is empty." Note that this also means that $B_m$ is empty for some finite $m$; otherwise $\bigcap_{n=1}^\infty B_n$ would be nonempty" $\endgroup$
    – pancini
    Commented Feb 15, 2020 at 8:35

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There is absolutely nothing wrong. He first assumed that $ B_n$ is not empty for each $n$ and arrived at a contradiction. From this he concluded that $B_m$ is empty for some $m$.

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  • $\begingroup$ So he assumed $B_n$ is not empty for any $n$, arrived at a contradiction and then he concludes that it is false that $x$ exists? $\endgroup$
    – Red Banana
    Commented Feb 15, 2020 at 8:11
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    $\begingroup$ Yes, the existence of $x$ is only under the assumption that $B_n$'s are non-empty. Once you arrive at a contradiction from this you no longer have such an $x$. @Billy Rubina $\endgroup$ Commented Feb 15, 2020 at 8:23
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    $\begingroup$ @BillyRubina: The fact that $K\subseteq \cup_{i=1}^{\infty} G_i$ and $H_j=\cup_{i=1}^{j}G_i$ together imply that $K\subseteq \cup_{i=1}^{\infty} H_i$. It's a logical deduction and not a part of assumption and therefore the contradiction can't falsify this. Method of contradiction uses some assumption and derives an absurdity which is then used to claim that the initial assumption is false. $\endgroup$
    – Paramanand Singh
    Commented Feb 15, 2020 at 11:17
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    $\begingroup$ @BillyRubina: yes you got that right. The proof proceeds in multiple stage: first reduce any covering to a countable covering (Lindelof Theorem). Then show that a finite covering also works by using Cantor intersection theorem. You can thus see that Cantor intersection theorem is thus on the same level as Heine Borel. $\endgroup$
    – Paramanand Singh
    Commented Feb 16, 2020 at 16:48
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    $\begingroup$ @BillyRubina : also note that Lindelof Theorem applies to any set but the Heine Borel applies only to closed and bounded sets. Thus any covering of a set can always be reduced to a countable covering, but a further reduction to finite covering is possible only for closed and bounded sets. $\endgroup$
    – Paramanand Singh
    Commented Feb 16, 2020 at 16:56

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