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I just learnt the statement "Let $S$ be a sequentially compact subset of $\mathbb{R}$. Then $S$ is compact." with its relevant proof.

My question is: Is it true to say "Let $S$ be a subset of $\mathbb{R}$ wich is not sequentially compact. Then $S$ is not compact."? If so, please prove/disprove it.

Thank you.

  • In general, the inverse of a statement is not true; however, I do not know specifically about the question you are asking--the inverse could be true in this case, but I do not know (perhaps more knowledgeable users will be able to answer this one). – Daniel W. Farlow Jan 31 '15 at 04:45
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    Sequential compactness and compactness are equivalent in metric spaces. – AMPerrine Jan 31 '15 at 04:46
  • Possible duplicate of this. Anyways answer is already there. But as AMPerrine stated the following are equivalent in metric space, compactness, sequential compactness and Bolzano-Weierstrass property. – creative Jan 31 '15 at 04:48
  • No no; my question is not to prove/disprove that "If $S$ is compact, then $S$ is a sequentially compact subset of $\mathbb{R}$.". I ask, does EVERY not-sequentially-compact results in not-compact? - And, my question is about sequences of real numbers in Real Analysis. –  Jan 31 '15 at 04:52
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    Have you ever heard of the contrapositive? – Matt Samuel Jan 31 '15 at 04:56
  • @Matt Samuel: Contrapositive means (?) that to answer my question I need to prove compactness $\implies $ sequential compactness, in real numbers in Real Analysis. If so, please let me know a reference to see its proof. –  Jan 31 '15 at 05:00
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    I would be surprised if it's not in your textbook. In any case, any proof for metric spaces will prove it for the real numbers. – Matt Samuel Jan 31 '15 at 05:04
  • Let $(x_n)$ be a sequence in $S$ with no convergent subsequence. Note $A={x_n\mid n\in\Bbb N}$ must be infinite. For each $s\in S$, there is an open ball $B_s$ centered at $s$ such that $B_s\cap A$ is finite. The collection ${B_s\cap S\mid s\in S}$ is an open cover of $S$ with no finite subcover (if one existed, the intersection of one of its elements with $A$ would be infinite). – David Mitra Jan 31 '15 at 09:06
  • @Matt Samuel: You are right! Later on, in the same textbook: Theorem (with proofs) - For a subset $S$ of $\mathbb{R}$, the following three assertions are equivalent to each other: i) $S$ is closed and bounded. ii) $S$ is sequentially compact. iii) $S$ is compact. –  Jan 31 '15 at 12:06

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