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1) Let $S =\{(−1)^n\; \mid\; n \in \mathbb{Z}\}$ . What is the greatest lower bound of $S$?

-1 is the Lower bound. But is it also the greatest lower bound? Or does it not exist?

Thanks.

And also

2) Let $S = \{p^2\; \mid\; p \in \mathbb{N}\; \text{is prime}\}$, this is a countable set, right?

ah11950
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John
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  • What happens if you take a number $\alpha > -1$ and assume it's a lower bound? Can you find an element of $S$ that is strictly smaller than $\alpha$? $S$ is a subset of the natural numbers, and therefore countable. – ah11950 Mar 25 '14 at 21:53
  • Yes, I can find an element which is -1, ( -1)^1 , which belongs to Z. Does it mean that -1 is the greatest lower bound, just because -1<alpha? – John Mar 25 '14 at 21:57
  • Indeed, you've shown that any number strictly greater than -1 isn't a lower bound, and you've (presumably) shown that -1 is a lower bound, thus it is the greatest lower bound. – ah11950 Mar 25 '14 at 21:58

1 Answers1

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As for 1$st$ question - it might be helpful to write explicitly content of $S$ as for the second notice that $S \subseteq \mathbb{N},$ where $\mathbb{N}$ is a set of all natural numbers.

borg
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  • S only consists of -1 and 1s. I'm only confused by the wording the greatest lower bound. Why should it be great if it's the smallest. So -1 is the GLB – John Mar 25 '14 at 21:59
  • Try to picture process of establishing of GLB in two steps: first take all lower bounds and then take the greatest of them (in case the set of lower bound is finite). You may also benefit from having a look at this. – borg Mar 25 '14 at 22:02