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$\mathbb{R}^2 \setminus \mathbb{Q} \times \mathbb{Q}$

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Riemann
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$\mathbb{R}^2 \setminus \mathbb{Q} \times \mathbb{Q}$

#1

Post by Riemann » Fri Jul 07, 2017 6:35 am

Is the set $\mathcal{S} = \mathbb{R}^2 \setminus \mathbb{Q} \times \mathbb{Q}$ complete? Give a brief explanation.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Ram_1729
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Re: $\mathbb{R}^2 \setminus \mathbb{Q} \times \mathbb{Q}$

#2

Post by Ram_1729 » Wed Dec 05, 2018 10:41 pm

No take a sequence $(2+\frac{\sqrt{2}}{n},\sqrt{2})\to (2,\sqrt{2})$
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Re: $\mathbb{R}^2 \setminus \mathbb{Q} \times \mathbb{Q}$

#3

Post by Riemann » Thu Dec 06, 2018 7:43 pm

Thank you Ram_1729. Exactly! It was an exam's question!
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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