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I found that the rings in real analysis are closed for finite union operations,but not closed for countable infinite union operations (only $\sigma$-rings are closed).This reminds me for group and ring structures in algebra,is them closed for infinite additions or multiplications ?

HeroZhang001
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th110256
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    'infinite addition' and 'infinite multiplication' are not defined on general group and ring structures. There can be structures that these operations are well defined, maybe something like topological groups. – Riemann Feb 17 '23 at 10:25

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It is not necessarily the case. Consider the field $\mathbb{Q}$ (which constitutes a ring, and also a group under addition). We have the result that

$$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$

The left-hand side is an infinite sum of rationals, but the right-hand side is clearly irrational, and hence we do not have closure under infinite addition.

Edit: I suppose it might not be as obvious as it appears at first glance that $\pi^2$ is irrational (though one sees a proof here). Another argument comes in from, if one prefers, Leibniz' formula: $$ \pi = 4 \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} $$

One can do this for multiplication as well, with the equivalent result

$$\prod_{\text{$p$ prime}} \frac{1}{1-p^{-2}} = \frac{\pi^2}{6}$$

Here, again, we have an infinite product of rationals on the left-hand side, but an irrational on the right hand side.

Edit: If likewise unconvinced on the matter of $\pi^2$, then one may use the Wallis product: $$ \pi = 2 \prod_{n=1}^\infty \frac{4n^2}{4n^2-1} $$

PrincessEev
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  • Irrationallity of $\pi^2$ is not really clear, since irrational number can have rational square, for instance $\sqrt{2}$. It requires proof. One may look for https://proofwiki.org/wiki/Pi_Squared_is_Irrational Also, there are rational series expansion for $\sqrt{2}$, using binomial series. https://math.stackexchange.com/questions/694699/infinite-series-for-sqrt-2 – Riemann Feb 17 '23 at 10:42
  • Similarly $e=\sum_{n\geq 0} 1/n!$. – Ben Feb 17 '23 at 10:47
  • A much simpler example is $\sum_1^{\infty}1$. This shows that the rationals, the reals, the complex numbers are not closed under "infinite addition". – Gerry Myerson Feb 17 '23 at 12:08
  • This also assumes that "infinite addition" has to mean the typical definition by convergence of partial sums. – Randall Feb 17 '23 at 14:42