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Normally what I know is that you can make a union or an intersection between 2 sets. In this expression Its a big union of a set. I'm asking about the meaning of such expression, What does it mean. What is the infinity sign means being at the top?

Things got even more complicated with seeing De Morgan's Law: enter image description here

Which uses the same expression of big U?

Anyone who can explain to me the expression or De Morgan's Laws would be much appreciated. Thanks

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The equation where you've enclosed the first part in red is the definition of the $\bigcup_{n=1}^{\infty} S_n$ notation.

It works just like sum nontation does: $$ \sum_{n=a}^{b} f(n) \quad\text{means}\quad f(a)+f(a+1)+\cdots+f(b-1)+f(b) $$ and $$ \bigcup_{n=a}^{b} f(n) \quad\text{means}\quad f(a)\cup f(a+1)\cup\cdots\cup f(b-1)\cup f(b) $$ $$ \bigcap_{n=a}^{b} f(n) \quad\text{means}\quad f(a)\cap f(a+1)\cap\cdots\cap f(b-1)\cap f(b) $$

When the upper limit is $\infty$ it means a union of infinitely many sets: $$ \bigcup_{n=a}^{\infty} f(n) \quad\text{means}\quad f(a)\cup f(a+1)\cup\cdots $$ whose precise meaning is defined in the explanation you quote.

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Think of the infinity sign on the top as playing the same role here as it does in a sum $\displaystyle\sum_{n=1}^\infty x_n$. – Harald Hanche-Olsen Sep 3 '13 at 13:32
One important difference with the $\sum_{n=1}^\infty f(n)$ notation is that $\bigcup_{n=1}^\infty S_n$ is always defined (provided the sets $S_n$ are), without any requirement of "convergence" as is the case for infinite sums. – Marc van Leeuwen Sep 3 '13 at 13:44
@MarcvanLeeuwen Sure. But even the sum is always defined (though possibly infinite) when the items being summed are all nonnegative. – Harald Hanche-Olsen Sep 3 '13 at 15:05
Please excuse me since I haven't used math or statistics for a very very long time. So What I understood from your example using summation(sigma) sign. That the upper part is the boundary of the range or the end condition and the lower part (n=1) is the start of the collection. So $\sum_{n=1}^\infty f(n)$ means that the summation of outputs for each parameter passed to f in this case n where n starts with one and ends with $\infty$. An example would be $\sum_{n=1}^3 f(n) = f(1) + f(2) + f(3)$, is that correct? – Omar A. Shaban Sep 7 '13 at 16:27
@OmarA.Shaban: Yes. – Henning Makholm Sep 7 '13 at 16:37

Considering de Morgan's laws, they become basic principles of handling negation in the presence of quantifiers in logic. Let's first state more formally $$ x\in\bigcup_{i\in\Bbb N}S_n \iff \exists n\in\Bbb N: x\in S_n \qquad\text{and}\qquad x\in\bigcap_{i\in\Bbb N}S_n \iff \forall n\in\Bbb N: x\in S_n. $$ Now the law $\left(\bigcup_{i\in\Bbb N}S_n\right)^c=\bigcap_{i\in\Bbb N}{S_n}^c$ becomes, remembering that set equality just means one is member of the left hand side if and only if one is member of the right hand side, $$ \lnot(\exists n\in\Bbb N: x\in S_n)\iff \forall n\in\Bbb N: \lnot(x\in S_n). $$ Similarly $\left(\bigcap_{i\in\Bbb N}S_n\right)^c=\bigcup_{i\in\Bbb N}{S_n}^c$ becomes $$ \lnot(\forall n\in\Bbb N: x\in S_n)\iff \exists n\in\Bbb N: \lnot(x\in S_n). $$ These are nothing more or less than the rules for handling negation of existentially or universally quantified phrases (in the particular case of quantification over$~\Bbb N$, but one could replace it by any set).

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