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I am trying to read the following two formulas in English but I am not certain how to, although I do understand their meaning.

Union:

$\bigcup_{i=1}^{n} A_{i} = \{x\, |\, \exists_i \in I(x \in A_i) \}$

The meaning of this is that the union of all sets $A_i$ is the set where a member exists in atleast one set in $A_i$.

Intersection:

$\bigcap_{i=1}^{n} A_{i} = \{x\, |\, \forall_i \in I(x \in A_i) \}$

The meaning of this is that the intersection of all sets $A_i$ is the set where a member exists in all sets in $A_i$.

For the union my best guess of how it is read is:

The union of set $A_i$ for $i\in I$ equals the set of members $x$ where there exists a set $i$ in all the sets $I$ where $x$ is a member of set $i$.

However I would like to the correct way of reading both of the formulae. Thanks.

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  • $\begingroup$ If I was reading it to myself or a fellow mathematician I would just say for the first "the union of the $A_i$" and for the second "the intersection of the $A_i$". $\endgroup$
    – almagest
    Commented Jun 7, 2016 at 19:45

2 Answers 2

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Well, first you have it wrong unless you say that $I=\{1,\dots,n\}$. Then the statement simplifies:

The union of $A_i$ for $i\in I$ is the set of all such $x$ that there exists $i\in I$ with $x\in A_i$.

The intersection of $A_i$ for $i\in I$ is the set of all such $x$ that for all $i\in I$ we have $x\in A_i$.

You're almost right, just all the sets is a very unfortunate (and actually wrong) thing to say here, $I$ is a single set (of indices).

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  • $\begingroup$ Thanks, that makes much more sense. And you are correct there was a mistake the formulae should have been $\bigcup_{i \in I}^{n} A_{i} = \{x\, |\, \exists_i \in I(x \in A_i) \}$ and $\bigcap_{i \in I}^{n} A_{i} = \{x\, |\, \forall_i \in I(x \in A_i) \}$. $\endgroup$
    – codefi
    Commented Jun 8, 2016 at 17:36
  • $\begingroup$ Just without the superscript $n$, please. $\endgroup$
    – yo'
    Commented Jun 8, 2016 at 17:50
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A better version is

$$\bigcup_{i=1}^n A_i=\{x|\text{ for some }j\in \{1,\ldots n\},x\in A_j\}$$

$$\bigcap_{i=1}^n A_i=\{x|\text{ for all }j\in \{1,\ldots n\},x\in A_j\}$$

and from there the "plain english" version follow naturally. The union includes anything in any of the sets. The intersection contains only those things in every set.

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