Let I be a nonempty set and a family of sets such that every element of the family is a subset of U.

$\mathcal F = \{A_i | i \in I\}$

I understand the meaning of this operation:

$$ \bigcap_{i \in I}A_i $$

That's the intersection of all the elements of the sets of the family $\mathcal F$.

But I don't truly understand what this other operation means and what's the relation between the above one. $$ \bigcap_{i=1}^{n}A_i $$

I understand that this operation is also an intersection but what I am trying to understand is the relation between $\bigcap_{i \in I}A_i$ and $\bigcap_{i=1}^{n}A_i$

$$ \bigcap_{i=1}^{n}A_i \subseteq\bigcap_{i \in I}A_i $$

Is the relation above true?

  • 1
    $\begingroup$ In the second case we have $I = \{ 1,2,\ldots, n \}$. $\endgroup$ – Mauro ALLEGRANZA Jun 23 '18 at 15:28

The notation $$\bigcap _{i=1}^nA_i$$ denotes the same as $$ \bigcap_{i\in I}A_i$$ for the special case $I=\{1,2,\ldots, n\}$.

  • $\begingroup$ So $\bigcap_{i\in I}A_i$ is the general case and $\bigcap _{i=1}^nA_i$ is a specific case for which we give concrete values to i? $\endgroup$ – adriana634 Jun 23 '18 at 15:40
  • $\begingroup$ @adriana634 Yes. One situation where we see both notations used simultaneously is when $I=\mathbb{N}$; then "$\bigcap_{i=1}^nA_i$" means "$\bigcap_{i\in \{1,2,..., n\}}A_i$," which is generally a superset of $\bigcap_{i\in I}A_i$. $\endgroup$ – Noah Schweber Jun 23 '18 at 15:42
  • $\begingroup$ @Noah Schweber But for example $\bigcap_{i=1}^{5}A_i$ means $I=\{1,2,\ldots, 5\}$, in this example $\bigcap_{i\in I}A_i$ is a superset of $\bigcap_{i=1}^{5}A_i$? I don't understand why you said $\bigcap_{i=1}^nA_i$ is a superset of $\bigcap_{i\in I}A_i$. $\endgroup$ – adriana634 Jun 23 '18 at 16:00
  • $\begingroup$ @adriana634 No. The intersection over a bigger index set yields a smaller set. E.g. if we take $I=\mathbb{N}$ and $A_i=\{i, i+1, i+2, ...\}$ for $i\in I$, then $\bigcap_{i=1}^5A_i=\{5, 6,7,...\}$ but $\bigcap_{i\in I}A_i=\emptyset$. More abstractly: $$I\supseteq J\implies \bigcap_{i\in I}A_i\subseteq \bigcap_{i\in J}A_i.$$ $\endgroup$ – Noah Schweber Jun 23 '18 at 16:02

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