0
$\begingroup$

What can you say about $\bigcap_{i\geq 1}A_i$ if for each $i$ we have $A\subset A_i$?

(A) $A\subset \bigcap_{i\geq 1}A_i$
(B) $\bigcap_{i\geq 1}A_i\subset A$
(C) There exists $i$ s.t. $A_i\subset A$

What exactly does $A$ mean - is that the union of all $A_i$ or the intersection of all $A_i$, or neither?

$\endgroup$
9
  • 5
    $\begingroup$ $A$ is just another set. $\endgroup$ Feb 11, 2022 at 0:36
  • 3
    $\begingroup$ $A$ is any set that is a subset of each of the $A_i$. In the future, posts the actual problem, not an image. Use mathjax: math.meta.stackexchange.com/questions/5020/… $\endgroup$ Feb 11, 2022 at 0:36
  • $\begingroup$ Right, but how does that A pertain to Ai? Like, if say, we assume Ai = [3,i], what would A be equal to? or are A and Ai separate entities? ^^^ above would A be the intersection of the subsets then? Because if A C Ai for all i values, wouldn't it have to satisfy all i values, and thus be an intersection of them all? $\endgroup$
    – user1021735
    Feb 11, 2022 at 0:37
  • $\begingroup$ $A \subset A_1, A \subset A_2, A \subset A_3, \cdots.$ Now, compare this inference against each of the $3$ offered choices, and see if the inference corresponds to one of them. For example, if $A \subset A_1$ and $A \subset A_2$, does this imply that $A \subset \left[A_1 \cap A_2\right]$? $\endgroup$ Feb 11, 2022 at 0:39
  • $\begingroup$ Oh so in that case, it would be b correct? because if A is a subset of all of them then A is a subset of the intersection of them? Sorry guys I am really new to set notation and all $\endgroup$
    – user1021735
    Feb 11, 2022 at 0:41

1 Answer 1

0
$\begingroup$

From what I can understand from the comments on this post $A$ is a subset of each $A_i$. I’ll take each of a, b and c and try to explain to my best knowledge:

a) It is true, because the intersection of all $A_i$’s will at the very least be $A$ itself since it is a subset of all of them. That means any element in $A$ is in $A_1,A_2$ and so on.

b) This one can’t be true. Let’s assume any $a\in \displaystyle\cap_{i\geq 1} A_i $, for which $a \notin A$, then that statement is not true. Plus at the very least, in accordance to what I mentioned in (a) you’d need to have an equality included since that’s the only way $$\displaystyle\cap_{i\geq 1}A_i \subseteq A $$ would be true.

(c) That is also untrue provided that $A$ doesn’t coincide with $A_i$ since we’re using “$A\subset A_i$” and not “$A\subseteq A_i$”.

I believe that’s all, but make sure to include some of your past effort next time or you’re risking downvotes and/or people not understanding what your problem is! 🥰

$\endgroup$

You must log in to answer this question.