To Prove $ \bigcap_{i \in I} A_i \in \bigcap_{i \in I} P(A_i) $

Question

$$ \bigcap_{i \in I} A_i \in \bigcap_{i \in I} P(A_i) $$ , $ I \neq \phi $

MY ATTEMPT

I use proof by Contradiction. Assume $ \bigcap_{i \in I} A_i \notin \bigcap_{i \in I} P(A_i) $$

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

i.e $ \{ x \} \notin ( P(A_1 ) \land P(A_2 \land... \land P(A_n)) )$

So $ \{x\} \not \subseteq A_i \forall i \in I$

So $ x \notin A_i \forall i \in I$

So $x \notin \bigcap_{i \in I} A_i $

Hence we arrive at contradiction

I don't see how you get ${x} \notin \bigcap_{i \in I} P(A_i)$. You seem to be saying $x\in A\notin B\implies x\notin B$ but that isn't so. (Take $x=1,\ A={1,2},\ B={1}$.) I'd prove it directly instead of by contradiction. Considering ${x}$ as you did, is the right idea. — saulspatz, Jun 20 '20 at 16:06
By the way, don't use $\land$ when you mean $\cup$; $\land$ is for the conjunction of propositions. Having said that, the behaviour's the same. — J.G., Jun 20 '20 at 16:54

Brian M. Scott · Accepted Answer · 2020-06-20T16:28:08.727

When you start by letting $x\in\bigcap_{i\in I}A_i$ and supposing that $x\notin\bigcap_{i\in I}\wp(A_i)$, you’re already getting off on the wrong foot: that would be a reasonable start if you were trying to show that $\bigcap_{i\in I}A_i$ was a subset of $\bigcap_{i\in I}\wp(A_i)$, but that isn’t what you want to show. You need to show that $\bigcap_{i\in I}A_i$ is an element of $\bigcap_{i\in I}\wp(A_i)$. In symbols, you’re setting out to try to prove that $\bigcap_{i\in I}A_i\color{red}{\subseteq}\bigcap_{i\in I}\wp(A_i)$, but what you need to prove is that $\bigcap_{i\in I}A_i\color{red}{\in}\bigcap_{i\in I}\wp(A_i)$.

Let’s back up for a minute and take a good look at the objects involved. In fact, let’s look at a very simple example. Suppose that $I=\{1,2,3\}$, so that you have sets $A_1,A_2$, and $A_3$. To be absolutely definite, let’s suppose that $A_1=\{1,2,4,5\}$, $A_2=\{2,3,4,5\}$, and $A_3=\{4,5,6\}$. Then

$$\begin{align*} \bigcap_{i\in I}A_i&=A_1\cap A_2\cap A_3\\ &=\{1,2,4,5\}\cap\{2,3,4,5\}\cap\{4,5,6\}\\ &=\{4,5\}\;. \end{align*}$$

This a subset of each of the sets $A_1,A_2$, and $A_3$, so like $A_1,A_2$, and $A_3$, it is a set of integers. Now what is $\bigcap_{i\in I}\wp(A_i)$?

$$\begin{align*} \wp(A_1)=&\big\{\varnothing,\{1\},\{2\},\{3\},\{4\},\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\},\\ &\{1,2,3\},\{1,2,4\},\{1,3,4\},\{2,3,4\},\{1,2,3,4\}\big\}\;, \end{align*}$$

$$\begin{align*} \wp(A_2)=&\big\{\varnothing,\{2\},\{3\},\{4\},\{5\},\{2,3\},\{2,4\},\{2,5\},\{3,4\},\{3,5\},\{4,5\},\\ &\{2,3,4\},\{2,3,5\},\{2,4,5\},\{3,4,5\},\{2,3,4,5\}\big\}\;, \end{align*}$$

and

$$\wp(A_3)=\big\{\varnothing,\{4\},\{5\},\{6\},\{4,5\},\{4,6\},\{5,6\},\{4,5,6\}\big\}\;,$$

and the intersection of these three sets is

$$\big\{\varnothing,\{4\},\{5\},\{4,5\}\big\}\;:$$

$\varnothing,\{4\},\{5\}$, and $\{4,5\}$ are the only sets of integers that are elements of all three power sets.

$\bigcap_{i\in I}A_i=\{4,5\}$ cannot possibly be a subset of $\bigcap_{i\in I}\wp(A_i)$: it’s the wrong kind of object. If it were a subset of $\bigcap_{i\in I}\wp(A_i)$, its elements would also be elements of $\bigcap_{i\in I}\wp(A_i)$. But the elements of $\{4,5\}$ are integers, while the elements of $\bigcap_{i\in I}\wp(A_i)$ are sets of integers.

$\{4,5\}$ can, however, be an element of $\bigcap_{i\in I}\wp(A_i)$, and indeed we see that it is:

$$\bigcap_{i\in I}\wp(A_i)=\big\{\varnothing,\{4\},\{5\},\color{red}{\{4,5\}}\big\}\;.$$

Now let’s go back and consider how to prove the result. You don’t need a proof by contradiction: you can show directly that $\bigcap_{i\in I}A_i\in\bigcap_{i\in I}\wp(A_i)$. For each $i\in I$, $\bigcap_{i\in I}A_i$ is a subset of $A_i$: $\bigcap_{i\in I}A_i\subseteq A_i$. By definition this means that $\bigcap_{i\in I}A_i\in\wp(A_i)$. Thus, $\bigcap_{i\in I}A_i\in\wp(A_i)$ for each $i\in I$, and that by definition means that $\bigcap_{i\in I}A_i$ is in the intersection of those power sets: $\bigcap_{i\in I}A_i\in\bigcap_{i\in I}\wp(A_i)$. And that’s what we wanted to prove.

Once you understand what’s going on here, you might try to prove the stronger result that

$$\bigcap_{i\in I}\wp(A_i)=\wp\left(\bigcap_{i\in I}A_i\right)\;.$$

$$ x\in\bigcap_{i\in I}A_i ,then, {x} \notin\bigcap_{i\in I}\wp(A_i) $$. Is this correct ? Given the assumption — Jessica Griffin, Jun 20 '20 at 16:47
@JessicaGriffin: I’m afraid not: if $x\in\bigcap_{i\in I}A_i$, then $x\in A_i$ for every $i\in I$, so ${x}\in\wp(A_i)$ for every $i\in I$, and therefore ${x}\in\bigcap_{i\in I}\wp(A_i)$. — Brian M. Scott, Jun 20 '20 at 16:49
Yes but i am saying if given the assumption i have opted for then ? — Jessica Griffin, Jun 20 '20 at 16:50
@JessicaGriffin: You mean the way that you began your argument in the question? That is completely on the wrong track and has no hope of going anywhere; you should simply forget about it and start over. — Brian M. Scott, Jun 20 '20 at 16:52

score 2 · Answer 2 · answered Jun 20 '20 at 16:12

2

Let $A=\bigcap_{i\in I}A_i$. By definition, $A\subseteq A_j$ for each $j\in I$ and thus $j\in I\implies A\in\mathcal P(A_j)$. $$\therefore A\in\bigcap_{i\in I}\mathcal P(A_i)$$

answered Jun 20 '20 at 16:12

Oliver Kayende

1,985

Thanks. i upvoted – Jessica Griffin Jun 20 '20 at 17:00

score 2 · Answer 3 · answered Jun 20 '20 at 16:16

2

Why not a direct proof? $$\bigcap_{i \in I}A_i \in \bigcap_{i \in I}P(A_i) \iff \forall k \in I: \bigcap_{i \in I} A_i \in P(A_k)$$ $$\iff \forall k \in I: \bigcap_{i \in I} A_i \subseteq A_k$$

and the latter statement is obvious.

answered Jun 20 '20 at 16:16

J. De Ro

21,438

Thanks. i upvoted – Jessica Griffin Jun 20 '20 at 17:00

To Prove $ \bigcap_{i \in I} A_i \in \bigcap_{i \in I} P(A_i) $

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