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I have been thinking of a problem that I could not get the most satisfying answer yet.

The problem I have been thinking of:

We know that the Null law says: $$A\cap \varnothing = \varnothing$$ but why $$\bigcap \varnothing = X$$ where $X$ is a universal set.

You may ask me, under what theory, since I believe there might be another theory I do not know? I am still unable to answer that, sorry. This question came to mind when I read the given link above. How is this even possible? Or did I miss something very obvious?

The reason I ask this problem is because my uncerebral, misguided thinking is that I could do this:

$$\begin{align} \bigcap_{i=1}^{n} \varnothing_i &= \varnothing_1 \cap \varnothing_2 \cap \bigcap_{i=3}^{n} \varnothing_i\\ &= \varnothing \cap \bigcap_{i=3}^{n} \varnothing_i \tag{by Null law}\\ &= \varnothing \cap \varnothing_3 \cap \cdots \cap \varnothing_{n-1} \cap \bigcap_{i=n}^{n} \varnothing_i \tag{by Null law}\\ &= \varnothing \end{align}$$

which I believe my thinking is absolutely incorrect and is a misguided thinking. Mayhap?

Could you elaborate your answer with mentioning the ZF as well?, since I read it from the link above that ZF does not have a universal set. I am sure I am just misunderstood and misguided. I am currently studying without a teacher, so it has been difficult for me. I added the "general topology" as the tag since I am reading a topology book by James R. Munkres where I am still at chapter 1. Please enlighten my confusion.

I did read this interesting topic:

Universe set and nullary intersection

However, I am afraid I am still confused as the answerer mentioned $\bigcap\varnothing = \varnothing$ in Set Theory by Kunen which I know nothing about.

Prem
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user516076
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    In the nullary intersection, it’s the index set that’s empty, not the sets in the intersection. – Pranay Dec 01 '24 at 02:51
  • @Pranay Thanks for your response, but what about what DanielWainfleet's comment about $\bigcap\varnothing=\varnothing$ in his answer post in the given link above? I'm more confused. Sorry. It doesn't have the index as well, so I assume the index is empty? – user516076 Dec 01 '24 at 02:57
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    I think with good notation you can avoid any confusion. Say $A_\alpha$’s are subsets of $X$, the universe set, with $\alpha$’s taken from an index set $I$. Then, if $A_\beta = \emptyset$ for some $\beta \in I$ and $J\subseteq I$ contains $\beta$, then $\bigcap_{\alpha\in J} A_\alpha = \emptyset$. This is the null law. On the other hand, the nullary intersection refers to $\bigcap_{\alpha\in\emptyset} A_\alpha = X$. – Pranay Dec 01 '24 at 03:08
  • Thank you very much. – user516076 Dec 01 '24 at 03:09
  • This answer might be relevant, too. – Ulli Dec 01 '24 at 05:39

1 Answers1

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The way I am seeing this is that the notation is "misleading" and we have to use the right interpretation to make the sensible claim.
What OP wants is (EX 5) , I am giving few other Examples to make it more intuitive.

SUMMARY : What the "misleading" notation $\cap\,\emptyset$ is trying to indicate : we are taking intersection of a list of Sets , though that list is itself Empty : We will have to get Universal Set.

Consider the similar concepts :

(EX 1)
$\sum_{i=1}^{i=3} x_i=x_1+x_2+x_3$
$\sum_{i=1}^{i=2} x_i=x_1+x_2$
$\sum_{i=1}^{i=1} x_i=x_1$
$\sum_{i=1}^{i=0} x_i=??$
We know that $\sum_{i=1}^{i=3} x_i=\sum_{i=1}^{i=2} x_i+x_3$ , likewise $\sum_{i=1}^{i=1} x_i=\sum_{i=1}^{i=0} x_i+x_1$ , which then makes us conclude that $\sum_{i=1}^{i=0} x_i=0$
Summary : Sum of Empty list is $0$

(EX 2)
Similarly , we can conclude that $\Pi_{i=1}^{i=0}=1$
Summary : Product of Empty list is $1$

(EX 3)
Special Case : We can conclude that $0!=1!=1$
Summary : Product of Empty list of consecutive integers is $1$

(EX 4)
When we have a list of Sets $S_1,S_2,S_3,S_4,S_5$ and we take the union , we will get a new set.
We can take the union of a smaller list and then we can include the left over Sets to get the same outcome.
$\cup_{i=1}^{i=5} S_5 = [ \cup_{i=1}^{i=4} S_i ] \cup S_5$
Naturally , we can continue that with smaller lists to get the same outcome.
$\cup_{i=1}^{i=2} S_5 = [ \cup_{i=1}^{i=1} S_i ] \cup S_2$
$\cup_{i=1}^{i=1} S_5 = [ \cup_{i=1}^{i=0} S_i ] \cup S_1$ , which gives us the fact that $[ \cup_{i=1}^{i=0} S_i ] = \emptyset$
Summary : Union of Empty list of Sets is $\emptyset$

(EX 5)
When we have a list of Sets $S_1,S_2,S_3,S_4,S_5$ and we take the intersection , we will get a new set.
We can take the intersection of a smaller list and then we can include the left over Sets to get the same outcome.
$\cap_{i=1}^{i=5} S_5 = [ \cap_{i=1}^{i=4} S_i ] \cap S_5$
Naturally , we can continue that with smaller lists to get the same outcome.
$\cap_{i=1}^{i=2} S_5 = [ \cap_{i=1}^{i=1} S_i ] \cap S_2$
$\cap_{i=1}^{i=1} S_5 = [ \cap_{i=1}^{i=0} S_i ] \cap S_1$ , which gives us the fact that $[ \cup_{i=1}^{i=0} S_i ] = X$ , where $X$ is the Universal Set.
$X \cap S_1 = S_1$ , where Universal Set $X$ will always work , no matter what $S_1$ is.
Summary : Intersection of Empty list of Sets is the Universal Set

This is what the "misleading" notation $\cap\,\emptyset$ is trying to indicate : we are taking intersection of a list of Sets , though that list is itself Empty : We will have to get Universal Set.

Tankut Beygu
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Prem
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