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I haven't found an answer yet that I can make complete sense of, so I'm asking again to try and clear it up if I have any misconceptions.

If one defines $\cap_{i\in I}\alpha_i = \{x:\forall i\in I, x\in\alpha_i\}$ then apparently if $I = \emptyset$ this definition yields the absolute universe. This is just stated as if it is clear why, though I cannot see why. If $i\notin I \forall i$ then there is no set $\alpha_i$ for any $x$ to be a member of...?

I know I must be misreading this, but I can't see how by so much...

Edit: Let this intersection be $Z$, for convenience.

Nethesis
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    Quick: Name an element that isn't in the intersection... – Jon Feb 15 '15 at 15:50
  • If the intersection is denoted $Z$, then if $x\notin Z$, we have $x\notin \alpha_i$ for some $i$.... – Nethesis Feb 15 '15 at 15:52
  • Except no such $i$ exists, so $x\in Z$. – Jon Feb 15 '15 at 15:56
  • But conversely if $x\in Z$ then $x\in \alpha_i$ for all $i$, and no such $i$ exists, so $x \notin Z$... – Nethesis Feb 15 '15 at 15:57
  • Can you find an $i$ for which $x$ is not an element of $\alpha_i$? – Jon Feb 15 '15 at 15:59
  • Can you find an $i$ for which $x$ is...? – Nethesis Feb 15 '15 at 16:01
  • Nope, I cannot seem to find any $i\in I$ such that $x\not\in \alpha_i$. By way of intuition, $A\supseteq A\cap B \supseteq A\cap B\cap C\supseteq \cdots$ so one would hope that the intersection of no sets contains any set. Hopefully this helps you accept the rigorous yet vacuous argument. – Jon Feb 15 '15 at 16:05
  • I'm feeling as if this cannot be proven either way to be honest...I accept that your argument is valid, but it seems that that gives a perfectly valid and yet contradictory argument at the same time – Nethesis Feb 15 '15 at 16:09
  • You haven't found an answer yet, but this question has been asked more than a handful of times before on this site alone. – Asaf Karagila Feb 15 '15 at 16:09
  • I know...and I'm beginning to understand why. It seems to be a difficult concept to wrap one's head around, though they can grasp the logic behind why $Z$ is the universe... – Nethesis Feb 15 '15 at 16:11

2 Answers2

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Any statement of the form $\forall i\in I, ...$ is true if $I$ is empty, so in particular the statement $\forall i\in I, x\in \alpha_i$ is true for all $x$. Perhaps the easiest way to see this, is that the negation of this statement is: $\exists i\in I$ such that $x\notin \alpha_i$, which is clearly false if are no $i$ to begin with.

Uncountable
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  • But if we have $\forall i\in I, x\in \alpha_i$, then $\exists i\in I$ such that $x\in \alpha_i$, too? I understand that I must be missing something, but I'm not sure of where the gap in my logic lies – Nethesis Feb 15 '15 at 16:04
  • @Nethesis No, universality does not presuppose existence. – amWhy Feb 15 '15 at 16:05
  • But if $x$ must be a member of some set $\alpha_i$ in order to be a member of $Z$ (the intersection), and no such set $\alpha_i$ exists, then does that not mean that $x\notin Z$ must also, paradoxically, hold? – Nethesis Feb 15 '15 at 16:07
  • In order for $x$ to be in this intersection, it must be also in...nothing...? – Nethesis Feb 15 '15 at 16:08
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Given $\cap_{i\in I}\alpha_i = \{x:\forall i\in I, x\in\alpha_i\}$. If I, the family of sets is empty then any x would satisfy the condition $\forall i\in I, x\in\alpha_i$ vacuously because there is no $\alpha_i$. Therefore $\cap_{i\in I}\alpha_i=$Universe if $I=\phi$. Unfortunately, there is no universal set in Zermelo Frankel set theory. We can overcome this problem by restricting the sets to a universal set U. In this case nullary intersection would be U.

Harmonic Explorer
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