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I noticed in my discrete math textbook that the Rosen makes reference to the universal set, denoted $U$, and establishes various results, like the complement of a set $A$ with respect to $U$, or the identity law $A \cap U = A$. But in what sense are we calling $U$ the universal set so that such a set is valid in elementary set theory? In other words, why is it legitimate in the author's case to speak of $U$ where it seems to be implied somehow that $U$ is not the set of all sets? Do I have to state that $U$ is specific universal set as to avoid confusion as to whether or not it is the set of all sets (but still calling the set $U$ rather than a concrete universe like $\mathbb{Z}$)?

With that question proposed, if we let $U$ be specific universal set (not the set of all sets), and define $\emptyset_1$ and $\emptyset_2$ to be two empty sets, then we can conclude that $\emptyset_1 = \emptyset_2$ by noting

$$U = {\emptyset_1}^c$$ $$U = {\emptyset_2}^c$$

implies that

$$U^c = ({\emptyset_1}^c)^c = \emptyset_1$$ $$U^c = ({\emptyset_2}^c)^c = \emptyset_2$$

Because $U^c = U^c$, then $\emptyset_1 = \emptyset_2$.

Of course, another way to do this proof would be to show that the two empty sets are subsets of each other.

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In those examples, we are working in the so-called "naive" set theory; in this case, the "universal" set is simply a collection of objects assumed as the domain where the variables range.

It can be the set $\mathbb N$ of natural numbers, or the set $Humanity$ of all human beings.

It is not intended to be the set of "everything".

With this approach, we can treat the elementary properties of sets : the so-called algebra of sets.

For example, if the domain (or universe) $U$ is the set $\mathbb N$ of naturals and $A$ is the set of odd numbers, then $A^c$ is the complement of $A$ "in $U$", i.e. the set of even numbers.


Regarding the "unicity" of $\emptyset$, this is straightforward.

We know that :

for all $x$, $x \notin \emptyset$.

But we have also the principle of extensionality for sets :

if for all $x, x \in A \leftrightarrow x \in B$, then $A=B$.

Applying it to $\emptyset_1$ and $\emptyset_2$, we clearly have :

for all $x, x \in \emptyset_1 \leftrightarrow x \in \emptyset_2$

because both are empty.

Thus, we conclude with :

$\emptyset_1 = \emptyset_2$.

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  • $\begingroup$ Ah, so proving that the two empty sets are subsets of each other demonstrates their equality. Definitely easier. Just for hiccups and giggles, is the proof that I wrote valid? $\endgroup$ – A. Lovelace Sep 26 '15 at 19:29
  • $\begingroup$ @A.Lovelace - yes, it is. $\endgroup$ – Mauro ALLEGRANZA Sep 26 '15 at 19:51
  • $\begingroup$ thank you for your assistance, sir $\endgroup$ – A. Lovelace Sep 26 '15 at 19:57

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