Why does "Naive Set Theory" by Halmos allow the universal set despite admitting its non-existence? Halmos openly says that the universe doesn't exist:

there is no universe

But several pages later Halmos says:

In order to record the basic facts about complementation as simply as possible, we assume nevertheless (in this section only) that all the sets to be mentioned are subsets of one and the same set E and that all complements (unless otherwise specified) are formed relative to that E. In such situations (and they are quite common) it is easier to remember the underlying set E than to keep writing it down, and this makes it possible to simplify the notation. An often used symbol for the temporarily absolute (as opposed to relative) complement of A is A′. 

The following rule from the book killed all my hopes that E isn't the universal set:

This seems like blatant self-contradiction with previous statement that there is no universe. I hope to resolve it, but I have almost no ideas what it can mean. My only guess is that it's some kind of "lie to children", when relatively simple lie is told because the teacher believes that his/her student(s) can't yet comprehend the whole truth.
P.S. In my other question I got comments that can probably shed some light on the problem. 

looking at the image you posted, I believe you are confusing two different scenarios. In many applications of set theory, there is a universal set at hand. For example, if we are studying natural numbers, the universal set for that purpose may be the set of all natural numbers. All of the rules for set complement assume that a temporary universal set has been chosen in that way. The other context is in studying formal set theory, such as ZFC. In this context, there is no single universal set that contains all sets.( Carl Mummert)

And another:

If your book isn't specifically on mathematical set theory, then the complement is usually taken with respect to a not explicitly stated super set U. E.g. the complement of the set of even numbers is probably meant to be the set of odd numbers, as the super set might be the natural numbers. This should be clear from the context. (M.Winter)

These comments give weak hope, but alas they don't seem to solve the conflict. If anything, they create new one: conclusion from them is in contradiction with statement that absolute coplement of E is the empty set.
For example, let's assume that E is equal to set of natural numbers N. It contains only natural numbers as its elements, but not sets of natural numbers (like it doesn't contain set {1,2,3} as its element). Thus complement of E wouldn't be empty as Halmos assures, it would contain such sets like {1,2,3}, {1,7,65382, 3235464567765}, etc. as its elements. Even worse, I doubt that complement of set E under such circumstances would even be a set in the first place, because it would contain everything that doesn't belong to set E, including set E and complement of set E itself. 
P.P.S. As Eric Wofsey pointed out, I missed consequences of phrase " all the sets to be mentioned are subsets of E". Indeed, then our complement set can't contain elements that aren't elements of E. In other words, our complement set of E must be subset of E! But at the same time they must have no any common elements. Complement of set E can't contain element that isn't element of set E, thus the only alternative is not contain any element at all, i.e. be the empty set because intersection of any set with empty set is empty set.
And of course, under such considerations E doesn't necessarily contain itself as its member.
@EricWofsey Thanks Eric! I think the question is solved. Halmos doesn't really allow the universe, I just got wrong conclusions from rule 

 A: Halmos simply means:

in a specific context, like e.g. real analysis, where we are working with the set $\mathbb R$ as "universe", we can safely use the notation $\{ 0, 1, 2, \pi \}^C$ meaning the complement (in $\mathbb R$) of the set $\{ 0, 1, 2, \pi \}$.

In fact, from a more "general" set-theoretic point of view, we are speaking of the set :


$\mathbb R \setminus \{ 0, 1, 2, \pi \}$.


This set exists in every theory proving (or assuming) the existence of $\mathbb R$.
If so, we have that the complement of $\mathbb R$ "relative to" the universe $\mathbb R$ will be the empty set :


$\mathbb R \setminus \mathbb R = \emptyset$.


A: Halmos never claims that $E$ is the universe, or that all sets are subsets of $E$.  Instead, he is just saying:

Every assertion in this section has an additional unstated hypothesis that all of the sets that are mentioned are subsets of $E$.  Moreover, in this section we use the notation $A'$ for the complement of $A$ relative to $E$, and refer to this as simply the "complement" of $A$.

This in particular applies to the equation $$E'=\emptyset$$ which appears in that section, so in this equation $E'$ stands for the complement of $E$ relative to $E$.
A: The complement here is not absolute, but to be understood relative to the chosen set $E$, that is, $S' = E\setminus S$. In particular, $E' = E\setminus E = \emptyset$.
For example, if $E=\mathbb N$, then you only look at sets containing nothing but natural numbers. Then the complement of such a set also contains nothing but natural numbers; in particular, not sets of natural numbers.
A: It says "all the sets to be mentioned are subsets of one and the same set $E.$"
It does not say "all sets are subsets of one and the same set $E.$"
Complementation is relative. If $E= \{ 1,2,3,4,5 \}$ and $F= \{4,5,6,7,8\}$ then the complement of $F$ relative to $E$ is $E\smallsetminus F = \{1,2,3\}.$
And if $E=\varnothing$ and $F=\{1,2,3\}$ then the complement of $F$ relative to $E$ is $E\smallsetminus F = \varnothing = E.$
