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Is there a rigid proof of the lemma? I've seen abstract statements and I do understand that it is true. It is very intuitive. However, I do not understand if there's a mathematical way (like in proper steps, not just an explanation).

Is there one? If so, how would you approach it?

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    $\begingroup$ I think calling this a lemma is a bit of an overstatement. It's just an immediate consequence of the definition. There is nothing in the empty set, so everything that is not in the empty set is just everything. Everthing is in the universal set, so there is no one thing not in the universal set, so the complement is empty. $\endgroup$
    – tomasz
    Sep 6, 2020 at 16:39
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    $\begingroup$ The idea of a universal set is potentially problematic (see eg the Russell Paradox). How are you defining a complement? $\endgroup$ Sep 6, 2020 at 16:39
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    $\begingroup$ Quite... We can talk about a complement with respect to a specific "universal set" such as the complement of $\{1,2\}$ with respect to the universal set $\{1,2,3,4,5\}$ being $\{3,4,5\}$... but when we do not specify what the universal set was then there is no way to adequately define the complement of $\{1,2\}$ is in general. There is no universal universal set. $\endgroup$
    – JMoravitz
    Sep 6, 2020 at 16:42
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    $\begingroup$ No, for any "universal" set, it's an immediate consequence of the definition, as @tomasz said. There's nothing shaky, or non-mathematical about it. $\endgroup$
    – saulspatz
    Sep 6, 2020 at 17:24
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    $\begingroup$ @Justanotherperson: Nonsense. An abstract argument can be a solid proof (for any reasonable definition of "solid proof"). $\endgroup$
    – TonyK
    Sep 6, 2020 at 17:25

2 Answers 2

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"like in proper steps, not just an explanation" The steps of a proof by definition is a statement of the definition and a demonstration that all the criteria apply.

Definition: $X^c = \{x \in U| x\not \in X\}$.

So $U^c = \{x\in U| x\not \in U\}=\{x|x \in U$ and $x\not \in U\}$. As $x \in U$ and $x\not \in U$ are contradictory statements this can never occur and so $U^c$ can have no elements and $U^c = \emptyset$.

And $\emptyset^c = \{x \in U| x\not \in X\}$. As no $x \in \emptyset$ then $x \in U \implies x \not \in \emptyset$ so $\emptyset^c=\{x\in U| x\not \in X\} = \{x \in U\} = U$.

That is the mathematical way.

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There's two steps in the "proof" of the statement (the quotes here are because one could argue if this can be considered as a proof or not); defining well your objects and then letting the definitions spit out the result. In this case we need three basic notions: universal set, complement of a set with respect to another set and empty set.

The universal set is just some set $\Omega$ that we fix, meaning that all elements and subsets which we will consider from now onwards will belong and be contained in $\Omega$, respectively. The complement of a subset $X \subseteq\Omega$ is defined as $$X^c= \{ x \in \Omega : x \notin X\}.\tag{$\dagger$}$$ Note that I have specified that $X$ is a subset of our universal set $\Omega$, so the notation in $(\dagger)$ seems slightly misleading since it hides this fact; note however that we've established before that all subsets are taken with respect to $\Omega$, so this is not a real issue. This is way we call $\Omega$ universal, so that when we argue about such set theoretic operations we can shorten our arguments by not repeating each time that the complement of $X$ is taken with respect to $\Omega$, as in our example.

Now, recalling that $\varnothing$ is the (unique) set which has no elements in it, the argument would go as follows:

$$\begin{align} \Omega^c &= \{ x \in \Omega : x \notin \Omega \} &\text{ by definition of complement}\\ &= \varnothing &\text{by definition of }\varnothing\end{align}$$

If you want to be really picky, you could justify even further the last equality as follows.

Trivially $\varnothing \subseteq \{ x \in \Omega : x \notin \Omega \}$, so it is left for us to prove the other inclusion. Assume for contradiction it is false, i.e. there is $a \in \{ x \in \Omega : x \notin \Omega \}$ such that $a \notin \varnothing$. Then, since $\{ x \in \Omega : x \notin \Omega \} \subseteq \Omega$, we have that $a \in \Omega$; but since $a \in \{ x \in \Omega : x \notin \Omega \}$ we also have that $a \notin \Omega$, so our assumption is false and equality follows.

Showing that $\varnothing^c = \Omega$ is analogous. As you see, the problem here is not in the argument itself, but in how you are defining the objects you are working with.

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