📂Set Theory

What is Resonance?

Theorem

Given the propositions $p$, contradiction $c$, and $A_{\alpha} \subset X$, the following hold:

• [1] Vacuous Truth: $c \implies p$
• [2] Union: $\displaystyle \bigcup_{\alpha \in \emptyset} A_{\alpha} = \emptyset$
• [3] Intersection: $\displaystyle \bigcap_{\alpha \in \emptyset} A_{\alpha} = X$

Explanation

For instance, in the statement “God is dead”, if God does not exist, the premise is already flawed. If God doesn’t exist, then it implies $0$ gods have died, so it is trivially true regardless of who truly died or lived. Similarly, “God is alive” is also true under the assumption God does not exist, as it attempts to verify the existence of $0$ god(s), making it necessarily true.

Thus, if the assumption is a contradiction, the claim becomes true regardless of its content, known as Vacuous Truth or sometimes referred to as Tautology. Although there are other forms of tautologies, $c \implies p$ is one that’s particularly hard to accept. Thinking about it in the context of sets, taking either the Union or Intersection of $0$ subsets of a given set can make it easier to understand. $\displaystyle \sum_{n=0}^{0} n = 0$ might make it appear more intuitive.

Proof

Vacuous Truth

To prove that $c \to p$ is true whether $p$ is true or false, note that since $x \to y \equiv \lnot ( x \land \lnot y )$, $$c \to p \equiv \lnot ( c \land \lnot p )$$ Whether $p$ is true or false, taking the logical AND with $c$ results in false, so $$\lnot ( c \land \lnot p ) \equiv \lnot c$$ Since the negation of contradiction $c$ is always true, $c \to p$ is true regardless of $p$.

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Union

To prove for all $x \in X$ that $\displaystyle x \notin \bigcup_{\alpha \in \emptyset} A_{\alpha}$ holds, $$\begin{align*} x \notin \bigcup_{\alpha \in \emptyset} A_{\alpha} \iff & \lnot \left( x \in \bigcup_{\alpha \in \emptyset} A_{\alpha} \right) \\ \iff & \lnot ( x \in A_{\alpha_{0}} \text{ for some } \alpha_{0} \in \emptyset ) \\ \iff &x \notin A_{\alpha} \text{ for all } \alpha \in \emptyset \\ \iff & \alpha \in \emptyset \to x \notin A_{\alpha} \end{align*}$$ If the set $\emptyset$ has an element, it contradicts the definition of an empty set, hence $\alpha \in \emptyset$ is false. According to [1] Vacuous Truth, $\alpha \in \emptyset \to x \notin A_{\alpha}$ is true, and thus $\displaystyle x \notin \bigcup_{\alpha \in \emptyset} A_{\alpha}$, which is equivalent, is also true.

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Intersection

To prove for all $x \in X$ that $\displaystyle x \in \bigcap_{\alpha \in \emptyset} A_{\alpha}$ holds, $$\begin{align*} x \in \bigcap_{\alpha \in \emptyset} A_{\alpha} \iff & x \in A_{\alpha} \text{ for all } \alpha \in \emptyset \\ \iff &\alpha \in \emptyset \to x \in A_{\alpha} \end{align*}$$ Similarly, if set $\emptyset$ has an element, it contradicts the definition of an empty set, making $\alpha \in \emptyset$ false. According to [1] Vacuous Truth, $\alpha \in \emptyset \to x \in A_{\alpha}$ is true, and thus $\displaystyle x \in \bigcap_{\alpha \in \emptyset} A_{\alpha}$, which is equivalent, is also true.

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