Unions & Intersections

Some of the more interesting exercises (a purely subjective qualification). Easy concepts, difficult language. Boo.

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Theorem:

The following are equivalent:
- the union of C and (the union of the B-alphas, over the alphas in lambda,)
- the union of (the union of C and the B-alpha), over the alphas in lambda.

Proof:

$$\begin{align*} p(x, C, \{B_\alpha\}_{\alpha\in\Lambda})& = \underbrace{x \in \left\{C \cup \left[\bigcup_{\alpha\in\Lambda}B_\alpha\right] \right\}}\\[1em] & = \underbrace{(x \in C)} \vee \underbrace{\left( x \in \left[\bigcup_{\alpha\in\Lambda}B_\alpha\right] \right)}\\[1em] & = \underbrace{ (x \in C)} \vee \underbrace{(\exists(B_\alpha), x\in B_\alpha) }\\[1em] & = \underbrace{ \exists(C\cup B_\alpha), x \in (C\cup B_\alpha) }\\[1em] &=\underbrace{x\in \left\{\bigcup_{\alpha\in\Lambda}(C\cup B_\alpha)\right\}}\\[1em] \end{align*}$$

Less than seven hours till first alarm. I really should get out of the office.

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Theorem:

The following are equivalent:
- the intersection of C and (the intersection of the B-alphas, over the alphas in lambda,)
- the intersection of (the intersection of C and the B-alpha), over the alphas in lambda.

Proof:

$$\begin{align*} p(x, C, \{B_\alpha\}_{\alpha\in\Lambda})& = \underbrace{x \in \left\{C \cap \left[\bigcap_{\alpha\in\Lambda}B_\alpha\right] \right\}}\\[1em] & = \underbrace{(x \in C)} \wedge \underbrace{\left( x \in \left[\bigcap_{\alpha\in\Lambda}B_\alpha\right] \right)}\\[1em] & = \underbrace{ (x \in C)} \wedge \underbrace{(\forall(B_\alpha), x\in B_\alpha) }\\[1em] & = \underbrace{ \forall(C\cap B_\alpha), x \in (C\cap B_\alpha) }\\[1em] &=\underbrace{x\in \left\{\bigcap_{\alpha\in\Lambda}(C\cap B_\alpha)\right\}}\\[1em] \end{align*}$$

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De Morgan's law:

Theorem:

Where all $(A_\alpha)$s are subsets of U, the following are equivalent:
- the complement in U of (the union over the alphas in lambda, of the A-alphas)
- the intersection over the alphas in lambda of (the complements in U, of the A-alphas)

Proof:

$$\begin{align*} p(x, U, \{A_\alpha\}_{\alpha\in\Lambda})& = x\in\left[\bigcup_{\alpha\in\Lambda} A_\alpha\right]^{\mathsf{\complement}}_{U}\\[1em] &=(x\in U)\wedge (\underbrace{\forall A_\alpha, A_\alpha \in U)}\wedge \underbrace{\neg(\exists A_\alpha, x\in A_\alpha)} \\[1em] &=(x\in U)\wedge (\underbrace{\forall A_\alpha, A_\alpha \in U)}\wedge \underbrace{(\forall A_\alpha, x\in (A_\alpha)^{\complement}_U)} \\[1em] &= x\in \left[\bigcap_{\alpha\in\Lambda} (A_\alpha)^{\mathsf{\complement}}_{U}\right]\\[1em] \end{align*}$$

Theorem:

Where all $(A_\alpha)$s are subsets of U, the following are equivalent:
- the complement in U of (the intersection over the alphas-in-lambda, of the A-alphas)
- the union over the alphas-in-lambda of (the complements in U, of the A-alphas)

Proof:

$$\begin{align*} p(x, U, \{A_\alpha\}_{\alpha\in\Lambda})& = x\in\left[\bigcap_{\alpha\in\Lambda} A_\alpha\right]^{\mathsf{\complement}}_{U}\\[1em] &=(x\in U)\wedge (\underbrace{\forall A_\alpha, A_\alpha \in U)}\wedge \underbrace{\neg(\forall A_\alpha, x\in A_\alpha)} \\[1em] &=(x\in U)\wedge (\underbrace{\forall A_\alpha, A_\alpha \in U)}\wedge \underbrace{(\exists A_\alpha, x\in (A_\alpha)^{\complement}_U)} \\[1em] &= x\in \left[\bigcup_{\alpha\in\Lambda} (A_\alpha)^{\mathsf{\complement}}_{U}\right]\\[1em] \end{align*}$$

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Theorem:

Proof: