CMSC250

Set Proofs and Countability

Set Proofs

Countability

Set Proofs

Note: Set theory Proofs are universal

(So disproving means finding a counterexample)

Subsets and Equality

Recall: $A \subseteq B \Leftrightarrow (\forall x, x \in A \Rightarrow x \in B)$

Recall: Set Difference is $A - B = A \cap B^c$

Prove: Set Difference is $A - B \subseteq A \cap B^c$

$x \in A - B\Rightarrow x \in A \land x \not\in B$

$x \not\in B\Rightarrow x \in B^c$

$x \in A \land x \in B^c\Rightarrow x \in A \cap B^c$

Subsets and Equality

Note: Disproving a subset means having a counterexample

Venn diagram helps, but we want concrete example

Disprove: $A \cup B \subseteq A^c \cup B$

$A \cup B$

$A^c \cup B$

Disprove: $A \cup B \subseteq A^c \cup B$

$U = \{1,2,3,4\}, A = \{1,3\}, B = \{2,3\}$

$A \cup B = \{1,2,3\}, A^c \cup B = \{2,3,4\}$

Subsets and Equality

Recall: $A = B \Leftrightarrow (A \subseteq B \land B \subseteq A)$

$(\forall x, x\in A \Rightarrow x \in B) \land (\forall y,y \in B \Rightarrow y \in A)$

$(\forall x \in A, x \in B) \land (\forall y \in B, y \in A)$

Recall: Absorbtion Law: $A \cup (A \cap B) = A$

Prove: Absorbtion Law: $A \cup (A \cap B) = A$

First: Prove $A \cup (A \cap B) \subseteq A$

$x \in A \cup (A \cap B)\Rightarrow x \in A \lor x \in A \cap B$

$x \in A \cap B\Rightarrow x \in A \land x \in B$

$x \in A \cap B\Rightarrow x \in A$

$x \in A$

Note: This is cases

Recall: Cases

$\begin{array}{ll} & p \lor q\\ & p \Rightarrow r\\ & q \Rightarrow r\\ \therefore & r \end{array}$

Note: This is cases

Recall: Cases

$\begin{array}{ll} & (x \in A) \lor (x \in A \cap B)\\ & (x \in A) \Rightarrow (x \in A)\\ & (x \in A \cap B) \Rightarrow (x \in A)\\ \therefore & (x \in A) \end{array}$

Prove: Absorbtion Law: $A \cup (A \cap B) = A$

Second: Prove $A \subseteq A \cup (A \cap B)$

$x \in A, x \in A, A \subseteq A$

$x \in A\Rightarrow x \in A \cup (A \cap B)\Rightarrow A \subseteq A \cup (A \cap B)$

Note: It is important to prove both directions

Prove: Absorbtion Law: $A \cup (A \cap B) = A$

Second: Prove $A \subseteq A \cup (A \cap B)$

$x \in A\Rightarrow x \in A \cup (A \cap B)\Rightarrow A \subseteq A \cup (A \cap B)$

Note: It is important to prove both directions

Recall: Demorgan's Law: $(A \cap B)^c = A^c \cup B^c$

Prove: Demorgan's Law: $(A \cap B)^c = A^c \cup B^c$

First: Prove $(A \cap B)^c \subseteq A^c \cup B^c$

$x \in (A \cap B)^c \Rightarrow x \not\in (A \cap B)$

$x \not\in (A \cap B) \Rightarrow x \not\in A \lor x \not\in B$ (or neither)

$x \not\in A, x \in A^c \Rightarrow x \in A^c \cup B^c$

$x \not\in B, x \in B^c \Rightarrow x \in A^c \cup B^c$

Prove: Demorgan's Law: $(A \cap B)^c = A^c \cup B^c$

Second: Prove $A^c \cup B^c \subseteq (A \cap B)^c$

$x \in A^c \cup B^c\Rightarrow x \in A^c \lor x \in B^c$ (or both)

$x \in A^c, x \not\in A, \Rightarrow x \not\in (A \cap B)$

$x \in B^c, x \not\in B, \Rightarrow x \not\in (A \cap B)$

$x \not\in (A \cap B)\Rightarrow x \in (A \cap B) \Rightarrow A^c \cup B^c \subseteq (A \cap B)^c$

Note: Cases again

Subsets and Equality

Note: Disproving set equality means having a counterexample

Venn diagram helps, but we want concrete example

Disprove: $A \cup (C - B) = (A \cup C) - B$

$A \cup (C-B)$

$(A \cup C) - B$

Disprove: $A \cup (C - B) = (A \cup C) - B$

$U = \{1,2,3,4,5,6,7,8\}, A = \{1,4,5,7\}, B = \{2,4,6,7\}$$C = \{3,5,6,7\}$

$A \cup (C-B) = \{1,3,4,5,7\}, (A \cup C) - B = \{1,3,5\}$

Emptiness

Difficult to show something that does not exist

(Trying to show nonexistence is hard)

Difficult to show something that does not exist

The power of Contradiction

Prove: $A \cap \emptyset = \emptyset$

Assume: $A \cap \emptyset \neq \emptyset$

$\exists x \in A \cap \emptyset$

$\exists x \in A \cap \emptyset, x \in A \land x \in \emptyset$

$x \in \emptyset$ Contradiction

Conditionals

Set conditionals are just arguments with sets

Recall: Valid argument is one where premises imply conclusion

Prove: If $A \subset B$ and $B \subseteq C^c$ then $A \cap C = \emptyset$

$\begin{array}{ll} & A \subset B\\ & B \subseteq C^c\\ \therefore & A \cap C = \emptyset \end{array}$

Prove: If $A \subset B$ and $B \subseteq C^c$ then $A \cap C = \emptyset$

$\forall x, x \in A \Rightarrow x \in B$

$\forall x, x \in B \Rightarrow x \in C^c$

$\forall x, x \in A \Rightarrow x \in C^c$

$\exists x \in B, x \not\in A$

$\exists x \in C^c, x \not\in A, \Rightarrow A \subset C^c$

$\forall x, x \in A\Rightarrow x \not\in C$

$A \cap C = \emptyset$

Prove: If $A \subseteq (B \cup C)$, $B \subseteq D$ and $C \subseteq E$ then $A \subseteq (D \cup E)$

$\begin{array}{ll} & A \subseteq (B \cup C)\\ & B \subseteq D\\ & C \subseteq E\\ \therefore & A \subseteq (D \cup E) \end{array}$

Prove: $((A \cap B) = A) \Leftrightarrow A \in \mathcal{P}(B)$

Countability

Recall:$\vert A\vert$

$\vert \{1,2,3,4,5,6\}\vert$

$\vert \{\}\vert$

$\vert \{\{1,2,3\}\{1,2\}\}\vert$

$\vert \mathbb{Z}\vert$

$\vert \mathbb{R}\vert$

$\vert \mathbb{Z}^{evens}\vert$

Consider $\mathbb{Z}^+$

$1,2,3,4,5,6,...$

Consider $\mathbb{Z}^+$

$1,2,3,4,5,6,...$

$\mathbb{R}?$

$0,0.1,0.5,1,...???$

A set is countable if we can match the elements to a 1-1 corespondence to $\mathbb{N}$.

ie. We can find an injective function $f:\mathbb{N} \mapsto S$

A set is countable if we can match the elements to a 1-1 corespondence to $\mathbb{N}^+$.

ie. We can find an injective function $f:\mathbb{N}^+ \mapsto S$

A set is uncountable if it's not countable

{positive evens} are countable

$\begin{array}{llc} \mathbb{N}^+& &\text{Positive Evens}\\\hline 1 & \mapsto & 2\\ & & \\ & & \\ & & \\ \end{array}$

{positive evens} are countable

$\begin{array}{llc} \mathbb{N}^+& &\text{Positive Evens}\\\hline 1 & \mapsto & 2\\ 2 & \mapsto & 4\\ & & \\ & & \\ \end{array}$

{positive evens} are countable

$\begin{array}{llc} \mathbb{N}^+& &\text{Positive Evens}\\\hline 1 & \mapsto & 2\\ 2 & \mapsto & 4\\ 3 & \mapsto & 6\\ & & \\ \end{array}$

{positive evens} are countable

$\begin{array}{llc} \mathbb{N}^+& &\text{Positive Evens}\\\hline 1 & \mapsto & 2\\ 2 & \mapsto & 4\\ 3 & \mapsto & 6\\ 4 & \mapsto & 8\\ \end{array}$

index $\times$ 2

$\mathbb{Z}$ are countable

$\begin{array}{llc} \mathbb{N}^+& &\mathbb{Z}\\\hline 1 & \mapsto & 0\\ & & \\ & & \\ & & \\ & & \\ \end{array}$

$\mathbb{Z}$ are countable

$\begin{array}{llc} \mathbb{N}^+& &\mathbb{Z}\\\hline 1 & \mapsto & 0\\ 2 & \mapsto & 1\\ & & \\ & & \\ & & \\ \end{array}$

$\mathbb{Z}$ are countable

$\begin{array}{llc} \mathbb{N}^+& &\mathbb{Z}\\\hline 1 & \mapsto & 0\\ 2 & \mapsto & 1\\ 3 & \mapsto& -1\\ & & \\ & & \\ \end{array}$

$\mathbb{Z}$ are countable

$\begin{array}{llc} \mathbb{N}^+& &\mathbb{Z}\\\hline 1 & \mapsto & 0\\ 2 & \mapsto & 1\\ 3 & \mapsto & -1\\ 4 & \mapsto & 2\\ & & \\ \end{array}$

$\mathbb{Z}$ are countable

$\begin{array}{llc} \mathbb{N}^+& &\mathbb{Z}\\\hline 1 & \mapsto & 0\\ 2 & \mapsto & 1\\ 3 & \mapsto & -1\\ 4 & \mapsto & 2\\ 5 & \mapsto & -2\\ \end{array}$

even index: $\frac{i}{2}$, odd index: $\frac{i-1}{-2}$

$\mathbb{Q}^+$ are countable

$x \in \mathbb{Q} \Rightarrow \exists a,b \in \mathbb{Z}, x = \frac{a}{b} \land b \neq 0$

Numerator: $\{1,2,3,4,5,...\}$

Denominator: $\{1,2,3,4,5,...\}$

$\begin{array}{c|ccccc} n/d & 1 & 2 & 3 & 4 & \ldots\\\hline 1 & & & & &\\ 2 & & & & &\\ 3 & & & & &\\ 4 & & & & &\\ \vdots & & & & &\\ \end{array}$

$\begin{array}{c|ccccc} n/d & 1 & 2 & 3 & 4 & \ldots\\\hline 1 & 1/1& 2/1& 3/1&4/1 & \ldots\\ 2 & 1/2& 2/2& 3/2& 4/2& \ldots\\ 3 & 1/3& 2/3& 3/3& 4/3&\ldots \\ 4 & 1/4& 2/4& 3/4& 4/4&\ldots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ \end{array}$

visited = []

visited = [1/1]
start at top left

visited = [1/1,1/2]
start at top left               
Go down one

visited = [1/1,1/2,2/1]
start at top left               
Go down one
Go up-right diagonal

visited = [1/1,1/2,2/1,3/1]
start at top left               
Go down one
Go up-right diagonal
Go right one

visited = [1/1,1/2,2/1,3/1,2/2,1/3]
start at top left               
Go down one
Go up-right diagonal
Go right one
Go down-left diagonal

visited = [1/1,1/2,2/1,3/1,2/2,1/3,
start at top left               
Go down one
Go up-right diagonal
Go right one
Go down-left diagonal
Repeat from line 3

visited = [1/1,1/2,2/1,3/1,2/2,1/3,
start at top left               
Go down one
Go up-right diagonal
Go right one
Go down-left diagonal
Repeat from line 3

visited = [1/1,1/2,2/1,3/1,2/2,1/3,
start at top left               
Go down one
Go up-right diagonal
Go right one
Go down-left diagonal
Repeat from line 3

Countability

What about $\mathbb{R}$

Consider [0,1]

$\begin{array}{llc} \mathbb{N}& &\mathbb{R}\\\hline 1 & \mapsto & 0.a_1a_2a_3a_4a_5\ldots\\ 2 & \mapsto & 0.b_1b_2b_3b_4b_5\ldots\\ 3 & \mapsto & 0.c_1c_2c_3c_4c_5\ldots\\ 4 & \mapsto & 0.d_1d_2d_3d_4d_5\ldots\\ \vdots& \vdots&\vdots \\ \end{array}$

New number $x$: $0.a_1b_2c_3d_4\ldots$

Add 1 to each digit, or loop to 0

differs the $k^{th}$ number by the $k^{th}$ digit

No matter how close the 1st and 2nd numbers are, I can find a value between them

$\frac{a+b}{2}$