Episode 06 – Set-builder notation European section – Season 2

Episode 06 – Set-builder notation

European section – Season 2

{2k+1:k ∈Z}

{2k+1:k ∈Z}

−7,−3, 5, 17

{2k+1:k ∈Z}

−7,−3, 5, 17 {n:n∈Z∧12

n ∈Z}

{2k+1:k ∈Z}

−7,−3, 5, 17

{n:n∈Z∧12 n ∈Z} 3, 6,−12, 1

{2k+1:k ∈Z}

−7,−3, 5, 17

{n:n∈Z∧12 n ∈Z} 3, 6,−12, 1

{n²:n∈N∩[0,5]}

{2k+1:k ∈Z}

−7,−3, 5, 17

{n:n∈Z∧12 n ∈Z} 3, 6,−12, 1

{n²:n∈N∩[0,5]} 0, 1, 9, 16

{2k+1:k ∈Z}

−7,−3, 5, 17

{n:n∈Z∧12 n ∈Z} 3, 6,−12, 1

{n²:n∈N∩[0,5]} 0, 1, 9, 16

{3k:k ∈N}

{2k+1:k ∈Z}

−7,−3, 5, 17

{n:n∈Z∧12 n ∈Z} 3, 6,−12, 1

{n²:n∈N∩[0,5]} 0, 1, 9, 16

{3k:k ∈N} 6, 21, 336, 4272

{x ∈R:3x ∈Z}

{x ∈R:3x ∈Z}

7

3,−17,−⁵²₃, 0

{x ∈R:3x ∈Z}

7

3,−17,−⁵²₃, 0

{a+bi:a∈R∧b∈R}

{x ∈R:3x ∈Z}

7

3,−17,−⁵²₃, 0

{a+bi:a∈R∧b∈R} 3+5i,√

2−i, 17,−1+πi

{x ∈R:3x ∈Z}

7

3,−17,−⁵²₃, 0

{a+bi:a∈R∧b∈R} 3+5i,√

2−i, 17,−1+πi {7k+1:k ∈Z}

{x ∈R:3x ∈Z}

7

3,−17,−⁵²₃, 0

{a+bi:a∈R∧b∈R} 3+5i,√

2−i, 17,−1+πi {7k+1:k ∈Z}

−20, 1, 50, 778

{x ∈R:3x ∈Z}

7

3,−17,−⁵²₃, 0

{a+bi:a∈R∧b∈R} 3+5i,√

2−i, 17,−1+πi

{7k+1:k ∈Z}

−20, 1, 50, 778

{n∈N:∃m∈Z,m²=n}

{x ∈R:3x ∈Z}

7

3,−17,−⁵²₃, 0

{a+bi:a∈R∧b∈R} 3+5i,√

2−i, 17,−1+πi

{7k+1:k ∈Z}

−20, 1, 50, 778

{n∈N:∃m∈Z,m²=n} 0, 9, 81, 196

{x ∈R:x²=x}

{x ∈R:x²=x} 0, 1

{x ∈R:x²=x} 0, 1

{x ∈R:∃p,q∈Z,q6=0∧xq=p}

{x ∈R:x²=x} 0, 1

{x ∈R:∃p,q∈Z,q6=0∧xq=p}

12

7,−3², 42, ⁻₂²⁸

{x ∈R:x²=x} 0, 1

{x ∈R:∃p,q∈Z,q6=0∧xq=p}

12

7,−3², 42, ⁻₂²⁸

{r ∈R:∃k ∈Z,r²=k}

{x ∈R:x²=x} 0, 1

{x ∈R:∃p,q∈Z,q6=0∧xq=p}

12

7,−3², 42, ⁻₂²⁸

{r ∈R:∃k ∈Z,r²=k} 5,√

2,−13,√ 13

{x ∈R:x²=x} 0, 1

{x ∈R:∃p,q∈Z,q6=0∧xq=p}

12

7,−3², 42, ⁻₂²⁸

{r ∈R:∃k ∈Z,r²=k} 5,√

2,−13,√ 13

{n∈Z:|n| ≤5}

{x ∈R:x²=x} 0, 1

{x ∈R:∃p,q∈Z,q6=0∧xq=p}

12

7,−3², 42, ⁻₂²⁸

{r ∈R:∃k ∈Z,r²=k} 5,√

2,−13,√ 13

{n∈Z:|n| ≤5}

−4,−2, 0, 3

{x ∈R:|x| ≤5}

{x ∈R:|x| ≤5}

√2,−π, 2.256,−4.99

{x ∈R:|x| ≤5}

√2,−π, 2.256,−4.99 {x ∈R:∀y ∈N,x ≤y}

{x ∈R:|x| ≤5}

√2,−π, 2.256,−4.99 {x ∈R:∀y ∈N,x ≤y}

−5,−2π,−⁴7,−0.0001

{x ∈R:|x| ≤5}

√2,−π, 2.256,−4.99

{x ∈R:∀y ∈N,x ≤y}

−5,−2π,−⁴7,−0.0001

{(x,y) :x ∈N∧y ∈N∧ x+y =7}

{x ∈R:|x| ≤5}

√2,−π, 2.256,−4.99

{x ∈R:∀y ∈N,x ≤y}

−5,−2π,−⁴7,−0.0001

{(x,y) :x ∈N∧y ∈N∧ x+y =7} (2,5);(1,6);(4,3);(3,4)

{x ∈R:|x| ≤5}

√2,−π, 2.256,−4.99

{x ∈R:∀y ∈N,x ≤y}

−5,−2π,−⁴7,−0.0001

{(x,y) :x ∈N∧y ∈N∧ x+y =7} (2,5);(1,6);(4,3);(3,4)

{x ∈R:x ≤ −3∨ x ≥5}

{x ∈R:|x| ≤5}

√2,−π, 2.256,−4.99

{x ∈R:∀y ∈N,x ≤y}

−5,−2π,−⁴7,−0.0001

{(x,y) :x ∈N∧y ∈N∧ x+y =7} (2,5);(1,6);(4,3);(3,4)

{x ∈R:x ≤ −3∨ x ≥5}

−5,−√

19, 17.25, 5

{n∈Z:∀m∈Z,n>m}

{n∈Z:∀m∈Z,n>m}

∅

{n∈Z:∀m∈Z,n>m}

∅

{n∈N:∄m∈N− {1,n}, n m ∈N}

{n∈Z:∀m∈Z,n>m}

∅

{n∈N:∄m∈N− {1,n}, n m ∈N} 17, 5, 43, 199

{n∈Z:∀m∈Z,n>m}

∅

{n∈N:∄m∈N− {1,n}, n m ∈N} 17, 5, 43, 199

{(x,y) :x ∈N∧y ∈N∧ xy ≤6}

{n∈Z:∀m∈Z,n>m}

∅

{n∈N:∄m∈N− {1,n}, n m ∈N} 17, 5, 43, 199

{(x,y) :x ∈N∧y ∈N∧ xy ≤6} (2,2);(1,5);(3,1);(1,1)

{n∈Z:∀m∈Z,n>m}

∅

{n∈N:∄m∈N− {1,n}, n m ∈N} 17, 5, 43, 199

{(x,y) :x ∈N∧y ∈N∧ xy ≤6} (2,2);(1,5);(3,1);(1,1)

{(x,y) :x ∈N∧y ∈N∧ y =2x}

{n∈Z:∀m∈Z,n>m}

∅

{n∈N:∄m∈N− {1,n}, n m ∈N} 17, 5, 43, 199

{(x,y) :x ∈N∧y ∈N∧ xy ≤6} (2,2);(1,5);(3,1);(1,1)

{(x,y) :x ∈N∧y ∈N∧ y =2x} (3,6);(12,24);(5,10);(131,262)

Set-builder notation

{. . .}: A set.

:or|: “such that”.

Sets of numbers

N : The whole numbers (natural numbers and zero).

Z : The integers.

D : The decimal numbers.

Q : The rational numbers R : The real numbers.

R^⋆: The non-zero real numbers.

R⁺: The positive real numbers.

R⁻: The negative real numbers.

Set, subsets and elements

∩: The intersection of two sets.

∪: The union of two sets.

∈: “is an element of the set”

⊂: “is a subset of the set”

Logical operators

¬: The negation of a proposition, “not”.

∧: The conjunction of two propositions, “and”.

∨: The disjunction of two propositions, “or”.

∀: “For all”

∃: “There exists”

∄: “There doesn’t exist”

Russell’s paradox

Let R be the set of all sets that don’t contain themselves, that is

Russell’s paradox

Let R be the set of all sets that don’t contain themselves, that is

R={X :X is a set and X 6∈X}.

Russell’s paradox

Let R be the set of all sets that don’t contain themselves, that is

R={X :X is a set and X 6∈X}.

Then R either is or is not an element of itself.

Russell’s paradox

Let R be the set of all sets that don’t contain themselves, that is

R={X :X is a set and X 6∈X}.

Then R either is or is not an element of itself.

If R is not an element of itself,

Russell’s paradox

Let R be the set of all sets that don’t contain themselves, that is

R={X :X is a set and X 6∈X}.

Then R either is or is not an element of itself.

If R is not an element of itself, then R is a set that doesn’t contain itself,

Russell’s paradox

Let R be the set of all sets that don’t contain themselves, that is

R={X :X is a set and X 6∈X}.

Then R either is or is not an element of itself.

If R is not an element of itself, then R is a set that doesn’t contain itself, and so R is an element of R,

Russell’s paradox

Let R be the set of all sets that don’t contain themselves, that is

R={X :X is a set and X 6∈X}.

Then R either is or is not an element of itself.

If R is not an element of itself, then R is a set that doesn’t contain itself, and so R is an element of R, which is precisely the

contrary of what we supposed.

Russell’s paradox

Let R be the set of all sets that don’t contain themselves, that is

R={X :X is a set and X 6∈X}.

Then R either is or is not an element of itself.

If R is not an element of itself, then R is a set that doesn’t contain itself, and so R is an element of R, which is precisely the

contrary of what we supposed.

Conversely, if R is an element of itself,

Russell’s paradox

Let R be the set of all sets that don’t contain themselves, that is

R={X :X is a set and X 6∈X}.

Then R either is or is not an element of itself.

If R is not an element of itself, then R is a set that doesn’t contain itself, and so R is an element of R, which is precisely the

contrary of what we supposed.

Conversely, if R is an element of itself, then R does not contain itself,

Russell’s paradox

Let R be the set of all sets that don’t contain themselves, that is

R={X :X is a set and X 6∈X}.

Then R either is or is not an element of itself.

If R is not an element of itself, then R is a set that doesn’t contain itself, and so R is an element of R, which is precisely the

contrary of what we supposed.

Conversely, if R is an element of itself, then R does not contain itself, so we also get to a contradiction.

Russell’s paradox

Let R be the set of all sets that don’t contain themselves, that is

R={X :X is a set and X 6∈X}.

Then R either is or is not an element of itself.

If R is not an element of itself, then R is a set that doesn’t contain itself, and so R is an element of R, which is precisely the

contrary of what we supposed.

Conversely, if R is an element of itself, then R does not contain itself, so we also get to a contradiction.

Russell’s paradox

Let R be the set of all sets that don’t contain themselves, that is

R={X :X is a set and X 6∈X}.

Then R either is or is not an element of itself.

If R is not an element of itself, then R is a set that doesn’t contain itself, and so R is an element of R, which is precisely the

contrary of what we supposed.

Conversely, if R is an element of itself, then R does not contain itself, so we also get to a contradiction.

So R cannot be a set, or our definition of a set is not satisfactory.