131a week 6 discussion - subsequences and countabilityazhou/teaching/20s/131a-week...alan zhou...

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

131A Week 6 Discussionsubsequences and countability

Alan Zhou

May 5, 2020

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Uses of subsequences

I One way to show a bounded sequence diverges is to findtwo subsequences which converge to different limits.I On the other hand, if a sequence converges, every

subsequence converges to the same limit.

I If E is a closed and bounded subset of R (i.e. acompact subset of R) and we want to show that thereis a point x ∈ E with some property P, one way to dothis is to define a sequence (xn) in E whose points “getcloser and closer to having property P.” We cannotthen say “the limit of this sequence has property P,”one of the reasons being that the limit may notexist. However, by the Bolzano-Weierstrass theorem,(xn) has a subsequence which converges, and the limitof this subsequence (hopefully) has property P.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Uses of subsequences

I One way to show a bounded sequence diverges is to findtwo subsequences which converge to different limits.

I On the other hand, if a sequence converges, everysubsequence converges to the same limit.

I If E is a closed and bounded subset of R (i.e. acompact subset of R) and we want to show that thereis a point x ∈ E with some property P, one way to dothis is to define a sequence (xn) in E whose points “getcloser and closer to having property P.” We cannotthen say “the limit of this sequence has property P,”one of the reasons being that the limit may notexist. However, by the Bolzano-Weierstrass theorem,(xn) has a subsequence which converges, and the limitof this subsequence (hopefully) has property P.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Uses of subsequences

I One way to show a bounded sequence diverges is to findtwo subsequences which converge to different limits.I On the other hand, if a sequence converges, every

subsequence converges to the same limit.

I If E is a closed and bounded subset of R (i.e. acompact subset of R) and we want to show that thereis a point x ∈ E with some property P, one way to dothis is to define a sequence (xn) in E whose points “getcloser and closer to having property P.” We cannotthen say “the limit of this sequence has property P,”one of the reasons being that the limit may notexist. However, by the Bolzano-Weierstrass theorem,(xn) has a subsequence which converges, and the limitof this subsequence (hopefully) has property P.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Uses of subsequences

I One way to show a bounded sequence diverges is to findtwo subsequences which converge to different limits.I On the other hand, if a sequence converges, every

subsequence converges to the same limit.

I If E is a closed and bounded subset of R (i.e. acompact subset of R) and we want to show that thereis a point x ∈ E with some property P, one way to dothis is to define a sequence (xn) in E whose points “getcloser and closer to having property P.”

We cannotthen say “the limit of this sequence has property P,”one of the reasons being that the limit may notexist. However, by the Bolzano-Weierstrass theorem,(xn) has a subsequence which converges, and the limitof this subsequence (hopefully) has property P.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Uses of subsequences

I One way to show a bounded sequence diverges is to findtwo subsequences which converge to different limits.I On the other hand, if a sequence converges, every

subsequence converges to the same limit.

I If E is a closed and bounded subset of R (i.e. acompact subset of R) and we want to show that thereis a point x ∈ E with some property P, one way to dothis is to define a sequence (xn) in E whose points “getcloser and closer to having property P.” We cannotthen say “the limit of this sequence has property P,”one of the reasons being that the limit may notexist.

However, by the Bolzano-Weierstrass theorem,(xn) has a subsequence which converges, and the limitof this subsequence (hopefully) has property P.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Uses of subsequences

I One way to show a bounded sequence diverges is to findtwo subsequences which converge to different limits.I On the other hand, if a sequence converges, every

subsequence converges to the same limit.

I If E is a closed and bounded subset of R (i.e. acompact subset of R) and we want to show that thereis a point x ∈ E with some property P, one way to dothis is to define a sequence (xn) in E whose points “getcloser and closer to having property P.” We cannotthen say “the limit of this sequence has property P,”one of the reasons being that the limit may notexist. However, by the Bolzano-Weierstrass theorem,(xn) has a subsequence which converges, and the limitof this subsequence (hopefully) has property P.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Subsequences of subsequences

TheoremLet (xn) be a sequence and ` ∈ R. If every subsequence hasa further subsequence which converges to `, then xn → `.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Subsequences of subsequences

TheoremLet (xn) be a sequence and ` ∈ R. If every subsequence hasa further subsequence which converges to `, then xn → `.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Countable and uncountable sets

I By countable, we mean “finite or countably infinite (inbijection with N).”

I Some countably infinite sets:I N (by definition)I Z (proof later)I Q (proof later)I the set of all intervals [a, b] with a, b ∈ QI the set of all finite subsets of N

I Some uncountable sets:I R (proof later)I the set of all irrational numbersI the set of all subsets of N (Cantor diagonalisation)

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Countable and uncountable sets

I By countable, we mean “finite or countably infinite (inbijection with N).”

I Some countably infinite sets:

I N (by definition)I Z (proof later)I Q (proof later)I the set of all intervals [a, b] with a, b ∈ QI the set of all finite subsets of N

I Some uncountable sets:I R (proof later)I the set of all irrational numbersI the set of all subsets of N (Cantor diagonalisation)

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Countable and uncountable sets

I By countable, we mean “finite or countably infinite (inbijection with N).”

I Some countably infinite sets:I N (by definition)

I Z (proof later)I Q (proof later)I the set of all intervals [a, b] with a, b ∈ QI the set of all finite subsets of N

I Some uncountable sets:I R (proof later)I the set of all irrational numbersI the set of all subsets of N (Cantor diagonalisation)

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Countable and uncountable sets

I By countable, we mean “finite or countably infinite (inbijection with N).”

I Some countably infinite sets:I N (by definition)I Z (proof later)

I Q (proof later)I the set of all intervals [a, b] with a, b ∈ QI the set of all finite subsets of N

I Some uncountable sets:I R (proof later)I the set of all irrational numbersI the set of all subsets of N (Cantor diagonalisation)

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Countable and uncountable sets

I By countable, we mean “finite or countably infinite (inbijection with N).”

I Some countably infinite sets:I N (by definition)I Z (proof later)I Q (proof later)

I the set of all intervals [a, b] with a, b ∈ QI the set of all finite subsets of N

I Some uncountable sets:I R (proof later)I the set of all irrational numbersI the set of all subsets of N (Cantor diagonalisation)

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Countable and uncountable sets

I By countable, we mean “finite or countably infinite (inbijection with N).”

I Some countably infinite sets:I N (by definition)I Z (proof later)I Q (proof later)I the set of all intervals [a, b] with a, b ∈ Q

I the set of all finite subsets of NI Some uncountable sets:

I R (proof later)I the set of all irrational numbersI the set of all subsets of N (Cantor diagonalisation)

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Countable and uncountable sets

I By countable, we mean “finite or countably infinite (inbijection with N).”

I Some countably infinite sets:I N (by definition)I Z (proof later)I Q (proof later)I the set of all intervals [a, b] with a, b ∈ QI the set of all finite subsets of N

I Some uncountable sets:I R (proof later)I the set of all irrational numbersI the set of all subsets of N (Cantor diagonalisation)

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Countable and uncountable sets

I By countable, we mean “finite or countably infinite (inbijection with N).”

I Some countably infinite sets:I N (by definition)I Z (proof later)I Q (proof later)I the set of all intervals [a, b] with a, b ∈ QI the set of all finite subsets of N

I Some uncountable sets:

I R (proof later)I the set of all irrational numbersI the set of all subsets of N (Cantor diagonalisation)

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Countable and uncountable sets

I By countable, we mean “finite or countably infinite (inbijection with N).”

I Some countably infinite sets:I N (by definition)I Z (proof later)I Q (proof later)I the set of all intervals [a, b] with a, b ∈ QI the set of all finite subsets of N

I Some uncountable sets:I R (proof later)

I the set of all irrational numbersI the set of all subsets of N (Cantor diagonalisation)

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Countable and uncountable sets

I By countable, we mean “finite or countably infinite (inbijection with N).”

I Some countably infinite sets:I N (by definition)I Z (proof later)I Q (proof later)I the set of all intervals [a, b] with a, b ∈ QI the set of all finite subsets of N

I Some uncountable sets:I R (proof later)I the set of all irrational numbers

I the set of all subsets of N (Cantor diagonalisation)

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Countable and uncountable sets

I By countable, we mean “finite or countably infinite (inbijection with N).”

I Some countably infinite sets:I N (by definition)I Z (proof later)I Q (proof later)I the set of all intervals [a, b] with a, b ∈ QI the set of all finite subsets of N

I Some uncountable sets:I R (proof later)I the set of all irrational numbersI the set of all subsets of N (Cantor diagonalisation)

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Getting countable sets

I If f : A→ B is an injective function and B iscountable, then A is countable.I Subsets of countable sets are countable.

I If g : B → A is a surjective function and B iscountable, then A is countable.

I If A and B are countable, then

A×B = {(a, b) | a ∈ A and b ∈ B}

is countable.

I If {Ai}i∈I is a collection of countable sets indexed by acountable set I, then⋃

i∈IAi = {a | a ∈ Ai for some i ∈ I}

is countable.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Getting countable sets

I If f : A→ B is an injective function and B iscountable, then A is countable.

I Subsets of countable sets are countable.

I If g : B → A is a surjective function and B iscountable, then A is countable.

I If A and B are countable, then

A×B = {(a, b) | a ∈ A and b ∈ B}

is countable.

I If {Ai}i∈I is a collection of countable sets indexed by acountable set I, then⋃

i∈IAi = {a | a ∈ Ai for some i ∈ I}

is countable.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Getting countable sets

I If f : A→ B is an injective function and B iscountable, then A is countable.I Subsets of countable sets are countable.

I If g : B → A is a surjective function and B iscountable, then A is countable.

I If A and B are countable, then

A×B = {(a, b) | a ∈ A and b ∈ B}

is countable.

I If {Ai}i∈I is a collection of countable sets indexed by acountable set I, then⋃

i∈IAi = {a | a ∈ Ai for some i ∈ I}

is countable.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Getting countable sets

I If f : A→ B is an injective function and B iscountable, then A is countable.I Subsets of countable sets are countable.

I If g : B → A is a surjective function and B iscountable, then A is countable.

I If A and B are countable, then

A×B = {(a, b) | a ∈ A and b ∈ B}

is countable.

I If {Ai}i∈I is a collection of countable sets indexed by acountable set I, then⋃

i∈IAi = {a | a ∈ Ai for some i ∈ I}

is countable.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Getting countable sets

I If f : A→ B is an injective function and B iscountable, then A is countable.I Subsets of countable sets are countable.

I If g : B → A is a surjective function and B iscountable, then A is countable.

I If A and B are countable, then

A×B = {(a, b) | a ∈ A and b ∈ B}

is countable.

I If {Ai}i∈I is a collection of countable sets indexed by acountable set I, then⋃

i∈IAi = {a | a ∈ Ai for some i ∈ I}

is countable.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Getting countable sets

I If f : A→ B is an injective function and B iscountable, then A is countable.I Subsets of countable sets are countable.

I If g : B → A is a surjective function and B iscountable, then A is countable.

I If A and B are countable, then

A×B = {(a, b) | a ∈ A and b ∈ B}

is countable.

I If {Ai}i∈I is a collection of countable sets indexed by acountable set I, then⋃

i∈IAi = {a | a ∈ Ai for some i ∈ I}

is countable.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Countability of Z and Q

I Z is countable:N 0 1 2 3 4 · · ·Z 0 −1 1 −2 2 · · ·

I Z× N+ is countable.

I Q is countable: The function Z× N+ → Q given by(m,n) 7→ m/n is surjective.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Countability of Z and Q

I Z is countable:N 0 1 2 3 4 · · ·Z 0 −1 1 −2 2 · · ·

I Z× N+ is countable.

I Q is countable: The function Z× N+ → Q given by(m,n) 7→ m/n is surjective.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Countability of Z and Q

I Z is countable:N 0 1 2 3 4 · · ·Z 0 −1 1 −2 2 · · ·

I Z× N+ is countable.

I Q is countable: The function Z× N+ → Q given by(m,n) 7→ m/n is surjective.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Uncountability of R

Our general strategy for proving that R is uncountable willproceed as follows:

I It suffices to show that [0, 1] is uncountable. Supposefor the sake of contradiction that it is countable, and let(xn) be an enumeration of [0, 1]. (An enumeration of aset X is a sequence which contains every element of X.)

I Suppose we have sets I1, I2, . . . ⊂ [0, 1] with theproperty that for every n, we have xn 6∈ In. Thenxn 6∈

⋂∞j=1 Ij for any n.

I If we can choose the sets In so that the intersection isnon-empty, then we will have found an element of [0, 1]which is not equal to any xn, a contradiction. Thus ourgoal is to show that this construction is possible.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Uncountability of R

Our general strategy for proving that R is uncountable willproceed as follows:

I It suffices to show that [0, 1] is uncountable. Supposefor the sake of contradiction that it is countable, and let(xn) be an enumeration of [0, 1]. (An enumeration of aset X is a sequence which contains every element of X.)

I Suppose we have sets I1, I2, . . . ⊂ [0, 1] with theproperty that for every n, we have xn 6∈ In. Thenxn 6∈

⋂∞j=1 Ij for any n.

I If we can choose the sets In so that the intersection isnon-empty, then we will have found an element of [0, 1]which is not equal to any xn, a contradiction. Thus ourgoal is to show that this construction is possible.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Uncountability of R

Our general strategy for proving that R is uncountable willproceed as follows:

I It suffices to show that [0, 1] is uncountable.

Supposefor the sake of contradiction that it is countable, and let(xn) be an enumeration of [0, 1]. (An enumeration of aset X is a sequence which contains every element of X.)

I Suppose we have sets I1, I2, . . . ⊂ [0, 1] with theproperty that for every n, we have xn 6∈ In. Thenxn 6∈

⋂∞j=1 Ij for any n.

I If we can choose the sets In so that the intersection isnon-empty, then we will have found an element of [0, 1]which is not equal to any xn, a contradiction. Thus ourgoal is to show that this construction is possible.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Uncountability of R

Our general strategy for proving that R is uncountable willproceed as follows:

I It suffices to show that [0, 1] is uncountable. Supposefor the sake of contradiction that it is countable, and let(xn) be an enumeration of [0, 1].

(An enumeration of aset X is a sequence which contains every element of X.)

I Suppose we have sets I1, I2, . . . ⊂ [0, 1] with theproperty that for every n, we have xn 6∈ In. Thenxn 6∈

⋂∞j=1 Ij for any n.

I If we can choose the sets In so that the intersection isnon-empty, then we will have found an element of [0, 1]which is not equal to any xn, a contradiction. Thus ourgoal is to show that this construction is possible.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Uncountability of R

Our general strategy for proving that R is uncountable willproceed as follows:

I It suffices to show that [0, 1] is uncountable. Supposefor the sake of contradiction that it is countable, and let(xn) be an enumeration of [0, 1]. (An enumeration of aset X is a sequence which contains every element of X.)

I Suppose we have sets I1, I2, . . . ⊂ [0, 1] with theproperty that for every n, we have xn 6∈ In. Thenxn 6∈

⋂∞j=1 Ij for any n.

I If we can choose the sets In so that the intersection isnon-empty, then we will have found an element of [0, 1]which is not equal to any xn, a contradiction. Thus ourgoal is to show that this construction is possible.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Uncountability of R

Our general strategy for proving that R is uncountable willproceed as follows:

I It suffices to show that [0, 1] is uncountable. Supposefor the sake of contradiction that it is countable, and let(xn) be an enumeration of [0, 1]. (An enumeration of aset X is a sequence which contains every element of X.)

I Suppose we have sets I1, I2, . . . ⊂ [0, 1] with theproperty that for every n, we have xn 6∈ In.

Thenxn 6∈

⋂∞j=1 Ij for any n.

I If we can choose the sets In so that the intersection isnon-empty, then we will have found an element of [0, 1]which is not equal to any xn, a contradiction. Thus ourgoal is to show that this construction is possible.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Uncountability of R

Our general strategy for proving that R is uncountable willproceed as follows:

I It suffices to show that [0, 1] is uncountable. Supposefor the sake of contradiction that it is countable, and let(xn) be an enumeration of [0, 1]. (An enumeration of aset X is a sequence which contains every element of X.)

I Suppose we have sets I1, I2, . . . ⊂ [0, 1] with theproperty that for every n, we have xn 6∈ In. Thenxn 6∈

⋂∞j=1 Ij for any n.

I If we can choose the sets In so that the intersection isnon-empty, then we will have found an element of [0, 1]which is not equal to any xn, a contradiction. Thus ourgoal is to show that this construction is possible.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Uncountability of R

Our general strategy for proving that R is uncountable willproceed as follows:

I It suffices to show that [0, 1] is uncountable. Supposefor the sake of contradiction that it is countable, and let(xn) be an enumeration of [0, 1]. (An enumeration of aset X is a sequence which contains every element of X.)

I Suppose we have sets I1, I2, . . . ⊂ [0, 1] with theproperty that for every n, we have xn 6∈ In. Thenxn 6∈

⋂∞j=1 Ij for any n.

I If we can choose the sets In so that the intersection isnon-empty, then we will have found an element of [0, 1]which is not equal to any xn, a contradiction.

Thus ourgoal is to show that this construction is possible.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Uncountability of R

Our general strategy for proving that R is uncountable willproceed as follows:

I It suffices to show that [0, 1] is uncountable. Supposefor the sake of contradiction that it is countable, and let(xn) be an enumeration of [0, 1]. (An enumeration of aset X is a sequence which contains every element of X.)

I Suppose we have sets I1, I2, . . . ⊂ [0, 1] with theproperty that for every n, we have xn 6∈ In. Thenxn 6∈

⋂∞j=1 Ij for any n.

I If we can choose the sets In so that the intersection isnon-empty, then we will have found an element of [0, 1]which is not equal to any xn, a contradiction. Thus ourgoal is to show that this construction is possible.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Uncountability of RNested Intervals TheoremLet I1 ⊃ I2 ⊃ · · · be nested compact intervals, and writeIn = [an, bn] with an, bn ∈ R and an ≤ bn. If bn − an → 0,then I =

⋂∞j=1 consists of exactly one point.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Uncountability of R

Let (xn) be an enumeration of [0, 1], and define I0 = [0, 1/3]or I0 = [2/3, 1], chosen so that x0 6∈ I0.

Suppose n ∈ N is given and that we have defined a compactinterval In so that xn 6∈ In. We can then choose a compactinterval In+1 ⊂ In so that xn+1 6∈ In+1 and the length ofIn+1 is 1/3 that of In. (How?)

Following the reasoning in the “general strategy” slide, thiscompletes the proof of uncountability of R.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Uncountability of R

Let (xn) be an enumeration of [0, 1], and define I0 = [0, 1/3]or I0 = [2/3, 1], chosen so that x0 6∈ I0.

Suppose n ∈ N is given and that we have defined a compactinterval In so that xn 6∈ In.

We can then choose a compactinterval In+1 ⊂ In so that xn+1 6∈ In+1 and the length ofIn+1 is 1/3 that of In. (How?)

Following the reasoning in the “general strategy” slide, thiscompletes the proof of uncountability of R.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Uncountability of R

Let (xn) be an enumeration of [0, 1], and define I0 = [0, 1/3]or I0 = [2/3, 1], chosen so that x0 6∈ I0.

Suppose n ∈ N is given and that we have defined a compactinterval In so that xn 6∈ In. We can then choose a compactinterval In+1 ⊂ In so that xn+1 6∈ In+1 and the length ofIn+1 is 1/3 that of In.

(How?)

Following the reasoning in the “general strategy” slide, thiscompletes the proof of uncountability of R.

131A Week 6Discussion

Alan Zhou

Subsequences

Countability

Uncountability of R

Let (xn) be an enumeration of [0, 1], and define I0 = [0, 1/3]or I0 = [2/3, 1], chosen so that x0 6∈ I0.

Suppose n ∈ N is given and that we have defined a compactinterval In so that xn 6∈ In. We can then choose a compactinterval In+1 ⊂ In so that xn+1 6∈ In+1 and the length ofIn+1 is 1/3 that of In. (How?)

Following the reasoning in the “general strategy” slide, thiscompletes the proof of uncountability of R.