131A Week 6 Discussion Alan Zhou Subsequences Countability 131A Week 6 Discussion subsequences and countability Alan Zhou May 5, 2020
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.