Tao Real Analysis-8th chapter-Infinite Set

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This chapter is by far the most laborious, the second laborious is the 3rd chapter set theory
8.3/8.4/8.5 back exercises did not do, feel more isolated, if you need to go back to look at it 8.1 The number of possible

If the set is limited, then the true subset cannot have the same cardinality and contradiction according to 3.6.14 (c).
An infinite set necessarily contains a subset of this subset to n double-shot. Assumed to be a:f (0), F (1),... A:f (0), F (1),..., then b:f (1), F (2),... B:f (1), F (2),... is a true subset, and there is a double shot: A->b F (n+1).

Method 1
Using exercise 4.4.2 The infinite descent principle is the simplest because the set cannot have no minimum value.
Method 2
Induction, can be to M induction, when M is 0, then obviously 0 is, inductive hypothesis for M, there is a minimum of n, then for m+1, if n belongs to the set, then is N, if n does not belong to the set, then is the m+1, prove complete.
Method 3
Using the principle of minimum upper bound, theorem 5.5.9,
Consider the collection {x:x≤m, m∈x} \{x:x\le m,\ m\in x\},0 belong to this collection, this set is not empty, so there must be a minimum upper bound n, the minimum upper bound must be natural number, otherwise N+εn+\varepsilon is also the upper bound. This natural number must belong to set X, otherwise n+1 is also the upper bound.

Infinite set (?), disprove, if finite set, then set X for this finite set plus all the AM a_m less than n, is a finite set
(?) An a_n increment sequence
So obviously you have to prove it.
(?) Previous (?) The direct result
(?) An∈x A_n\in X According to the definition of an A_n

8.1.4 no hint.
A is a subset of N (possibly a true subset). When F is limited to a, then for any m≠m m\ne m, Am≠an a_m\ne A_n, otherwise assuming m<n m, then the definition of N and a contradiction, should not appear in a, prove the single shot, but also need to prove the map. For any element f (n) in F (n), if f (n) is not equal to F (m) M<n F (m) m, then N is in a, f (n) also appears in F (A), otherwise there is m<n,f (m) =f (n) m,m in A and f (n) =f (m).

X can be counted, then there is a double-shot g (n) =x of N to X, then f∘g F\CIRC g is a function, which is done according to the 8.1.8 proof.

A maximum number, if a is limited, then obviously. If A is infinite, then the number, the existence of a to n double-shot, double-shot is a single shot.
If a is limited, then the certificate is automatically completed. If A is infinite, then f (a) is a subset of N, according to 8.1.6,f (a) can be counted and mapped, the proof is complete.

The feeling hint has been proved, because H (N) is a function, and the inference 8.1.9 can be used.

F (m,n): = (n,m) for double-shot


Diagonal method
0 1 2 3 4 5
3 8.2 Summing on an infinite set

Proof of theorem 8.2.2
The proof process mainly uses the Proposition 6.3.8, from limited to infinite.
The second equation can be introduced from Proposition 7.4.3 because of the wireless and absolute convergence, and the corollary 8.1.13 proves that NXN can be counted.
Why 1
Why the finite set and less than L, according to the hint, G (X) bounded bound, and L is the upper bound.
Why is that enough?
Because that's the definition of real numbers. Exercise

Lemma 8.2.3
If X is limited, then two propositions are set up naturally.
Consider the infinite number of x, so if f (x) is absolutely convergent, then it must converge to L. Similar theorem 8.2.2 proof, there is ∑x∈a|f (x) |≤l \sum _{x\in a}|f (x) |\le L, according to the Proposition 6.3.8 proof is completed.
If Sup∑x∈a|f (x) |<∞sup{\sum_{x\in a}|f (x) |}, then necessarily less than a positive real number l, according to the Proposition 6.3.8, the proof is completed.


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