Theorem 2.13 of Principles of Mathematical Analysis

[math771]

This question is directed toward people who have a copy of the text--in other words, I'm too lazy to retype the whole theorem and proof (sorry).
First, I'm having trouble simply understanding the meaning of certain parts of the theorem itself. I'm taking Bn to include only n-tuples (i.e. no (n-1)-tuples) that include once and only once every ak belonging to A. On this interpretation, what distinguishes the different n-tuples of Bn is only the order of their members. Is this correct?
Second, in the proof, there is the statement: B1=A. In other words, the set {(a1)} is equal to the set {a1}. Is this always true when the ordered pair consists of only one member?
Third, the proof notes that the elements of Bn are of the form (b, a) where b belongs to B(n-1) and a belongs to A (and does not belong to B(n-1), I ask myself). However, this appears to imply that all elements of Bn are 2-tuples, which would contradict my reading of the statement of the theorem (1) that Bn includes only n-tuples. What have I got wrong?
The proof's strangest statement to me is: "For every fixed b, the set of pairs (b, a) is equivalent to A." Note: I understand the phrase "is equivalent to" and do not require an explanation of that concept. I simply fail to see what sort of function makes a 1-1 correspondence between the set of pairs (b, a) and A--or to phrase it a different way, why there is such a function.

Thank you!

I'm new to this website, so please let me know if this is the wrong place for this question.

 

[micromass]

Hi math771!

This question is directed toward people who have a copy of the text--in other words, I'm too lazy to retype the whole theorem and proof (sorry).
First, I'm having trouble simply understanding the meaning of certain parts of the theorem itself. I'm taking Bn to include only n-tuples (i.e. no (n-1)-tuples) that include once and only once every ak belonging to A. On this interpretation, what distinguishes the different n-tuples of Bn is only the order of their members. Is this correct?

Hmm, not sure what you're saying here. We simply have

$$B_n=\{(a_1,...,a_n)~\vert~a_i\in A,\forall i\}$$

Thus B_n consists of n-tuples (indeed, no (n-1)-tuples) in which every element of the tuple is an element of A.
What distinguishes the n-tuples is the order of their elements and which elements they are. For example, if $A=\{0,1\}$, then

$$B_2=\{(0,0),(0,1),(1,0),(1,1)\}$$

Second, in the proof, there is the statement: B1=A. In other words, the set {(a1)} is equal to the set {a1}. Is this always true when the ordered pair consists of only one member?

Equal is a bit too much here, and what Rudin says is not 100% nice and correct. Here he identifies (a) and a. For example, if A={0,1}, then

$$B_1=\{(0),(1)\}$$

So you see that B₁ is not the same as A, but it actually is up to identification. In either case, it is easily seen to have as much elements of A.

Third, the proof notes that the elements of Bn are of the form (b, a) where b belongs to B(n-1) and a belongs to A (and does not belong to B(n-1), I ask myself). However, this appears to imply that all elements of Bn are 2-tuples, which would contradict my reading of the statement of the theorem (1) that Bn includes only n-tuples. What have I got wrong?

Again, Rudin is not 100% nice and correct here. He identifies things of the following kind:

$$((a_1,a_2),a_3)=(a_1,a_2,a_3)$$

Rigourously, the former is a 2-tuple and the latter a 3-tuple, so they aren't equal. But Rudin identifies them anyway. For example, when A={0,1}, then we have the following two sets:

$$\{((0),0),((0),1),((1),0),((1),1)\}~\text{and}~\{(0,0),(0,1),(1,0),(1,1)\}$$

these sets aren't equal, but they are up to identification. In any case, it is easily seen to have the same amount of elements.

The proof's strangest statement to me is: "For every fixed b, the set of pairs (b, a) is equivalent to A." Note: I understand the phrase "is equivalent to" and do not require an explanation of that concept. I simply fail to see what sort of function makes a 1-1 correspondence between the set of pairs (b, a) and A--or to phrase it a different way, why there is such a function.

The bijection is

$$f(a)=(b,a)$$

 

[math771]

Thank you for the detailed response! I would like to verify just one point. Correct me if I'm wrong: Rudin is saying that Bn is countable because a set with the same number of elements that closely resembles Bn--namely the set of pairs (b, a) where b belongs to B(n-1) and a belongs to A (but does not belong to B(n-1), I suspect)--is countable.

 

[micromass]

Thank you for the detailed response! I would like to verify just one point. Correct me if I'm wrong: Rudin is saying that Bn is countable because a set with the same number of elements that closely resembles Bn--namely the set of pairs (b, a) where b belongs to B(n-1) and a belongs to A (but does not belong to B(n-1), I suspect)--is countable.

Exactly!

On another note, I don't quite understand why Rudin doesn't follow the standard notations for this. The set B_n is usually by mathematicians as Aⁿ.