Question about countable sets from Rudin's Analysis book

Question

I am starting to read Baby Rudin's book and there is a section devoted to countable sets in which he said that a set $ A $ is countable if there exists a $ 1 $-$ 1 $ correspondence between $ A $ and $ \mathbb{Z}^{+}. $ So my question is suppose that I have an infinite set $ A, $ I pick one element $ s \in A $ and map $ 1 $ to $ s, $ then I pick another element $ t \in A $ and map $ 2 $ to $ t, $ and so on, then I will get a $ 1 $-$ 1 $ correspondence between $ A $ and $ \mathbb{Z}^{+} $ eventually, meaning $ A $ is countable. I know this is incorrect since it will mean that every set is countable, but I don't seem to see the idea in the concept of countable sets, so can someone help me clarifying this?

Rudin also proved this theorem which has been asked here. The theorem said that "the union of countably many countable sets is countable." My second question is where in the proof did he use the condition that each set $ E_{i} $ is countable? Also, suppose that I change the statement into "the union of many countable sets is countable," then this statement is no longer true, but where in the proof that he used this condition that it has to be the "the union of $ \textbf{countably} $ many countable sets?"

Answer 1

"I pick one element $s \in A$ and map $1$ to $s$, then I pick another element $t \in A$ and map $2$ to $t$, and so on": not quite. You see, and so on need not apply at all! Have you seen any proof of a set being "uncountable"? I'll give an example:

Let $\{\{a_n\}\}$ be the set of all sequences $\{ a_n\}$ with $a_i \in \{ 0,1\} $ for all $i$. Then $\{\{a_n\}\}$ is uncountable.

Proof: Let us try to do what you said: "I pick one element $s \in A$ and map $1$ to $s$, then I pick another element $t \in A$ and map $2$ to $t$, and so on". Now, we will write the $s$ you picked, then the $t$, then so on, one below the other like this: $$ 1\quad0 \quad0 \quad0 \quad1 \quad0\quad1\quad\ldots $$ $$ 0\quad1 \quad0 \quad1 \quad0 \quad1\quad1\quad\ldots $$ $$ 1\quad0 \quad0 \quad0 \quad0 \quad0\quad1\quad\ldots $$ $$ 0\quad1 \quad1 \quad1 \quad0 \quad1\quad1\quad\ldots $$ in the order that they were numbered. Now, if $\{\{a_n\}\}$ were countable, then all elements of $\{\{a_n\}\}$ should be on some line of this list. I will explicitly generate an element that is not in any line.

For doing this, take the diagonal in your grid above. It reads $1\quad 1 \quad 0 \quad 1 \quad \ldots$ Now, what we are goin to do is flip every element of this diagonal, so that $1$ becomes $0$ and $0$ becomes $1$. Thus, we will get a new sequence $0\quad0\quad1\quad0\quad\ldots$

Claim: This sequence does not occur on any line!

Proof: Because of the flipping, on the nth line in the nth position, the value on the grid is different from the nth value of the flipped sequence. Hence the sequence is different from all the lines.

But this is a new sequence! Hence the contradiction.

The problem with and so on, is that you may leave out elements like the one sequence left out above.

By the way, another way to look at "countable" is to say : I can list these items. To give an example, in the proof that union of countably many sets in countable, we are able to list down the elements of each $E_i$, right? This is possible only if they are countable, otherwise, if one of the $E_i$ is uncountable, you cannot list it down by mapping it to the integers, because by definition of uncountability, that is impossible.

Answer 2

Here is the standard contradiction to your argument: suppose you have a bijection $f:Z^+\rightarrow P(Z^+)$. Consider $P(Z^+)$ the set whose elements are subsets of $Z^+$. Let $S$ be the subset of $Z^+$ defined by $S=\{s\in Z^+$ such that $s$ is not an element of $f(s)\}$. Suppose that $S=f(n), n\in Z^+$, if $n\in S$ contradiction since we have assumed that $s$ is not in $f(s)$ for $s\in S$. If $n$ is not in $S$, then $n\in S$ by definition since $S=\{s\in Z^+$ $s$ is not in $f(s)\}$. Contradiction.

Answer 3

If $A$ isn't countable, your procedure won't define a function $A \to Z^+$, merely a function from a countable subset of $A$ to $Z^+$. Think about it: the only thing that guarantees that your procedure will eventually reach every element of $A$ is precisely the fact that $A$ is countable.

To answer your second question, we need a countable number of $E_i$ to list the elements in an infinite array.

Answer 4

"I am starting to read Baby Rudin's book and there is a section devoted to countable sets in which he said that a set A is countable if there exists a 1-1 correspondence between A and Z+. So my question is suppose that I have an infinite set A, I pick one element s∈A and map 1 to s, then I pick another element t∈A and map to t, and so on, then I will get a 1-1 correspondence between A and Z+ eventually, meaning A A is countable. I know this is incorrect since it will mean that every set is countable, but I don't seem to see the idea in the concept of countable sets, so can someone help me clarifying this?" Your error is in assuming that you can pick one item after another and so get all members of the set. That is equivalent to the set being "countable".

Suppose your set was the set of all real numbers from 0 to 1, inclusive. You can choose the first number as, say, 0, the next number as 1, the next number as 1/2 the next number 1/3, then 1/4, 1/5, etc. but you would never get all real numbers that way.

Answer 5

As you have considered f:N--A is a function and let s€A and we send 1 to s;let's consider another term t and send 2 to t;and so on....but this doesn't imply that it's a surjective function.You are considering at a time one element(s,t...etc) so you are taking that element as a image for some natural no.But this doesn't imply that whatever element is chosen from A has a preimage in N.This case can be considered when we take an infinite subset of A.If we take an infinite subset if A then by setting preimage of s,t,...from N can't gurantee the mapping covers the whole infinite set.Lets consider an example....F:N--R is a real valued function.Now let T={x:x is not an algebraic number}.Then we can send 1 to e,2 to π,.....but this steps doesn't imply that the whole set is covered by F.That is we are not able to found a surjective function from N to T.By uncountably proof of R,we see that actually we can't do this.