# Help understanding a proof

1. Apr 10, 2014

### trash

I'm reading about countable and uncountable sets, I found the following statement: "The set of the functions from $\mathbb{Z}$ to $[0,1]$ is uncountable" with the following proof: "To see that, suppose the set countable having the list $\{f_1,f_2,\dots\}$ and define $f(x) = f_n(1/n)$ if $x=1/n$ and $f(x)=0$ if $x\neq 1/n$ for any n".

Could someone explain this proof further?. It seems to me that he is trying to construct a function that is different from every $f_i$, but I don't see how the new function is necessarily different from every $f_i$, can't we have the possibilty that all the functions have the same values at $1/n$ for every $n$ but they are different for other values?.

Last edited: Apr 10, 2014
2. Apr 10, 2014

### jbunniii

The proof doesn't make any sense. The function $f$ being constructed does not even appear to be defined on the correct domain, $\mathbb{Z}$.

Try the following instead. Suppose that we have an enumeration $\{f_1, f_2, \ldots\}$. We now construct $f : \mathbb{Z} \rightarrow [0,1]$ that is not in the list.

First, let $g : \mathbb{Z} \rightarrow \mathbb{Z}^{+}$ be any bijection. For each $n \in \mathbb{Z}$, set $f(n)$ equal to any value in $[0,1]$ other than $f_{g(n)}(n)$. If we want to be concrete (to avoid invoking the axiom of choice), put
$$f(n) = \begin{cases} 1 & \text{ if }f_{g(n)}(n) = 0 \\ 0 & \text{ otherwise} \\ \end{cases}$$
Now given any $m \in \mathbb{Z}^+$, we see that $f$ cannot equal $f_m$ because it differs from $f_m$ at the point $n = g^{-1}(m)$.

Last edited: Apr 10, 2014
3. Apr 10, 2014

### trash

Thanks a lot.
But that was my mistake. The proposition is about the set of functions from $[0,1]$ to $\mathbb{Z}$

4. Apr 10, 2014

### jbunniii

OK, the proof is still incorrect as stated. It would be OK if it said

Perhaps for concreteness, something like

5. Apr 10, 2014

### PeroK

Assuming you know [0, 1] is uncountable:

$$f_a(a) = 1 \ and \ f_a(x) = 0 \ otherwise$$
Defines an uncountable set of functions: one for each a in [0, 1]

6. Apr 10, 2014

### trash

Thanks a lot, now that makes sense.
Still, for the last example if $f_1(1)=1$ wouldn't be $f(1)=2$ which exceeds the domain?, maybe something like $f(x)=[f_n(1/n)]/2$ would work better?.

7. Apr 10, 2014

### jbunniii

I thought you wanted to map from $[0,1]$ to $\mathbb{Z}$. So the value of $f(x)$ can be as big as we like, but you can't divide by 2, because the result may not be an integer.

8. Apr 10, 2014

### trash

Yes, you're right. I'm sorry, I read it too fast and didn't think it through