Is the Set of Functions from Z to [0,1] Truly Uncountable?

[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 possibility that all the functions have the same values at 1/n for every n but they are different for other values?.

 

[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 $n = g^{-1}(m)$.

 

[trash]

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

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

 

[jbunniii]

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

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

$f(x) = \text{ any value except } f_n(1/n)$ if $x=1/n$

Perhaps for concreteness, something like

$f(x) = f_n(1/n) + 1$ if $x=1/n$

 

[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]

 

[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?.

 

[jbunniii]

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?.

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.

 

[trash]

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.

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