I have to prove whether or not this set is countable: the functions from $\mathbb Z$ to $\mathbb R$ such that $f(n)=0$ except for a finite number of $n \in \mathbb Z$.
I think this set is uncountable. So, if $f$ belongs to this set, there is an $n_0$ such that $f(n)=0$ $ \forall $ $n>n_0 $. I tried to prove that the set of functions for which this is true for a certain $n_0$ is uncountable but I'm having a hard time finding a function to do so... Thanks


HINT Find an injection from $\Bbb R$ into the set of functions by considering only functions which satisfy $g(n)=0$ for all $n\neq0$.

  • $\begingroup$ does this work: f from the set of functions which obtain a non zero value at n=0 to R. f(g)=g(0)? $\endgroup$ – Rachel Apr 3 '14 at 1:01
  • $\begingroup$ Yes, it works just fine. $\endgroup$ – Asaf Karagila Apr 3 '14 at 1:02
  • $\begingroup$ I'm not sure it's an injection though, because even if the functions are different they can have the same value at 0. $\endgroup$ – Rachel Apr 3 '14 at 1:03
  • $\begingroup$ (1) You can use a similar idea to define a surjection from the set of functions onto $\Bbb R$, that alone should suffice to prove the set is uncountable; (2) the injection is in the other direction, note that functions that have at most one nonzero element in their range at the point $0$ must be equal iff the agree on that element. $\endgroup$ – Asaf Karagila Apr 3 '14 at 1:09
  • $\begingroup$ ok got it, thanks! $\endgroup$ – Rachel Apr 3 '14 at 1:17

Explicitly, you could note that $\forall r\in\mathbb{R}$, the function that is $r$ at zero and 0 everywhere else is in the set. There are obviously uncountably many such functions.


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