$A=\{f \mid f:\mathbb{Z}_+ \to \{0,1\}\}$ is uncountable 
Consider the set $A=\{f \mid f:\mathbb{Z}_+ \to \{0,1\}\}.$
  I need to show that it is uncountable.

I was trying to find a bijection between $A$ and $\mathbb{R}$
or if I can show that there is no injection from $A$ to $\mathbb{Z}_+$ then also it'll work !
 A: 
The situation cries out for Cantor's diagonal argument:

Consider any function $s:\mathbb N\to A$ and, for each $n$, call $f_n=s(n)$. Define $f:\mathbb N\to\{0,1\}$ by $f(n)=1-f_n(n)$ for every $n$. Then $f$ is in $A$ but not in $s(\mathbb N)$ (why?). Hence $s$ is not a surjection. This proves that $A$ is uncountable.
A: Think of binary representation of real numbers in $[0,1]$. 
(You can write every real number in $[0,1]$ as
$$x=\sum_{n=1}^\infty \frac{a_n}{2^n}$$
where $a_n\in\{0,1\}$. 
This will give you the idea for a surjection from $A$ to $[0,1]$.
A: By $\mathbb{Z}_+$ you mean $\mathbb{N} = \{1, 2, \ldots\}$? Then
$$\sum_{n\in\mathbb{N}} f(n)2^{-n} \in [0, 1] \subset \mathbb{R}$$
And for the injection $[0,1] \to A$ chose $$T(x)[n] = \lfloor x\cdot 2^n \text{ mod } 2 \rfloor$$
Then you have $|A| = |[0,1]| = |\mathbb{R}|$ q.e.d.
A: You can identify $A$ with the power set of $\mathbb Z_+$. What do you know about the cardinality of a power set?
A: Consider $g:A\to P(\mathbb{Z_+})$, $ g(f)=\cup_{n:f(n)=1}\{n\}$. This is a bijection between $A$ and $P(\mathbb{Z_+})$, and $|P(\mathbb{Z_+})|=|\mathbb{R}|$.
