Show that both sets are infinite 
Suppose $\lbrace f_i : i \in \Bbb N\rbrace \subseteq \lbrace 0, 1\rbrace^{\Bbb N}$. Prove that there exists $g \in \lbrace 0,1\rbrace^{\Bbb N}$ such that for every $i \in \Bbb N$, the set $\lbrace n\in\Bbb N : g(n) = f_i(n) \rbrace$ and the set $\lbrace n\in\Bbb N : g(n) \ne f_i(n) \rbrace$ are both infinite.

I tried to attack the problem from different angles but I just couldn't find a suitable function $g$.
I am looking only for guidance or hints. Please don't post full answers. 
Thanks.
 A: HINT: Make $g(4 \cdot k+1) = f_0(4 \cdot k + 1)$...

 and $g(4 \cdot k + 3) \ne f_0(4 \cdot k + 3)$.

...

 This satisfies the condition for $f_0$ and all the even numbers are still available.  

Can you see how to continue?
A: Every positive $n $ can be written uniquely as $(\sum_{i=1}^k i)+j;0\le j\le k $.  
So use that.
Or if you want use any bijection, $k $, between $\mathbb N\times \mathbb N $ and $\mathbb N $.  
For every $i $, there are an infinite number of $k(m,i)=n $.  If we let $g (n=k(m,i))=f_i (n)$ there will be an infinite such $n$, (all $k(\mathbb N\times\{i\}))$ will satisfy.  
Just need to expand this to the infinite inequality sets.
Let $j:\{0,1\}\times \mathbb  N^2\rightarrow \mathbb N $ be a bijection.
Define $g (j(0,k,m))=f_m (j(0,k,m))$ for each $m $ there will be infinite values where $g (n)=f_m (n)$
Define $g (j(1,k,m)= f_m (j(1,k,m))+1\mod 2$.  Then for each $m$ there will be infinite values where $g (n)\ne f_m (n)$.
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If you want to define a precise $g$ consider.
$j: \{0,1\}\times \mathbb N \times \mathbb N\rightarrow \mathbb N$ via $j(a, k,m) = 2((m-1) + \sum_{i=1}^{i+(m-1)} i)- a$.
$j$ can be proven to be a bijection[$*$].
Define $g(n = j(a,k,m)):= f_m(n)+ a\mod 2$. 
For each $m$ there will be an infinite number of $A_m = \{n| n= f(0,k,m); k \in \mathbb N\}$ (all elements of $A_m$ are even by the way) and for all $n \in A_m$, $g(n) = f_m(n)$.  And for each $m$ there will be an infinite number of $B_m = \{n| n = f(1,k,m); k \in N\}$ (all elements of $B_m$ will be odd by the way) and for all $n \in B_m$, $g(n) \ne f_m(n)$.
[$*$] That $j: \{0,1\}\times \mathbb N \times \mathbb N\rightarrow \mathbb N$ should be obvious.  $k \ge 1; m-1 \ge 0;m,n \in \mathbb N$ so $k + (m-1) \ge 1$ and $\sum_{i= 1}^{k+ (m-1)} i \in \mathbb N$ and $m-1 \ge 0$ so $(m-1) +\sum_{i= 1}^{k+ (m-1)}i \in \mathbb N$ and $(m-1) +\sum_{i= 1}^{k+ (m-1)}i \ge 1$ so as $a \le 1; a\in \mathbb N$ then $j(a, k,m) = 2((m-1) + \sum_{i=1}^{i+(m-1)} i)- a\ge  1$ and is a natural number.
Surjective: If $n \in \mathbb N$ then either $n$ is even and $n = 2j$ for some natural $c$ or $n$ is odd and $n = 2c - 1$ for some natural $c$.
The sequence $1 = \sum_{i=1}^1 i < 1 + 2 = \sum_{i =1}^{2}i <..... < \sum_{i=0}^v i < \sum_{i=0}^v i + (v+ 1) = \sum_{i=0}^{v+1} i < ....$ spans the range of all natural numbers.  So there exists a natural $v$ so that $\sum_{i=1}^v i \le c < \sum_{i=1}^{v+1} i$.
Let $m = 1+c-\sum_{i=1}^v i$. Then $1 \le m < v+1$ so $k = v - (m-1) \ge 1$. and $c = \sum_{i=1}^{v=k+(m-1)}i + (m-1)$ and $n = 2((m-1) + \sum_{i=1}^{i+(m-1)} i)- a= j(a,k,m)$ where $a = 0$ if $n$ is even and $a = 1$ if $n$ is odd$.  
So $j$ is surjective.
$j$ is injective:
$j(0,b,c)$ is even and $j(1,d,e)$ is odd. so if $j(a,k,m) = j(a',k',m')$ then $a= a'$.
If $b + c < d+ e$ then  $j(a,d,e)= 2((e-1) + \sum_{i=1}^{d+e}i) - a = 2((e-1) + \sum_{i=1}^{b+c} i + \sum_{i=b+c+1}^{d+e}i) -a$
$\ge 2((e-1) + d+e + \sum_{i=1}^{b+c}i) - a$
$> 2(c1 +\sum_{i=1}^{b+c}i) - a = j(a,,b,c)$.
So if $j(a,k,m) = j(a',k',m')$ then $k+m = k' + m'$
If $d < c$ then $j(a,v-d, d) = 2((d-1) + \sum_{i=1}^{v-d + (d-1)=v-c+(c-1)}i) - a$
$<  2((c-1) + \sum_{i=1}^{v-d + (d-1)=v-c+(c-1)}i) - a=j(a,v-c,c)$
So if $j(a,,k,m) = j(a',k',m')$ then $a = a'$; $k+m = k'+m'$ ; $m = m'$ and so $m= m'$
So $j$ is injective.
A: For each $n$, let's call $A_n$ the set of $i$ on which we'd like $g(i)= f_n(i)$, and $B_n$ the set of $i$ on which we'd like $g(i)\ne f_n(i)$. Ideally we'd just like all of the $A_n$, $B_n$ to be disjoint from one another, that way we're free to just set $g$ to be whatever we want on those sets, with no chance of our definitions interfering with one another.
So you just need to find a (countable) infinite family of subsets of $\mathbb N$, $C_n$, all of which are disjoint, and then you can set $A_1 = C_1, B_1 = C_2, A_1 = C_3, B_1 = C_4$, and so on.
Can you find a countable family of pairwise disjoint subsets of $\mathbb N$?
