Countable and uncountable number Is there a function that has an uncountable number of continuity points and an uncountable number of discontinuity points in any neighborhood contained in the interval $(0,1)$?
 A: A set $G\subseteq \mathbb R$ is $G_\delta$ (that is, a countable intersection of open sets) if and only if there exists a function $f\!:\mathbb R\to\mathbb R$ that is continuous at each point of $G$ and discontinuous at each point of its complement.
The result you are requesting follows from noticing that there are $G_\delta$ sets $G$ such that both $G$ and its complement have uncountable intersection with any open interval. 

First, the set of points of continuity of a function is $G_\delta$: $f$ is continuous at $x$ if and only if $x$ is in the intersection of the open sets $U_n$, where $$ U_n=\{y\in\mathbb R:\exists\delta>0\,\forall x,z\in(y-\delta,y+\delta)\,(|f(x)-f(z)|<1/n)\}. $$
Second, any $G_\delta$ set $G$ is the set of continuity points of a function. To see this, 
(a) note first that if $U$ is an open interval one can easily build $f$ that is continuous (even 0) outside of $U$ and discontinuous everywhere in $U$. From this, one easily gets the same for any open set $U$. 
(b) Now, given a $G_\delta$ set $G$ whose complement does not contain any interval, write $G$ as $\bigcap_n U_n$ where the complement $C_n$ of $U_n$ is closed nowhere dense, consider $$f(x)=\sum_n2^{-n}\chi_{C_n}(x)$$ and note that $f$ is continuous at each point of $G$ and discontinuous at any point in its complement. 
(c) The result easily follows for arbitrary $G_\delta$ sets $G$ by combining (a) and (b).
Third, there are $G_\delta$ sets $G$ such that $G$ and its complement have uncountable intersection with any open set. For instance, let $(q_n:n\ge0)$ enumerate the rationals. For $i,j\in\mathbb N$ let $$I_{i,j}=\left(q_i-\frac1{2^{i+j}},q_i+\frac1{2^{i+j}}\right),$$ $G_j=\bigcup_i I_{i,j}$ and $G=\bigcap_j G_j$. The set $G$ is $G_\delta$, by construction. 
Note that the measure of $G_j$ is at most $\frac1{2^j}\sum_i\frac2{2^i}=\frac4{2^j}$, so $G$ has measure 0. It follows that the complement of $G$ has full measure and therefore meets every interval in an uncountable set. 
At the same time, the complement of $G$ is the countable union of the sets $G_j^c$, each of which is nowhere dense, so $G$ is comeager and also hits every interval in an uncountable set.

The construction above is quite flexible and additional properties can be required of $f$, as pointed out in the comments.
A: Yes, there are many such functions.
For example let $$f(x)=0, \text{ if } x\ge 1/2 $$
$$f(x)=1, \text{ if } x\notin Q, x<1/2$$
$$f(x)=-1, \text{ if } x\in Q, x<1/2$$
A: Sure. If $x\in (0,1/2)$, let $f(x)=0$. So $f$ is continuous on this interval. On $[1/2,1)$, we can define $f$ to be discontinuous everywhere. For instance, $f=1_{\mathbb{Q}}$ would do. 
