Continuity of discrete valued function Is the ceiling function continuous when considered as a function from real numbers to integers (with discrete topology), and what is the formal argument for the proof?
Do we have general results about that kind of functions?
 A: No, it's not continuous. In the discrete space $\mathbb{Z}$, the set $\{0\}$ is open. But $f^{-1}(\{0\}) = (-1,0]$, which is not open in $\mathbb{R}$ with the usual topology. Since the inverse image of an open set is not necessarily open, the function $f$ is not continuous.
Added. In the comments you ask whether there is a nonconstant continuous discrete valued function with domain the reals. The answer is "no": if the function is not constant, then there exist at least two values, $a\neq b$. If $Y$ is the target space, then $Y-\{b\}$, $\{a\}$ is a disconnection of the target space. Then $f^{-1}(Y-\{b\})$ and $f^{-1}(\{a\})$ would have to be open, disjoint, and their union would be $\mathbb{R}$. But $\mathbb{R}$ is connected, so this is impossible.
A: To say that a function $f:X\to Y$ is continuous, where $X$ and $Y$ are topological spaces, is to say that for every open set $U\subseteq Y$, $f^{-1}[U]$ is an open set in $X$. For each $n\in\Bbb Z$, $\{n\}$ is an open set in $\Bbb Z$ with the discrete topology, but if $f:\Bbb R\to\Bbb Z:x\mapsto\lceil x\rceil$, then $f^{-1}[\{n\}]=(n-1,n]$, which is not an open set in $\Bbb R$.
Much the same thing happens with the floor function $g:\Bbb R\to\Bbb Z:x\mapsto\lfloor x\rfloor$, only now we have $g^{-1}[\{n\}]=[n,n+1)$; again this is not closed.
In fact, the only functions from $\Bbb R$ to $\Bbb Z$ that are continuous are the constant functions. On the one hand, constant functions are always continuous. On the other hand, if $f:\Bbb R\to\Bbb Z$ is not constant, its range is not a connected set. However, $\Bbb R$ is connected, and continuous functions preserve connectedness, so non-constant functions from $\Bbb R$ to $\Bbb Z$ cannot be continuous.
