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?
Continuity of discrete valued function
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0It depends on the topology on $\mathbb{R}$. It is continuous with the lower limit topology on $\mathbb{R}$, but not with the 'usual' topology. – 2012-06-19
2 Answers
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.
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0@Brian: I thought I had fixed that. Thanks. – 2019-02-16
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.