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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?

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The preimages of the open(!) sets $\{n\}$ aren't open in $\mathbb R$, hence the function isn't continuous. –  martini Jun 19 '12 at 15:39
    
It depends on the topology on $\mathbb{R}$. It is continuous with the lower limit topology on $\mathbb{R}$, but not with the 'usual' topology. –  copper.hat Jun 19 '12 at 15:53
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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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Any example of a continuous discrete valued function (reals to integers)? –  Liz Jun 19 '12 at 15:43
    
A non trivial one... –  Liz Jun 19 '12 at 15:45
    
@Liz: No; any nonconstant continuous function from $\mathbb{R}$ to a discrete space would yield a disconnection of $\mathbb{R}$. I've added this to my post. –  Arturo Magidin Jun 19 '12 at 15:48
    
@Brian: I thought I had fixed that. Thanks. –  Arturo Magidin Jun 19 '12 at 16:16
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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.

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