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How much is $\lceil\frac{1}{\infty}\rceil$ ?

On one hand, $\frac{1}{\infty}=0$, so its ceiling is also $0$.

On the other hand, for all $x\geq 1$, $\lceil\frac{1}{x}\rceil = 1$, so, when $x$ goes to infinity, the function should remain with the same value...

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You have already answered your question. If it means one thing, then it's that thing. If it means another thing, then it's the other thing. What is the context or purpose for using this notation? – Jonas Meyer Jul 31 '13 at 21:20
The accepted answer here addresses these concerns. – Emily Jul 31 '13 at 21:22
@JonasMeyer: the context is this: I am writing a paper about rectangles whose aspect ratio is at most R. I proved a certain theorem about rectangles, where the expression ⌈1/R⌉ appears. The theorem is also true for general rectangles (i.e. $R=\infty$), but only if $⌈1/\infty⌉=1$. So, I want to know if I can assume that $⌈1/\infty⌉=1$. – Erel Segal-Halevi Aug 1 '13 at 19:25
undefined because $\infty$ is not a number, if write it in the form of a limit then it's meaningful. – Alt Mar 19 '14 at 5:54
up vote 8 down vote accepted

It's always dangerous to write $\infty$ in calculations. You have to be sure what you mean with writing $\infty$. In this case, you have two possibilities: $$\lim_{n\to\infty}\left\lceil \frac{1}{n} \right\rceil=1$$ or $$\left\lceil \lim_{n\to\infty} \frac{1}{n} \right\rceil=0.$$ You can not switch a function and a limit without further explanation. Compare with the important problem of the analysis where they try to switch a limit and an integral (the reason why the Lebesgue-integral is invented).

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This is a great example of why we can't pass limits inside of non-continuous functions. Notice the following:

$$1=\lim_{n\to\infty}1=\lim_{n\to\infty}\left\lceil\frac{1}{n}\right\rceil\\0=\lceil 0\rceil=\left\lceil\lim_{n\to\infty}\frac{1}{n}\right\rceil$$

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No, $\frac1\infty$ is not $0$: it is undefined. So, therefore, is $\left\lceil\frac1\infty\right\rceil$.

Your argument that it ought to be $1$ is also incorrect: it’s based on an unconscious assumption that the ceiling function is continuous. The same argument would say that since $1=\lim_{x\to 0^+}(x+1)$, and since $\lceil x+1\rceil\ge 2$ for all $x>0$, with $\lceil x+1\rceil=2$ for all small $x>0$, therefore $\lceil x+1\rceil$ ought to be $2$. But of course it isn’t: it’s $1$. The ceiling function isn’t continuous from the right at integers.

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It is undefined unless one defines it, which in some contexts is a good idea. – Jonas Meyer Jul 31 '13 at 21:24

Great question! The main problem anyone's going to have answering it is that the fraction $\frac1\infty$ isn't defined: $\infty$ isn't part of the ordered field of real numbers, and you can't just divide $1$ by $\infty$. Moreover, the function $\lceil.\rceil$ is only usually defined on the real number line, so even in strange fields like the hyperreal numbers (I think), where things like $\frac1\infty$ do make sense, it's a lot harder to make sense of $\lceil\frac1\infty\rceil$.

Here's one way you might try to make sense of it. You say that $\frac1\infty=0$. In mathematics, we don't say that, but we do say that $\lim_{x\to\infty}\frac1x=0$. '$\lim_{x\to\infty}$' means that we consider the behaviour of $\frac1x$ as $x$ becomes arbitrarily large. As we make $x$ larger, $\frac1x$ gets closer and closer to $0$: in fact, we can make it as close to $0$ as we like by choosing $x$ large enough. So $\lim_{x\to\infty}\frac1x=0$. Therefore, $\lceil\lim_{x\to\infty}\frac1x\rceil=0$.

But we could also interpret $\lceil\frac1\infty\rceil$ as the mathematical expression $\lim_{x\to\infty}\lceil\frac1x\rceil$. Now we are considering the behaviour of $\lceil\frac1x\rceil$ as $x$ becomes arbitrarily large. This time, the behaviour is much simpler: as long as $x\ge1$, $\lceil\frac1x\rceil=1$. Therefore, $\lim_{x\to\infty}\lceil\frac1x\rceil=1$.

Because we can't swap the order of taking limits and taking the ceiling function, we say that the ceiling function $\lceil y\rceil$ is discontinuous at the point $y=0$. Discontinuity means that there is a 'jump' in the graph of the function: you can't draw it without taking your pen off the page:

Notice the 'jump' at the point $0$ in this graph of the ceiling function.

The opposite of 'discontinuous' is continuous. A function $f$ is continuous at a point $a$ if $f(a)=\lim_{x\to a}f(x)$. Functions which are continuous everywhere include all polynomials, and lots of other lovely functions like $\sin$, $\cos$ and the Bessel functions, but the ceiling function is discontinuous at $0$, which is why we can't give meaning to $\lceil\frac1\infty\rceil$.

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what about the Complex plan is $1/\infty$ defined ? – Eli Elizirov Jul 31 '13 at 21:45
@EliElizirov: moving to the complex plane does not make $1/\infty$ defined. Complex numbers are just $x+iy$, where $x$ and $y$ are real numbers: since $1/\infty$ (presumably) has no imaginary part, that would mean that $1/\infty$ was defined in the real numbers, which is - as I have said - not the case. – Donkey_2009 Jul 31 '13 at 22:26
I meant like C^ which is C + the infinity point so $1/\infty$ must be defined no ? – Eli Elizirov Aug 1 '13 at 7:18
No - $\mathbb C\cup\{\infty\}$ is not a field, and $1/\infty$ is not defined. The image of the Möbius map $1/z$ at the point $\infty$ is defined to be $0$, but that is not the same thing. – Donkey_2009 Aug 1 '13 at 8:08

To elaborate on the comment above by @Jonas Meyer, there are number systems extending $\mathbb{R}$ which contain infinite numbers. If the symbol "$\infty$" is interpreted as referring to such a positive number, then $\frac1\infty$ is a positive infinitesimal. In some of these number systems such as the hyperreals, there is a principle that allows one to extend functions to the larger number system. In particular, the floor and ceiling functions extend in this way, and one neccesarily has that $\lceil\frac1\infty\rceil$ is indeed 1.

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$\frac{1}{\infty}=lim_{x\to\infty}\frac{1}{x}=lim_{x\to 0^+}x=0$ and ceil function is right discontinuous at $\mathbb{Z}$ especially at zero,now We have $\lceil\frac{1}{\infty}\rceil=0=\lceil\lim_{x\to 0^+}x\rceil\neq lim_{x\to 0^+}\lceil x\rceil=1$.

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It depends on if $\infty$ is $+\infty$ or $-\infty$, for the former one, the ceiling is $1$ and for the latter one its $0$.

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