# Limit of function of a set of intervals labeled i_n in [0,1]

Suppose we divide the the interval $[0,1]$ into $t$ equal intervals labeled $i_1$ upto $i_t$, then we make a function $f(t,x)$ that returns $1$ if $x$ is in $i_n$ and $n$ is odd, and $0$ if $n$ is even.

What is $\lim_{t \rightarrow \infty} f(t,1/3)$?
What is $\lim_{t \rightarrow \infty} f(t,1/2)$?
What is $\lim_{t \rightarrow \infty} f(t,1/\pi)$?
What is $\lim_{t \rightarrow \infty} f(t,x)$?

joriki clarification in comments is correct, does $\lim_{t \rightarrow \infty} f(t,1/\pi)$ exist, is it 0 or 1 or (0 or 1) or undefined? Is it incorrect to say that is (0 or 1)?

Is there a way to express this:
$K=\lim_{t \rightarrow \infty} f(t,x)$

K, without limit operator ?

I think to say K is simply undefined is an easy way out. Something undefined cant have properties. Does K have any properties? Is K a concept?

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$1$. and $2$. are not well defined as for any $k$, $f(3k,\frac{1}{3}) = 1$ but also $0$, same for $f(2k,\frac{1}{2})$. Same for $4$. when $x \in \mathbb{Q}$. – milcak Feb 21 '11 at 10:56
Question needs to be rewritten. For example, is $t=n$ and/or is there a doubly indexed sequence of subintervals? – Did Feb 21 '11 at 14:18
@Didier Piau: No, $n$ is the index of the interval in which $x$ lies, i.e. $1\le i \le t$. I don't think there's a problem in that regard; only in that the boundary points are in two different intervals, as milcak pointed out. – joriki Feb 21 '11 at 14:24
@Didier Piau: Yes, sorry, $n=i$, and yes, $f(t,\cdot)=1_{U(t)}$. I think the intended meaning of "labeling the equal intervals $i_1$ to $i_t$" was to label them in order, i.e. $i_j=[(j-1)/t,j/t]$. That still leaves the problem of the boundary points being assigned two different function values. – joriki Feb 21 '11 at 15:56
I don't get this question at all, I would appreciate it if OP would edit it for clarity. If I understand the question properly, the OP seems to think there is some reason these limits could exist, but I don't see why. – Matt Calhoun Feb 21 '11 at 16:35

## 2 Answers

There is no limit for any $0<x<1$.

(a) $f(t, 1/3) = 1$ for $t$ of the form $6n+1$ or $6n+2$ and $0$ for $t$ of the form $6n+4$ or $6n+5$, and on the boundary in other cases. For example $\frac{2n}{6n+1} < \frac{1}{3} < \frac{2n+1}{6n+1}$ and $\frac{2n+1}{6n+4} < \frac{1}{3} < \frac{2n+2}{6n+4} .$

(b) $f(t, 1/2) = 1$ for $t$ of the form $4n+1$ and $0$ for $t$ of the form $4n+3$ and on the boundary in other cases

(c) If $f(t, 1/\pi) = 1$ then $f(t+3, 1/\pi) = 0$ or $f(t+4, 1/\pi) = 0$ and similarly if $f(t, 1/\pi) = 0$ then $f(t+3, 1/\pi) = 1$ or $f(t+4, 1/\pi) = 1$.

(d) If $f(t, x) = 1$ then $f\left(t+\lfloor{1/x1}\rfloor , x \right) = 0$ or $f\left(t+\lceil{1/x1}\rceil , x \right) = 0$ and similarly if $f(t, x) = 0$ then $f\left(t+\lfloor{1/x1}\rfloor , x \right) = 1$ or $f\left(t+\lceil{1/x1}\rceil , x \right) = 1$.

So there is no convergence and so no limit.

If instead you explicitly gave boundary cases the value $1/2$ (only necessary for rational $x$) and took the partial average of $f(s,x)$ over $1 \le s \le t$, then the limit of the average as $t$ increases would be $1/2$.

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Let $n(t, x)$ be the index of the interval into which $x$ falls when $[0,1)$ is divided into $t$ identical intervals, $$[0,1) = \bigcup_{i=1}^{t} \left[\frac{i-1}{t}, \frac{i}{t}\right).$$ Then $(n-1)/t \le x < n/t$, so $$n(t,x) = \lfloor{tx + 1}\rfloor.$$ Clearly $n(t+1,x)-n(t,x) \le 1$ for $x<1$; that is, $n(t,x)$ cannot skip any values. On the other hand, for $x>0$, $n(t,x)$ grows without bound as $t\rightarrow\infty$. Combining these two facts, we see that for $x>0$, $n(t,x)$ is both even and odd infinitely often, hence $f(t,x)$ is equal to both $0$ and $1$ infinitely often, and hence $\lim_{t\rightarrow\infty}f(t,x)$ does not exist. At the remaining point, $x=0$, the limit does exist: $n(t,0)$ is identically equal to $1$, which is odd, so $f(t,0)$ is identically equal to $1$, and $\lim_{t\rightarrow\infty}f(t,0)=1$.

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