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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 $x \in \mathbb{Q}$.2011-02-21
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    Question needs to be rewritten. For example, is $t=n$ and/or is there a doubly indexed sequence of subintervals?2011-02-21
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    @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.2011-02-21
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    @joriki ($n=i$?) Sooo, let me try to rephrase: for each $t$, one is given $t$ intervals $i_n(t)$ indexed by $1\le n\le t$, and $f(t,\cdot)=\mathbb{1}_{U(t)}$ where $U(t)$ is the union of the odd numbered intervals $i_{2n+1}(t)$. Is this correct? If it is, one should also explain how the intervals $i_n(t)$ are chosen since the answer to the questions the OP asks seems to depend on whether these are ordered or chosen at will.2011-02-21
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    @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.2011-02-21
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    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.2011-02-21
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    @joriki Cool, now the questions are clear. Two remarks. 1. One can deal with the boundary points in any way one likes, this will not change the answers. 2. One should not use the $\lim$ operator unless one is sure that convergence there is. (P.S.: joriki=kakemonsteret?)2011-02-21
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    @Matt: Indeed. :-)2011-02-21
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    @Didier Piau: joriki$\neq$kakemonsteret -- I was just trying to be helpful :-) I agree with your two remarks.2011-02-21
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    any $x\in[1,0]$ will be in both an even-numbered interval and an odd-numbered interval infinitely often so that $\limsup_{t\to\infty}f(t,x)=1$ and $\liminf_{t\to\infty}f(t,x)=0$ for all $x$.2011-02-24
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    Why is there an enormous bounty on this question? It has been answered correctly...?2011-02-24

2 Answers 2

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There is no limit for any $0

(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 $\lim_{t\rightarrow\infty}f(t,0)=1$.