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?
 A: 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 $1/2$.   
A: 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 $\lim_{t\rightarrow\infty}f(t,0)=1$.
