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Suppose $I$ is an interval $[a,b]$. It is noted that $a$ and $b$ are real integers. Divide the interval into $n$ parts with step size $h=(b-a)/n$. Clearly all the points $a$, $a+h$, $a+2h$,....$a+(n-1)h$ are rational numbers. Consider P be a set of all these points. As $n$ goes to infinity the cardinality of the set P goes to infinity. The interval is uncountably infinite whereas the set P is countable.

Question is for large n, can we say the set P is an approximation of the interval?

Basically I have a function which need to be defined over an interval. But I would like to define the map only on the set $P$. Does that ensure the map would be well defined (almost approximately) over the interval?

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Let $A_n=\{a+k\frac{b-a}{n}|k\in \{0,\ldots,n\}\}$

I assume your function, say $f$ is continuous.

  • Consider $A=\displaystyle\cup_{n=1}^\infty A_n$

It is clear that $A$ is dense in $[a,b]$.

Since $f$ is continuous, it is safe to define $f$ only over $A$ and use density to define it elsewhere.

  • If you wish to define $f$ over some $A_N$ for fixed $N$, you're going to run into troubles since $A_N$ is not dense.
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  • $\begingroup$ The function I had in my mind is continous everywhere and nowhere differentiable on $A_n$. Yes the function is defined for all $n$. Could you please let me know how to use density to define elsewhere? Actually the 'elsewhere' set is the set of irrationals. Am I Right? $\endgroup$ – Fukuzita Jul 8 '14 at 12:33
  • $\begingroup$ Hmm the irrationals are included in the "elsewhere" set, but there might be some rationals as well (not sure). To define the function out of $A$, set $x\notin A$. By density, there is a sequence $y_n$ of points in $A$ such that $y_n\to x$. By continuity of $f$, $\lim_{n\to \infty} f(y_n) =f(\lim y_n) =f(x) $. $\endgroup$ – Gabriel Romon Jul 8 '14 at 12:53
  • $\begingroup$ Great, I got it. Please drop an email to sarif.hassan@icts.res.in I would like to talk in more detail. $\endgroup$ – Fukuzita Jul 8 '14 at 12:59
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If your function is continuous the answer is yes, as $P$ is dense in $[a,b]$.

If you start by defining the function on $P$, as long as it is uniformly continuous on $P$, it can be extended to $[a,b]$.

Without continuity, in general this doesn't hold.

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You ask "can we say $P$ is an approximation of the interval", which is not really a well defined question. In order to talk about approximations, you must have some sort of distance defined.

For example, the number $1$ is some approximation of $\sqrt 2$, but the number $1.4$ is a better approximation of the same number, because $d(1,\sqrt 2) > d(1.4, sqrt 2)$ (here, the distance $d$ is, of course, $d(x,y)=|x-y|$.

In your case, you have not defined any distance measure between different sets, so you cannot say you are really approximating anything...


However, you then started talking about functions defined only on $P$. In that case, more can be said. In general, we still cannot say anything, because you can, for example, have the functions $$f(x)=\begin{cases}0 & \text{ if } x\in P\\10^9&\text{ if }x\notin P\end{cases},\\ g(x)=\begin{cases}0 & \text{ if } x\in P\\-10^9&\text{ if }x\notin P\end{cases}.$$

These two functions are equal on $P$ (and are equal to $0$ on $P$) but are otherwise far from being the same function.

You can, however, say a lot if you only look at continuous functions on your interval. Because $P$ is a dense subset of $[a,b]$, this means that every continuous function $f$ defined on $P$ also uniquely determines a continuous function $F$ on $[a,b]$ such that $F|_P=f$.

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