Rationals over an interval

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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...

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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 $F|_P=f$.