# Mapping The Countable Infinity of Integers Onto A Finite Segment of the Real Number Line

I was looking at the derivation of the Feynman path integral, recently. The teacher starts out with the path integral as an N (integer) dimensional integral, with N being the number of equally spaced divisions between the initial starting time of the path, $t_i$, and the final ending time of the path, $t_f$. He then lets N approach infinity so that every real number (every point along that segment of the time line continuum) between $t_i$ and $t_f$ is taken into account.

General concerns about the overall rigor of the path integral not withstanding, my question is, since N is an integer, then won't letting N approach infinity only take into account a countable infinity of points between $t_i$ and $t_f$, and hence not every real number between $t_i$ and $t_f$? Or, does an infinity of points along a finite segment of the real number line (produced by letting an integer number of equally spaced points on that segment approach infinity), produce a continuum of points (non-countable infinity of points) between them?

Maybe another way to ask is if you map the entire countable infinity of integer points of the real line onto a finite line segment of the real line, does every point of the real line segment have an integer mapped onto it?

I am thinking that the points on the real number line, along a finite segment is a non countable infinity (a continuum of points), so mapping a countable infinity (the integers) onto that will not cover all the points of that segment and hence the derivation of the path integral has at least that one flaw, if not others.

My background is science, not specifically math, so I am no expert on set theory. Therefore, please pardon my lack of sophistication in how I am asking the question or if this has already been answered but I cannot recognize the question or answer stated in a different form..

I don't know about Feynman path integrals, but from what you are saying it suffices to consider equally spaced partitions of a given interval, say $I:=[0,1]$.
In the first place we don't have a map $$f:\quad{\mathbb N}\to[0,1],\qquad n\mapsto t_n\ ,$$ but for each $N\in{\mathbb N}_{\geq1}$ a partition of $I$ into $N$ subintervals $I_k=[t_{k-1},t_k]$ of equal length $$t_k-t_{k-1}={1\over N}\ ,$$ hence $$t_k=k{1\over N}\qquad(0\leq k\leq N)\ .\tag{1}$$ In order to be more precise we would even have to denote the $t_k$ by $t_{N,\,k}$ $\>(0\leq k\leq N)$, since for each $N$ a new set of such dividing points is created.
"At each moment in time", i.e., for each fixed $N$ we only have finitely many, namely $N$, subintervals, but this number increases without bound as $N$ becomes ever larger. From $(1)$ it is clear that all dividing points $t_{N,\, k}$ occurring in this process are rational numbers; furthermore the index set organizing these $t_{N,\,k}$ is countable. But for any $N$ each individual irrational $\alpha\in[0,1]$ is included in one of the subintervals $I_{N,\,k}$.
• Alternatively, consider a physics problem involving a function of a real variable, such as position along a 1-D line, and discretize it to a set $N-1$ sub-intervals and approximately solve the problem using a summation of the function evaluated at those discrete points, where each term is multiplied by the length of the interval. if you let N approach infinity, you get a Riemann integral. But that integral will represent a summation where every real in not represented as a term, so the Riemann integral is not a true continuum generalization of the discrete sum? Yet it seems to work in physics. – David Apr 7 '18 at 18:26