Intuition for a physical real line vs. a physical "hyperreal line"

The assertion that $\Bbb R$ ‘has no holes’ is an informal paraphrase of the mathematically precise statement that $\langle\Bbb R,\le\rangle$ is a complete linear order. In other words, every non-empty $A\subseteq\Bbb R$ that is bounded above has a least upper bound in $\Bbb R$. This in no way prevents us from shoving new elements into $\Bbb R$. For example, I can take some object $p\notin\Bbb R$, let $X=\Bbb R\cup\{p\}$, and define a linear order $\preceq$ on $X$ by $x\preceq y$ iff

• $x,y\in\Bbb R$ and $x\le y$, or
• $x\in\Bbb R$, $y=p$ and $x\le 0$, or
• $x=p$, $y\in\Bbb R$, and $y>0$, or
• $x=y=p$.

This in effect inserts $p$ between $0$ and all of the positive reals. The various constructions of the hyperreals do something similar, but on a grand scale, surrounding each real number with a ‘cushion’ of infinitesimally different hyperreals, and moreover adding whole galaxies of infinite hyperreals at both ends of the line. This is very different from filling existing holes, which is what we do when we complete the rationals to form the reals.

When it is said that the line has no holes, one must be precise about what is meant. Here's one way of looking at it, maybe the simplest: Suppose you divide the real line into two non-empty subsets $A$ and $B$, such that every real number is in one or the other of those, and every number in $A$ is strictly less than every number in $B$. Then there must be a boundary: a number $c$ such that every number less than $c$ is in $A$ and every number more than $c$ is in $B$. The number $c$ itself could be in either of those two sets.

This is opposed to a situation like this: consider the set of all non-zero real numbers. You can divide those into two subsets, namely the negative numbers $A$ and the positive numbers $B$, in such a way that every non-zero real number is in one or the other of those, and every non-zero number in $A$ is strictly less than every number in $B$. But if you take any negative number $x$, you can find a larger negative number, e.g. $x/2$, and for any positive number $x$, there is a smaller positive number $x/2$. Thus no negative or positive number $c$ can serve as a boundary, so that every number less than $c$ is negative and every number greater than $c$ is positive.

Even if we didn't know that irrational numbers exist, we could prove that for every rational number whose square is less than $2$, there is a larger rational number whose square is less than $2$, and for every rational number whose square is more than $2$, there is a smaller rational number whose square is more than $2$. Thus the set of all rational numbers has a hole, and in fact has many holes.

The hyperreals also have holes. For example let $A$ be the union of the set of all non-positive hyperreals (including $0$) and the set of all positive infinitesimal hyperreals, and let $B$ be the complement of that set. There's no hyperreal $c$ that can serve as a boundary. One can see that as follows. If $c\in B$, then $c/2$, being less than $c$, would be in $A$, but that would mean half of a non-infinitesimal positive number is infinitesimal, so that can't work. But if $c\in A$, then $2c$, being greater than $c$, would be in $B$, thus non-infinitesimal. But then again, half of the non-infinitesimal number $2c$ would be infinitesimal, and again that can't work.

If you think a missing hyperreal means a hole in the reals, consider this: the set of all integers has no holes. Lots of reals are missing, but that doesn't create holes in the sense defined above. To see this, consider the two sets $A=\{\ldots,-3,-2,-1,0,1,2,3\}$ and $B=\{4,5,6,\ldots\}$. Either $3$ or $4$ can serve as a boundary as defined above, since every number less than $3$ or less than $4$ is in $A$ and every number greater than $3$ or greater than $4$ is in $B$. And all other ways of dividing the set in two that satisfy the requirements above also yields a boundary. So there are no holes.