Would it be possible (or even make any sense) to define a new type of number such that there were an infinite amount of these numbers between an infinitely small space?
For example, there could be an infinite amount of these new numbers between 1.0 and 1.0.
Another example:
There could be an infinite amount of these numbers that are neither greater than, less than, nor equal to 1.0.
On
Based on "There could be an infinite amount of these numbers that are neither greater than, less than, nor equal to 1.0." from the question, and "I'm sort of thinking that there could be a space in which new numbers could exist such that that space doesn't exist (where the space is 0 as you stated)." in a comment, it sounds like the OP is looking for something a little weirder than just "an infinitesimal in an ordered field", as Henning Makholm addressed well.
Something that I think might satisfy the OP can be found in the theory of combinatorial games, in which positions in games are assigned "values", some of which correspond to real numbers, and many of which do not.
Assuming you're unfamiliar with the theory, I'll try to introduce enough about a game to get at the sort of idea asked about in the question.
"Hackenstrings" is a game in two players ("Blue" and "Red") take turns hacking away at strings. An individual string will be comprised of "B"s, "R"s, and "E"s. On their turn, Blue gets to remove a B or an E (and everything to the right of that letter) and Right would get to remove an R or an E (and everything to the right of it). A player with no legal move on their turn (in any available string) loses.
For example, in the string "RBRBEB", Blue's legal moves would be to "R", "RBR", "RBRB" (hacking at the 'E'), or "RBRBE". Red's legal moves would be to "" (the empty string), "RB", or "RBRB". In the game with three strings, {"E","B","B"}, Red is sure to lose even if Blue goes first since Red has at most one available move and Blue has two at the start.
It turns out to be reasonable to assign values to Hackenstrings to help understand who will win in a game with multiple strings. The usual unit is "moves for Blue". "BB" has value 2, "RRR" has value -3, "" has value 0, etc. If the sum of the values of strings is positive, then Blue can win. If it's negative, then Red can win. If it's zero, then the player whose turn it is now loses.
It turns out we can get a sensible value for any string of Bs and Rs, although we get some dyadic fractions. For example, "BBBRRB" has value $2+\frac38$. (For more on why this sort of value is assigned, see my answer here.)
But what value should be assigned to "E"? The problem is that "E" is a first person win, which isn't covered by our "positive, negative, or zero" classification. "E" is [i]incomparable[/i] to "", and the value is traditionally denoted $*$ (read "star"). Even though $*$ isn't positive or negative or zero, it's still "infinitesimal" in that $-\dfrac{1}{2^n}<*<\dfrac{1}{2^n}$ for all $n$ (because Blue will win a game like {"E","BRRR"} no matter who goes first).
"EE" has a different value than "E" since {"E","E"} is a second-player win ($*+*=0$), but {"E","EE"} is a first player win. In fact, the game of Hackenstrings using just "E"s is complicated (it also goes by the name normal-play Nim), but the bottom line is that every string of "E"s has a different value. Traditionally, the value of "EE" is denoted $*2$, the value of "EEE" is denoted $*3$, etc.
This means that a value of the type you're looking for would be $1+*$, the value of the game {"E","B"}. Since first-player wins {"E","B","R"}, $1+*$ is neither equal to $1$, nor greater or less than $1$. However, it sort of fits in an infinitely small space since $1-\dfrac{1}{2^n}<1+*<1+\dfrac{1}{2^n}$ for all $n$. Also, there are infinitely many game values that fit in this space: $1+*,1+*2,1+*3,\ldots$
If you are interested in learning more about combinatorial game theory, there is a good list of resources at this answer by Andrés E. Caicedo. I particularly recommend the undergraduate text Lessons in Play: An Introduction to Combinatorial Game Theory.
On
This probably isn't the kind of answer you're looking for, but just for consideration: on the real line, looking at the fixed point of $1$, you only have one point. But then consider the complex plane- each complex number $1+xi$ projects onto the point $1$ on the real line, so we can think of it as being between $1$ and $1$, but the number of such points we can pick are not only infinite, but in fact uncountable, as for any real number $x$, we can pick $1+xi$ as a point between $1$ and $1$.
On
The answer is affirmative. In the hyperreal number system, between $0$ and a positive infinitesimal $\epsilon$ there are infinitely many other infinitesimals; for example $\epsilon^2$, $2\epsilon^2$, $3\epsilon^2, \ldots, H\epsilon^2$. Here $H$ is the infinite number $\frac{1}{\epsilon}$.
What you want is a number system with infinitesimals.
You can't get infinitely many numbers "between $1$ and $1$", but there are various ways of making a number system that contains an infinity of numbers that are all between $1$ and $1+\frac1n$ for every natural number $n$.
Two popular approaches are:
Algebraically: Decide on a formal constant $\varepsilon$ with you define to have the property that it is strictly between $0$ and $\frac1n$ for every $n$. You can then extend the ordering of the reals to numbers of the form $$ \frac{p(\varepsilon)}{q(\varepsilon)} $$ where $p$ and $q$ are polynomials with real coefficients and $q$ is not the zero polynomial. Computations with these numbers can be done following the usual rules for arithmetic on fractions and polynomials.
For example, we would have $$ 0 < \varepsilon^2 < \frac1{100000}\varepsilon < \varepsilon < 2\varepsilon < 2\varepsilon+\varepsilon^2 < \frac1{100000} < 1 < \cdots < 100^{100} < 1/\varepsilon < \cdots $$
What this amounts to is just a particular total ordering of the field of rational functions over $\mathbb R$.
Using formal logic: Non-standard analysis gives a way to construct a number system with infinitesimals which is in certain senses "better behaved" than the above algebraic ansatz. But it requires somewhat heavy technical preliminaries to understand how it works, and there are various subtle pitfalls one needs to learn to avoid when trying to apply it.