My background: I have a bachelor's CS degree and have never taken anything beyond part of a first course in abstract algebra - no real analysis or complex analysis. I learned about higher cardinalities than $R$ in my automata course.

The inverse of addition works on integers.

The inverse of multiplication forces us to get to the rationals, but that's still the same cardinality.

The inverse of exponentiation forces us out of the rationals and into complex numbers, but we're still at integer cardinality because it's algebraic.

Consider $a^x$ is transcendental at least for $a>1$ and when given imaginary variables (at $x=0$ it will look like a sine wave, probably different period than $sin(x)$, which is associated with e^x, along the imaginary part of y axis) needs a real cardinality.

Therefore, $x^x$, the next level up from exponentiation is going to be at least real-valued in cardinality, as well. By simply increasing the "level" of my operation, I bumped my way out of the integers.

My intuition says if I keep bumping up my operation, eventually I may have a function such that if I need a cardinality greater even than the reals to smooth it out.

Is there a "jagged" real-valued function that is "smooth" in cardinalities greater than the reals?

I'd say something is "smooth" if it is not "flat" and if it always approaches being "flatter" as you "zoom in". Flat would mean minimum distance and flatter would mean approaches without overshooting. Like $sin(x) * e^x$ and $e^x$ both "approach" 0 (in the the range necessary to contain all values towards infinity strictly decreases) as x grows arbitrarily negative, even though in the first case, it "overshoots" and crosses the $x$-axis.

Just like the reals are in between the integers that fill in the gaps of some functions, I wondered if there were numbers in between the reals to fill in the gaps of some functions, but I decided to try to word it a little more precisely.

In response to "How can a function on the reals be extended to a completely different set? And how do you measure it?":

Consider all finite non-empty sequences of $1$'s and $0$'s that start with a $1$, call it $S_1$. This is all positive integers base-2.

Now consider all proper subsets of that set, call it $S_2$.

Now consider $f_1$ that takes in an element of $S_2$ and returns a string whose $i$th digit is $1$ if the base-2 representation of $i$ is in $S_2$, $0$ otherwise.

Now consider the elements of $S_3 = \{ (s_1, f_1(s_2)) | s_2 \in S_2, s_1 \in S_1 \cap \{0\} \}$. This will correspond to all non-negative real numbers. The first thing in the tuple will be the part of the number to the left of the decimal. The second thing in the tuple will be the decimal part.

Is $S_1$ a subset of $S_3$? We can map $S_1$ to $S_3$ by saying elements $s_1$ in $S_1$ correspond to the elements in $S_3$ that have $s_2$ as the empty set and have $s_1$ as the first thing in their tuple. In other words integers get mapped to the reals that have all $0$'s as their decimal portion.

Now consider all proper subsets of $S_3$, call it $S_4$.

Now consider the elements of $S_5 = \{ (s_3, s_4) | s_3 \in S_3, s_4 \in S_4 \}$. This is our greater-than-the-reals set (non-negative reals), and we say the reals are a subset of this set by saying they correspond to the elements of $S_5$ where $s_4$ is the empty set.

I can give a partial definition of measuring by assuming $s_3$ is greater than $t_3$ and that $s_4$ is a superset of $t_4$ and say their distance is $(s_3 - t_3, s_4 - t_4)$. I don't know what to do if $s_3 > s_4$ and $t_3 \not \supset t_4$ because you need some way to "borrow" like in basic arithmetic if you try to remove a set that isn't there.

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    $\begingroup$ Your writing is a bit imprecise, but in your last sentence you essentially ask about there being gaps in the real numbers, and the answer is that there are not. This is due to the completeness of the real numbers. $\endgroup$ – Santeri Apr 15 '16 at 23:19
  • $\begingroup$ @Santeri "In the decimal number system, completeness is equivalent to the statement that any infinite string of decimal digits is actually a decimal representation for some real number." Well that only works if you assume your infinite string is $R$ in cardinality. If I choose $Z$ $0$'s and $1$'s I can put those in a real number (say the binary decimals in $[0, 1)$). I don't think I can fit $2^Z=R$ 0's and 1's into a real number in the same way I can't fit $Z$ 0's and 1's into $Q$ or $Z$. I would need $2^R$ to do that. Also completeness from that article seems like it's an axiom. $\endgroup$ – Words Like Jared Apr 15 '16 at 23:50
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    $\begingroup$ Simply requiring that exponentiation have an inverse doesn't necessitate moving to the real numbers. You just need to take the algebraic closure of the rationals, which is the field of algebraic numbers. This field is still countable. There is nothing inherently algebraic about the reals, and algebra will not lead you to them in a natural way. Most constructions either require an order relation or a topology. $\endgroup$ – Matt Samuel Apr 15 '16 at 23:56
  • $\begingroup$ @MattSamuel I see what you're saying. Division and multiplication of integers needs only 2 integers to represent. In general, as long as any finite sequence of operations (and finite number of operations) result in a finite number of integers to represent the result (and a countably infinite input set), it will still be the integers in terms of cardinality. I guess that doesn't change my question - I was just trying to construct a candidate example, although it may be generalized inverses of addition, multiplication, exponentiation, etc. may not take me there... $\endgroup$ – Words Like Jared Apr 16 '16 at 0:09
  • $\begingroup$ @Santeri I do see, now, how my writing even was more imprecise than I realized at the time. I was conflating the decimal notation of something (infinite length for irrationals) with other possible encodings, like the algebraic field works with. If I took a finite number of steps, $k$, to get to my result, I can represent those steps as a $k$ tuple which will be $Z$ in cardinality. $\endgroup$ – Words Like Jared Apr 16 '16 at 0:16

The functions of analysis have ranges that are within $\Bbb C$ (often $\Bbb R$), so you will not need a larger set to capture the range. Even if they are wildly discontinuous, the range still $\Bbb C$.

In your effort to make functions continuous by adding points, it seems you are motivated by the fact that any function on the integers can be made continuous by defining it properly on the reals between the integers. A critical difference is the topology of the sets. The integers are discrete, so you can define intervals like $(0.1,0.6)$ that do not contain any integers. You can then define the function to do what you want between the integers to make it continuous over the reals.

The reals are dense in themselves, so as long as you use the standard $\epsilon-\delta$ definition of continuity, you cannot make a discontinuous real function continuous by adding more points.

  • $\begingroup$ Towards the end of my question when I went to define how to represent the extra information necessary of these numbers by taking the cross product of the reals with its powerset; if you had $(a,b)$ and $(a,c)$, with $a \in \Bbb R$ and $b,c \in 2^{\Bbb R}$ then you'd have a range that doesn't contain any real numbers in between them. This is akin to $(0,1)$ and $(0,6)$ from your example. I would assume you're right on the "standard definition" - I was just wondering if a self-consistent "extension" could be made... and I ran into the problem of not always being able to define ordinality. $\endgroup$ – Words Like Jared Apr 17 '16 at 23:54
  • $\begingroup$ Now that I think about it, we could look at the countably finite sequence of digits after your decimal point as a countably infinite polynomial. Maybe the digits of your uncountably infinite number could be defined as an integral or summation or something... You would just need some way to make the larger reals not as signifiant/independent like digits not being as significant for higher order polynomial terms. $\endgroup$ – Words Like Jared Apr 17 '16 at 23:58
  • $\begingroup$ You have to think about the topology of your set as well as just defining it. Certainly you can embed the reals in a set of large cardinality, but now you have to figure out how to deal with open sets so you can talk about continuity. You also want the embedding to respect the real functions we are used to. I don't think you can make the reals discrete as a subset of the larger set while maintaining the density. $\endgroup$ – Ross Millikan Apr 18 '16 at 0:02
  • $\begingroup$ I had never heard of an open set before. I've never studied Topology. Could you give me a link to what you mean by "density"? $\endgroup$ – Words Like Jared Apr 18 '16 at 0:07
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    $\begingroup$ The reals being dense (so are the rationals) as an order just means that between any two there is another. Another use of density is to say that the rationals are dense in the reals which means there is a rational arbitrarily close to any real. Open sets are generalizations of open intervals in the reals. In metric spaces they contain an open ball around each of their points. $\endgroup$ – Ross Millikan Apr 18 '16 at 0:19

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