Extension of the concept of summation. I am not a mathematician and I am just asking this question out of my curiosity.
$Warning:$ This question might not have the right tags.
In an infinite series:$\sum_{n\in\mathbb{N}}a_n$, if you create a set $S_0=${$a_k|k\in\mathbb{N}$},then $Cardinality(S_0)=ℵ_0$
We know that we can write integrals($∫$) as riemann sums then it seems like adding infinitely many infinitesimal amounts and getting a finite value.
Like $\int_{C}a_n dn$ where $C$ is a continuous curve
Now if we let $S_1=${$x|x\in C$}(The set of all point in $C$)
Then $Cardinality(S_1)=\aleph_1$
So it seems like integral is an extension of regular summation to higher cardinality. Integral is the continuous counterpart of regular sums and regular sum is the discrete counterpart of integral.
So finally my question is that: Is it possible to extend the concept further to higher and higher cardinalities or is my question nonsense?
(I am not a mathematician that is why I couldn't describe my question in a precise way, but I still hope that my question is understandable and answerable)
 A: So basically you are asking if we can talk about integral in a set of cardinality $\aleph_{2}$ for instance. Not really actually, but we can still work out something although it won't be as you would expect it. For instance, you should know that you can define the integral operator by means of a series: that is the (naive) definition of $\textit{Lebesgue}$ integral. We approximate a function $f$ via a series of step functions and say the integral of $f$ is given by the maximum value the integral of such step functions series can have. Now, why do we do that? Because:

The cardinality of Lebesgue integrable function is $\aleph_{2}$, whereas Riemann integrable functions are $\aleph_{1}$.

Let's denote via $L^{1}$ such set. Then I suppose you'd like to write:
$$\int_{C}Fdm$$
where $C$ is a subset of $L^{1}$ and $F$ is an integrable function from $C$ to, for the sake of simplicity, $L^{1}$. Now, such writing is actually improper. We do not give priority to the subset $C$, but rather to functions that are in $C$ and the way $F$ tranforms them. Let's say $f\in C$ (function $f$ in $C$), with $\mathbb{R}$ as domain and codomain. Let $u$ another such function, and define:
$$F_{u}(f)=u\circ f$$
the composition. Such $F$ is said to be a $\textit{linear continous operator}$. We do not talk about $\textit{the integral of $F$}$, but rather the $\textit{norm}$ of $F$. Since now $u\circ f$ is a function which we can integrate, we can write:
$$\int_{\mathbb{R}}(u\circ f)(x)dx=\int_{\mathbb{R}}F_{u}(f)(x)dx$$
and define the norm of it as $\textit{the maximum value the above integral can have}$ (again a Naive definition, because we must talk about the norm of $f$ and say it is $1$ in order to define the norm of $F$). Given this setting, I imagine you want to write:
$$\int_{C}F_{u}(f)df$$
but this doesn't make any sense, or as far as I know it is not given any! Notice that the integral is a real number we associate to a function in a very specific way (as in the fundamental theorem of calculus for instance), so the above expression is actually pretty weird looking as it is not clear in which sense we are associating a real number to the function $F_{u}(f)$, but we can speak about the norm. In conclusion I would say that, in a functional analysis context, the concept of norm of an operator is more or less what you are asking for.
There are more and more general concepts and settings were the word "integral" is used, maybe in those settings the considered objects may have cardinality much higher than $\aleph_{2}$, but the interpretetion of such things is completely different from what we discussed here. Since you are not a mathematician I discourage you to look for such things, or te be more precise: I encourage you to study mathematics so that you will be able to "see" all the abstract objects you wish for!
A: Background
Firstly, I don't think being a mathematician or not matters in this case. Mathematics is always open to all interested in it. It is similar to how one doesn't have to be a professional athlete to play sports :)
Secondly, we should write $``|S_1| = \mathfrak{c}"$ instead of $``|S_1| = \aleph_1"$. $\mathfrak{c} = |\mathbb{R}|$ is the cardinality of the real numbers, while $\aleph_1$ is the first uncountable cardinal in the sense that for all set $S$ with $\aleph_0 < |S|$ we have $\aleph_1 \le |S|$. It is well-known that the continuum hypothesis $\mathfrak{c} = \aleph_1$ can neither be proved nor disproved.
Now let's consider

*

*How to sum uncountably many real numbers?

*Is there a general concept extending both continuous (Riemann) integral and discrete sum?

In your question, you didn't separate these two cases. But I would argue we have different answers to each of them.
Summing uncountably many real numbers
A standard way to sum uncountably many non-negative real numbers $(p_j)_{j \in J}$ where $J$ is an uncountable index set and $p_j \ge 0$, is to define the sum to be the supremum of all finite partial sums
$$\sum_{j \in J} p_j = \\ \sup \left\{p_{j_1} + p_{j_2} + \dots + p_{j_k} \mid p_{j_1}, p_{j_2}, \dots , p_{j_k} \in J, k \in \mathbb{Z}_{\ge 0}\right\}$$
Note that the sum always exists because supremum always exists as a non-negative extended real number in $[0, +\infty]$. Also, the index set $J$ can be of any cardinality. The definition allows us to add up as many real numbers as we like, even $|\mathscr{P}(\mathscr{P}({\mathbb{R}}))|$ many real numbers.
Now to sum uncountably many real numbers, we split $(a_j)_{j \in J}$ into positive part and negative part by defining $$p_j = \begin{cases}a_j \text{ if }a_j > 0\ \\ 0 \text{ otherwise}\end{cases}$$ and $$n_j = \begin{cases}-a_j \text{ if }a_j < 0\ \\ 0 \text{ otherwise}\end{cases}$$ By definition, $(a_j) = (p_j) - (n_j)$ and $p_j, n_j \ge 0$ for all $j \in J$. It now makes sense to define $\sum_{j \in J} a_j = \sum_{j \in J} p_j - \sum_{j \in J} n_j$. Note that the sum exists as an extended real number in $[-\infty, +\infty]$ exactly when we don't have a $``\infty - \infty"$ situation.
Now you may ask: if we already have a perfectly fine definition for uncountable sum, why aren't people studying uncountable sum?
The answer is: if the sum of uncountably many real numbers is finite, i.e. $-\infty < \sum_{j \in J} a_j < +\infty$, then all but countable many terms $a_j$ are zeros! So basically, we gain nothing by considering uncountable sum. If we want the sum to be finite, we can only have countably many non-zero terms. The proof can be found in The sum of an uncountable number of positive numbers. The proof works for non-negative real numbers but can easily be extended to the general case.
The above definition extends the idea of absolute convergence to uncountable sum. The idea of conditional convergence can also be extended to uncountable sum, and we get the same result that finite sum implies all but countably many terms are zeroes (see Sum of a series indexed by ordinals).
Extending both continuous (Riemann) integral and discrete sum
Integral feels like a "continuous sum", but how to make this precise? To keep things simple, let's restrict ourselves to integration of real-valued function.
Lebesgue integral comes to rescue! It generalizes Riemann integral on closed bounded interval to any measure space.
If we (Lebesgue) integrate on the measure space $(\mathbb{R}, \mathcal{L}, \mu)$ where $\mu$ is the Lebesgue measure on $\mathbb{R}$ and $\mathcal{L}$ is the Lebesgue measurable sets on $\mathbb{R}$, we get an extension of Riemann integral, in the sense that any Riemann integrable function $f: [a, b] \to \mathbb{R}$ is also Lebesgue integrable and both integrals have the same value (see General condition that Riemann and Lebesgue integrals are the same).
If we (Lebesgue) integrate on the measure space $(\mathbb{N}, \mathscr{P}(\mathbb{N}), \#)$ where $\#$ is the counting measure and $\mathscr{P}(\mathbb{N})$ is the power set of the natural numbers, then we get discrete sum (see Is Lebesgue integral w.r.t. counting measure the same thing as sum (on an arbitrary set)?).
For example, let $f, f_n: \mathbb{N} \to \mathbb{R}$ be defined by $f(k) = \frac{1}{k^2}$ and
$$f_n(k) = \begin{cases} \frac{1}{k^2} \text{ if } k \le n \\ 0 \text{ otherwise}\end{cases}$$
Then
$$f_n(k) = \sum_{m = 1}^{n} \frac{1}{m^2} \chi_{\{m\}}(k)$$
and thus
$$\begin{align} \int_\mathbb{N} f \mathrm{d}\#
&= \sup_{n \in \mathbb{N}} \left\{\int_\mathbb{N} f_n \mathrm{d}\#\right\} \\
&= \sup_{n \in \mathbb{N}} \left\{\sum_{m = 1}^{n} \frac{1}{m^2} \#(\{m\})\right\} \\
&= \sup_{n \in \mathbb{N}} \left\{\sum_{m = 1}^{n} \frac{1}{m^2}\right\} \\
&= \sum_{m = 1}^{+\infty} \frac{1}{m^2}\end{align}$$
