Why are Fourier series considered to be the subset of Fourier transform? It should have been other way around. Because a non periodic pulse is a subset of periodic pulse with period infinite. So Fourier transform (non periodic signals) is a subset of Fourier series (periodic signals).
[Math] Fourier transform vs Fourier series
fourier analysisfourier series
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Suppose we have a function $\tilde x(t)$ that is zero except on the interval $[-T_0/2,T_0/2]$ (on which $\tilde x(t) = x(t)$) and whose Fourier transform is given by $$ \widehat x(\omega) = \int_{-\infty}^\infty \tilde x(t) e^{-i\omega t}dt = \int_{-T_0/2}^{T_0/2} x(t) e^{-i\omega t}dt $$ Using $\widehat x(\omega)$, we would like to find the Fourier series for the $T_0$-periodic function that agrees with $x(t)$ on this interval. We note that the coefficients of the Fourier series for $x$ are given by $$ X_n = \frac{1}{T_0} \int_{-T_0/2}^{T_0/2} x(t)e^{-i (2 \pi n/T_0) t}\,dt $$ for any integer $n$. Notice the similarity! From here, you can derive $$ X_n = \frac{1}{T_0}\widehat{x}(2 \pi n/T_0) $$ Alternatively, let's say you wanted to look directly at $\mathcal F\{x(t)\}$. Note that $x(t) = \sum_{n = -\infty}^\infty X_n e^{i 2 \pi n/T_0}$. It follows that $$ \mathcal{F}\{x(t)\} = \sum_{n=-\infty}^\infty X_n \mathcal{F}\{e^{i 2 \pi n/T_0}\} = \sum_{n=-\infty}^\infty X_n \delta(\omega - 2 \pi n/T_0) $$
A Fourier series is only defined for functions defined on an interval of finite length, including periodic signals, as you can see from the definition of the Fourier coefficients (in the basis $\{e^{inx}\}_{n\in\mathbb{Z}}$) $$ a_n = \frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-inx}~dx. $$
You can't define an aperiodic signal on an interval of finite length (if you try, you'll lose information about the signal), so one must use the Fourier transform for such a signal.
In principle, one could take a Fourier transform of a periodic signal in the sense that one could extend the signal outside the interval $[-\pi,\pi]$ by zero. But the resulting Fourier transform would look like $$ \widehat{f}(p) = \frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-ipx}~dx $$ and this isn't particularly different from the Fourier coefficients. Moreover, being able to express a periodic signal as a discrete sum of frequencies is a stronger statement than expressing it as a continuous sum via the inversion formula.
Edit: In response to the comment.
I mean "stronger" rather loosely, not in the mathematical sense of one statement implying another.
Taking the Fourier transform of a periodic signal by extending it to $0$ outside $[-\pi,\pi]$ gives no information that the usual Fourier transform would not. Furthermore, the inversion formula still holds, and therefore we see that $f$ can be recovered from its Fourier transform as a continuous sum over all frequencies.
What is remarkable about periodic functions is that one does not actually need all this information to recover the function; out of uncountably many values $\widehat{f}(p)$, one only needs the integer values $\widehat{f}(n)$ (i.e. the Fourier coefficients) to reconstruct the function.
So in this sense, for a periodic function, there is more to say than what the Fourier transform alone provides.
Best Answer
The set of periodic functions is surely a subset of the set of all functions (most of which are aperiodic). So the Fourier series – which allow us to describe periodic functions differently – are a "subset" (in this sentence, the word "subset" is stranger than before) of the Fourier transform – which is a tool to encode an arbitrary function.
Alternatively, we may see the "subset" relationship in the frequency representation, too. Fourier series may be written as a special case of the Fourier transform in which only the frequencies that are multiply of the basic angular frequency $2\pi/T$ where $T$ is the periodicity are allowed. The Fourier transform of a periodic function is a very special kind of a function, a combination of delta-functions $$\tilde f(\omega) = \sum_{n\in{\mathbb Z}} c_n \delta(\omega-n\omega_0) $$ and functions that are a combination of delta-functions like that (determining Fourier series) are a subset of all functions, including distributions (which may be identified with the Fourier transform of the original function).