Can every canonical transformation be broken down into a large number of infinitesimal canonical transformations? Consider a canonical transformation from $(q,p)$ to $(Q,P)$ depending upon a continuous parameter $\alpha$ such that:
$$Q_i=Q_i(q,p,t,\alpha), \space P_i=P_i(q,p,t,\alpha)$$
where $q$ and $p$ represent the set of all $q_i$s and $p_i$s respectively.
From what I leant, The generator corresponding to the infinitesimal Canonical transformations has to be represented in the form:
$$F(q,P,t)=\sum q_iP_i +\alpha G(q,P,t). \tag{1}\label{1}$$
Because: Since ICTs have to be infinitesimally close to original coordinates, transformation equations have to be of the form:
$$Q_i=q_i+\alpha(something), \space P_i=p_i+\alpha(something) \tag{2}\label{2}$$
(where $\alpha$ is small) and this is made sure by eq(1) because $F=\sum q_iP_i$ corresponds to an identity transformation and hence eq(1) gives identity + something, just like ICT demands in eq(2).
Question: If we consider any general CT given by generator function $F$(need not be a function of $q,P$), depending on a continuous parameter $\alpha$; and we shrink $\alpha$ and neglect the second and higher order terms in $\alpha$, are we guaranteed to get it in the form of eq(1)? (see below for an example)Intuitively I wouldn't think so, because $F$ need not be a function of $q,P$; However, does it mean that when we shrink $\alpha$ the resultant GF doesn't correspond to an ICT? How can it be possible that a finite CT cannot be broken down into many ICTs?
PS:
example: if we consider a rotation of coordinate systems by an angle $\alpha$, the GF will be given by:
$$F=qP\sec(\alpha) -\frac 12 (q^2+P^2)\tan(\alpha).$$
For small angles:
$$F=qP -\frac 12 (q^2+P^2)\alpha$$
which is clearly in the form of eq. (1).
 A: *

*TL;DR: A finite CT can not always be composed from infinitely many infinitesimal CTs because the space of CTs is not always path connected for topological reasons, cf. this related Phys.SE post.


*Let us elaborate on OP's example. OP considers a CT in a 2D phase space $M=\mathbb{R}^2$ with a generator
$$ F_2(q,p)~=~\frac{\alpha}{2}q^2 + \beta qP + \frac{\gamma}{2}P^2 \tag{i}$$
of type 2,
where $\alpha,\beta,\gamma\in\mathbb{R}$ are 3 real parameters.


*It is straightforward to check that the corresponding CT is
$$\begin{align} \begin{bmatrix} Q \cr P \end{bmatrix}
~=~& A\begin{bmatrix} q \cr p \end{bmatrix}, \cr 
A~:=~& \begin{bmatrix}a & b\cr c & d  \end{bmatrix}
~=~\frac{1}{\beta}\begin{bmatrix} \beta^2-\alpha\gamma & \gamma\cr -\alpha & 1  \end{bmatrix}\cr
~\in~&Sp(2,\mathbb{R})
~=~SL(2,\mathbb{R}), \cr
1~=~&\det A~=~ad-bc.\end{align}\tag{ii}$$


*Notice that the CT (ii) is only well-defined if $\beta\neq 0$.


*Also notice that the identity CT $A={\bf 1}_{2\times 2}$ corresponds to $(\alpha,\beta,\gamma)=(0,1,0)$.


*This suggests that if we are using a single type 2 CT (i) then we can only deform continuously the identity CT $A={\bf 1}_{2\times 2}$ into a CT (ii) with positive $\beta$ value.


*In particular, OP considers a rotation
$$ A~=~ \begin{bmatrix}\cos\theta & \sin\theta\cr -\sin\theta & \cos\theta  \end{bmatrix}.\tag{iii} $$
It is easy to see that $\beta >0 \Leftrightarrow \cos\theta >0$. But any rotation (iii) is a CT. This suggests that we in this case can overcome the topological restriction $\beta >0$ by composing several type 2 CTs (i).
