Elliptic curves with infinitely many rational points but all of them of bounded absolute value Does there exist an elliptic curve of positive rank such that the coordinate of all the points is bounded by some number? How many such curves exist (curves are considered the same if there is a birational map between them)? If finitely many, can you list all of them? If infinitely many, does there exist a constant, $C$, such that all rational points on all such curves are bounded by it? (For each curve choose the affine model with integer coefficients such that the bound on the points is minimized)
 A: Note: In the previous version, I forgot the connectness and concluded that the boundness of $E(\mathbb{R})$ and $E(\mathbb{Q})$ is the same, which is a big mistake. So I correct the statement in this version.

If $E$ is an elliptic curve defined over $\mathbb{R}$, and $P\in E(\mathbb{R})$ is a point of infinite order, then the closure of $S:=\{ nP:n\in\mathbb{Z}\}$ in $E(\mathbb{R})$ is the connected components containing points in $S$, see this paper for the proof. Therefore, for an elliptic curve $E$ defined over $\mathbb{Q}$ with positive rank, the boundness of $E(\mathbb{Q})$ is closely related to the boundness of $E(\mathbb{R})$.
For instance, consider the elliptic curve
$$J:\begin{cases}x^2+y^2=1\\7x^2+z^2=1\end{cases}$$
with the base point $O=(0,1,1)$. Then $P=(12/37,35/37,19/37)\in J(\mathbb{Q})$ has infinite order, but $J(\mathbb{Q})\subset J(\mathbb{R})$ is bounded in $[-1,1]^3$.
But the boundness of $E(\mathbb{R})$ is not birational invariant, neither is $E(\mathbb{Q})$. In fact, every elliptic curve has Weierstrass model, which can not be bounded. So $J$ and its Weierstrass model do not coincide in boundness. We could also say that every elliptic curve defined over $\mathbb{R}$ has unbounded model. Moreover, for a Weierstrass curve $E$ defined over $\mathbb{Q}$ with positive rank, since the identity component of $E(\mathbb{R})$ is unbounded, we know that $\overline{E(\mathbb{Q})}$ is unbounded, so is $E(\mathbb{Q})$.
On the other hand, for a Weierstrass curve $E:y^2=x^3+ax+b$ defined over $\mathbb{Q}$, the functions $t=1/(1+x^2),u=x/(1+x^2)$ and $v=y/(1+x^2)$ all have bounded values on $E(\mathbb{R})$, and $\iota:P\in E\mapsto(t(P),u(P),v(P))$ gives a birational map onto its image
$$\iota(E):\begin{cases}u^2=t(1-t)\\v^2=u+(a-1)ut+bt^2\end{cases}$$
As $\iota(E(\mathbb{R}))$ is bounded, we can say that every Weierstrass curve (also every elliptic curve) has bounded model.
A: For curves in Weierstrass form $E/\Bbb{Q}:y^2=4x^3-ax-b$ then no.
This is because $C/\Bbb{C}:zy^2=4x^3-axz^2-bz^3\subset \Bbb{P^2(C)}$ is compact.
$E(\Bbb{Q})$  has infinitely many points $(x_n,y_n)\ne O$ will give that there are some $[x_n:y_n:1]\in C(\Bbb{C})$ arbitrary close to $[0:1:0]$ in the complex topology on $\Bbb{P^2(C)}$ ie. $\lim \inf_{n\to \infty}|1/y_n|=0$.
