A Lie subalgebra need not integrate to a closed subgroup

The torus Lie algebra is abelian, so every linear subspace is bracket closed; in particular $\mathfrak{h}$ is a Lie subalgebra. The derivative of the winding $i$ at $0$ has image $\mathbb{R}(1,\alpha )=\mathfrak{h}$, and its source $\mathbb{R}$ is connected. Thus [F2] makes $i(\mathbb{R})$ a connected immersed integral subgroup for $\mathfrak{h}$. It is not all of ${\mathbb{T}}^2$: its intersection with $\{1\}\times S^1$ is the countable set $\{(1,e^{2\pi i\alpha n}):n\in \mathbb{Z}\}$, not the whole circle. Since [F2] also makes it dense, it is not closed.

Suppose, for contradiction, that a closed Lie subgroup $K\subseteq {\mathbb{T}}^2$ has Lie algebra $\mathfrak{h}$. Let $K^0$ be the connected component of its identity. By [F3], $K^0$ is open and closed in $K$. It is a subgroup: multiplication maps the connected space $K^0\times K^0$ continuously into a connected subset containing the identity, and inversion does the same to $K^0$, so [F4] puts both images in the identity component. With the open submanifold structure, $K^0$ has $T_e{K}^0=T_{e}K=\mathfrak{h}$ and is a connected immersed Lie subgroup of ${\mathbb{T}}^2$.

Uniqueness in [F1], applied to steps 1.1 and 1.2, identifies $K^0$ as an immersed subgroup with $i(\mathbb{R})$. But $K^0$ is closed in $K$ by [F3] and $K$ is closed in ${\mathbb{T}}^2$ by the assumption in step 1.2, so $K^0=i(\mathbb{R})$ is closed in ${\mathbb{T}}^2$, contradicting step 1.1.

Therefore the irrational line $\mathfrak{h}$ is not the Lie algebra of any closed Lie subgroup, even though [F1] integrates it uniquely to the connected immersed winding. This refutes the claim. The argument assumes only the stated ${\mathrm{A}\mathrm{C}}_{\omega }$, inherited by the correspondence theorem.