Proving this is easy: Let $X$ and $Z$ be the two curves in $\Bbb R^3$. Assume $X$ and $Z$ intersect at a point $y$. Then, at most $\dim(T_y(X) + T_y(Z)) = 2$, where $T_x(X)$ and $T_z(Z)$ denotes the tangent space of $X$ at $x$ and $Z$ at $z$, respectively. Thus, they do not intersect transversally at a point of intersection $y$, because this would require $\dim(T_y(X) + T_y(Z)) = 3$.
I am looking for an example of transversal curves in $\Bbb R^3$ that do not intersect. Can someone help me out?
Let $Z$ be a submanifold of a manifold $Y$.
Definition: The smooth map $f:X \rightarrow Y$ where $X$ and $Y$ are smooth manifolds is said to be transversal to the submanifold $Z$ if $\operatorname{Image}(df_x) + T_y(Z) = T_y(Y)$ for each point $x$ in the preimage of $Z$.
Let $Y$ = $\mathbb{R}^3$. Let $Z$ be some curve in $\mathbb{R}^3$. Then, $T_z(Z) \cong \mathbb{R}$. Now I need to define the map $f$ so that $\operatorname{Image}(df_x) \cong \mathbb{R}^2$, but for any curve given by a map $f:\mathbb{R} \rightarrow \mathbb{R}^3$, $\operatorname{Image}(df_x) \cong \mathbb{R}^3$.
This is where I am stuck. Thanks
Any two nonintersecting curves are transversal. Given $f:X\to Y$ and $Z\subset Y$ such that the image of $f$ doesn't intersect $Z$, the preimage of $Z$ under $f$ is empty. Then the condition "$Image(df_x)+T_{f(x)}(Z) = T_{f(x)}Y$ for all $x\in f^{-1}(Z)$" is trivially satisfied, because $f^{-1}(Z)$ is empty.